Write a whole number that round to 2000 if we are rounding to the nearest thousaand

Answer:

[tex]-\frac{4}{3}[/tex](Your answer choices were really confusing, so I just put the plain answer)

Step-by-step explanation:

To find the slope of a line with just two points, we can use the formula [tex]\frac{y2-y1}{x2-x1}[/tex]. In this case, [tex]\frac{5-(-3)}{2-8}[/tex]. This is [tex]\frac{8}{-6}[/tex]. This reduces to [tex]-\frac{4}{3}[/tex]

Answer:

A. 8/7

Step-by-step explanation:

4/7 - (-4/7) --> 4/7 + 4/7

The two minutes turn into a plus, removing the bracket.

4/7 + 4/7 = 8/7

Add the two fractions together.

Answer:

A. 8/7

Step-by-step explanation:

The net number of yards lost by the cowboys football team is; 8 yards.

It follows from the task content that the number of yards gained or lost by the football team during each of the three plays is; gained 7 yards, lost 8 yards and then lost 7 more yard.

Hence, by polarising yard gain as positive and yard loss as negative; we have;

Net total = 7 - 8 -7 = -8

Hence, the team lost 8 yards in the total game.

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Answer: $350

Step-by-step explanation:

$700 divided by 2 is 350

The value of [tex]b_{1}[/tex] is 69/4h and value of [tex]b_{2}[/tex] is 43/4h if the area of trapezoid is 14 square units.

Given that area is equal to 1-2([tex]b_{1} -b_{2}[/tex])h and area is 14 square units.

We know that area of trapezoid is equal to (a+b)h/2.

We have to put the given equation equal to 14 to get our first equation.

14=1-2([tex]b_{1} -b_{2}[/tex])h

14=1-2[tex]b_{1}[/tex]h+2[tex]b_{2}[/tex]h

2[tex]b_{1}[/tex]h-2[tex]b_{2}[/tex]h=-13------------1

Now put the value of area in the formula of area of trapezoid.

14=[tex](b_{1} +b_{2})h/2[/tex]

[tex]b_{1}[/tex]h+[tex]b_{2}h[/tex]=28-----------------2

Multiply equation 2 by 2 and then subtract 2 from 1

[tex]2b_{1}h -2b_{2}h -2b_{1}h-2b_{2}h[/tex]=-13-56

-4[tex]b_{2}h[/tex]=-43

[tex]b_{2}[/tex]=43/4h

Put the value of [tex]b_{2}[/tex] in 2 to get the value of [tex]b_{1}[/tex].

[tex]b_{1} h+43/4h*h=28[/tex]

[tex]b_{1}[/tex]=69/4h

Hence The value of [tex]b_{1}[/tex] is 69/4h and value of [tex]b_{2}[/tex] is 51/4h if the area of trapezoid is 14 square units.

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Evaluating the function from the given graph, g(-9) = 1.

To find the value of a function for a given x-value using a given graph, trace the x-value against the corresponding y-value on the y-axis.

To evaluate the function from the graph given, for g(-9), find the y-value that corresponds to the x-value of -9.

The graph shows that when x = -9, the corresponding y-value is 1.

Therefore, g(-9) = 1.

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The 2 x 2 matrices of the transformations are

[tex]\left[\begin{array}{cc}-1&0\\0&1\end{array}\right][/tex], [tex]\left[\begin{array}{cc}0&1\\1&0\end{array}\right][/tex], [tex]\left[\begin{array}{cc}-1&0\\0&-1\end{array}\right][/tex] and [tex]\left[\begin{array}{cc}\lambda&0\\0&\lambda\end{array}\right][/tex]

The rule of reflection in the y-axis is

(x, y) ⇒ (-x, y)

This is represented as:

[tex]\left[\begin{array}{cc}a&b\\c&d\end{array}\right] \left[\begin{array}{c}x&y\end{array}\right] = \left[\begin{array}{c}-x&y\end{array}\right][/tex]

When solved using a graphing calculator, we have:

[tex]\left[\begin{array}{cc}-1&0\\0&1\end{array}\right] \left[\begin{array}{c}x&y\end{array}\right] = \left[\begin{array}{c}-x&y\end{array}\right][/tex]

Hence, the 2 x 2 matrix is [tex]\left[\begin{array}{cc}-1&0\\0&1\end{array}\right][/tex]

The rule of reflection in the line y = x is

(x, y) ⇒ (y, x)

This is represented as:

[tex]\left[\begin{array}{cc}a&b\\c&d\end{array}\right] \left[\begin{array}{c}x&y\end{array}\right] = \left[\begin{array}{c}y&x\end{array}\right][/tex]

When solved using a graphing calculator, we have:

[tex]\left[\begin{array}{cc}0&1\\1&0\end{array}\right] \left[\begin{array}{c}x&y\end{array}\right] = \left[\begin{array}{c}y&x\end{array}\right][/tex]

Hence, the 2 x 2 matrix is [tex]\left[\begin{array}{cc}0&1\\1&0\end{array}\right][/tex]

The rule of rotation through 180◦ about the origin is

(x, y) ⇒ (-x, -y)

This is represented as:

[tex]\left[\begin{array}{cc}a&b\\c&d\end{array}\right] \left[\begin{array}{c}x&y\end{array}\right] = \left[\begin{array}{c}-x&-y\end{array}\right][/tex]

When solved using a graphing calculator, we have:

[tex]\left[\begin{array}{cc}-1&0\\0&-1\end{array}\right] \left[\begin{array}{c}x&y\end{array}\right] = \left[\begin{array}{c}-x&y\end{array}\right][/tex]

Hence, the 2 x 2 matrix is [tex]\left[\begin{array}{cc}-1&0\\0&-1\end{array}\right][/tex]

The rule of the enlargement from the origin with scale factor λ.

(x, y) ⇒ (λx, λy)

This is represented as:

[tex]\left[\begin{array}{cc}a&b\\c&d\end{array}\right] \left[\begin{array}{c}x&y\end{array}\right] = \left[\begin{array}{c}\lambda x&\lambda y\end{array}\right][/tex]

When solved using a graphing calculator, we have:

[tex]\left[\begin{array}{cc}\lambda&0\\0&\lambda\end{array}\right] \left[\begin{array}{c}x&y\end{array}\right] = \left[\begin{array}{c}\lambda x&\lambda y\end{array}\right][/tex]

Hence, the 2 x 2 matrix is [tex]\left[\begin{array}{cc}\lambda&0\\0&\lambda\end{array}\right][/tex]

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Answer:

this mean in case of multiplying exponents with same base you should add the exponent and keep base same.

example:

[tex] {6}^{4} \times {6}^{5 } = {6}^{4 + 5} \\ = {6}^{9} [/tex]

Answer:

[tex]A=\frac{\sqrt{3}}{4}a^2[/tex]

Step-by-step explanation:

Since an equilateral has all of the sides equal, we can find the height of triangle using: [tex]a^2+b^2=c^2[/tex]. I attached a diagram which should explain how I got the dimensions of the three sides. Using the information from the diagram we get the equation:

[tex]h^2+(\frac{a}{2})^2=a^2[/tex]

Subtract a^2 from both sides

[tex]h^2=a^2-(\frac{a}{2})^2[/tex]

Take the square root of both sides

[tex]h = \sqrt{a^2-(\frac{a}{2})^2}[/tex]

If you know the area of a triangle, it's: [tex]\frac{1}{2}bh[/tex]. In this case the base=a, and the height is what we defined above. Using this we get:

[tex]A = \frac{a}{2}*\sqrt{a^2-(\frac{a}{2})^2}[/tex]

We can distribute the exponent over the division to get:
[tex]A = \frac{a}{2}*\sqrt{a^2-(\frac{a^2}{4})[/tex]

Now we can rewrite a^2 as 4a^2/4

[tex]A = \frac{a}{2}*\sqrt{\frac{4a^2}{4}-(\frac{a^2}{4})[/tex]

Now add the two fractions:

[tex]A = \frac{a}{2}*\sqrt{\frac{3a^2}{4}[/tex]

We can distribute the square root the division just like how we distributed the exponent 2, since the square root can be expressed as an exponent (1/2)

[tex]A = \frac{a}{2}*\frac{\sqrt{3a^2}}{\sqrt{4}}[/tex]

There's a radical identity that states: [tex]\sqrt[n]{a} * \sqrt[n]{b} = \sqrt[n]{a*b}[/tex]. We can use this to rewrite one radical as multiple radicals to simplify it:

[tex]A = \frac{a}{2}*\frac{\sqrt{a^2}*\sqrt{3}}{2}[/tex]

Simplify:

[tex]A = \frac{a}{2}*\frac{a*\sqrt{3}}{2}[/tex]

Now multiply the two fractions

[tex]A = \frac{a^2*\sqrt{3}}{4}[/tex]

This is the formula for the area of an equilateral triangle, but it is also often written as:
[tex]A=\frac{\sqrt{3}}{4}a^2[/tex]

This snow flake-like figure can be generated by rotating an end 60° five times around the center of the hexagon. There are two forms: (i) clockwise, (ii) counterclockwise.

Geometrically speaking, snow flakes represent regular hexagons. Regular hexagons can divided into six concentric regular triangles, whose central angles have a measure of 60°. This fractal figure can be generated by rotating 60° five times around the center.

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Answer:

Step-by-step explanation:

Since diameter x pi=3.14,this should be easy.

4x3=12

4x14/100=.56

12+.56=12.56

12.56 is the answer

Since the multiplication between two matrices is not commutative, then [tex]\vec A\, \times\,\vec I \ne \vec I \,\times \,\vec A[/tex], regardless of the dimensions of [tex]\vec A[/tex].

In linear algebra, we define the product of two matrices as follows:

[tex]\vec C = \vec A \,\times \vec B[/tex], where [tex]\vec A \in \mathbb{R}_{m\times p}[/tex], [tex]\vec B \in \mathbb{R}_{p\times n}[/tex] and [tex]\vec C \in \mathbb{R}_{m \times n}[/tex]     (1)

Where each element of the matrix is equal to the following dot product:

[tex]c_{ij} = \left[\begin{array}{cccc}a_{i1}&a_{i2}&\ldots&a_{ip}\end{array}\right]\,\bullet\,\left[\begin{array}{ccc}b_{1j}\\b_{2j}\\\vdots\\b_{pj}\end{array}\right][/tex], where 1 ≤ i ≤ m and 1 ≤ j ≤ n.     (2)

Because of (2), we can infer that the product of two matrices, no matter what dimensions each matrix may have, is not commutative because of the nature and characteristics of the definition itself, which implies operating on a row of the former matrix and a column of the latter matrix.

Such "arbitrariness" means that resulting value for [tex]c_{ij}[/tex] will be different if the order between [tex]\vec A[/tex] and [tex]\vec B[/tex] is changed and even the dimensions of [tex]\vec C[/tex] may be different. Therefore, the proposition is false.

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The value of x in 2^x - 2 when x = 0, x = 1 and x = 2 are -1, 0 and 2 respectively. option A

2^x - 2

when x = 0

2^x - 2

= 2^0 - 2

= 1 - 2

= -1

when x = 1

2^x - 2

=2^1 - 2

= 2 - 2

= 0

when x = 2

2^x - 2

= 2^2 - 2

= 4 - 2

= 2

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There are 1024 outcomes in the event

The given parameters are:

Device = coin

Number of flips = 10

A coin has 2 faces.

So, the number of outcomes is

Outcome = Faces^Flip

This gives

Outcome = 2^10

Evaluate

Outcome = 1024

Hence, there are 1024 outcomes in the event

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Answer:

[tex]1024[/tex]

Step-by-step explanation:

Each and every flip of our coin has two possible outcomes on the whole: it is either the heads or the tails. Besides, it resembles the bit in computing, right? Indeed, it has two options likewise: it can be either [tex]0[/tex] or [tex]1[/tex]. For now on I suggest thinking neither in heads nor in tails, but in bits. Suppose the heads is [tex]0[/tex], the tails is [tex]1[/tex]. Let us slightly simplify the condition, namely: suppose that the number of flips is [tex]3[/tex]. Then, our possible outcomes:

[tex]000 \\ 001 \\ 010 \\ 011 \\ 100 \\ 101 \\ 110\\ 111[/tex]

Take a look at the first number. In case it is [tex]0[/tex], we have [tex]4[/tex] outcomes because we can put either [tex]0[/tex] or [tex]1[/tex] not only on the second place, yet also on the third one. Abiding by the same logic, we also have [tex]4[/tex] outcomes when the first one is [tex]1[/tex]. In mathematics: [tex]2 * 2 * 2 = 2^3 = 8[/tex] outcomes.
As for the initial condition, it does not globally differ from our simplification, it just has more flips, hence it is not just a walk in the park to list all the possible outcomes anymore in this case, it is time-consuming otherwise. Even though, it is still not a big deal to count it: [tex]2 * ... * 2 = 2^{10} = 1024[/tex] outcomes.

Answer:

2x+2

Step-by-step explanation:

x²+2x-3

=2(x^2-1)+2*1(2x^1-1)

1. Consult the linked question. [tex]n(A\cap B\cap W) = \boxed{3}[/tex].

2. We have

[tex]n(A\cap B) = n(A\cap B \cap W) + n(A\cap B \cap W') \\\\ \implies 11 = 3 + n(A\cap B \cap W') \\\\ \implies n(A\cap B\cap W') = 8[/tex]

Similarly we can find

[tex]n(A\cap B' \cap W) = 3[/tex]

[tex]n(A'\cap B\cap W) = 6[/tex]

[tex]n(A\cap B'\cap W') = 16[/tex]

[tex]n(A'\cap B\cap W') = 8[/tex]

[tex]n(A'\cap B'\cap W) = 12[/tex]

Then the total number of students that applied to at least two of the universities is

[tex]\underbrace{n(A\cap B\cap W') + n(A\cap B'\cap W) + n(A'\cap B\cap W)}_{\text{only 2}} + \underbrace{n(A\cap B\cap W)}_{\text{all 3}} = \boxed{20}[/tex]

3. There's a small ambiguity here. Are we interested in students that applied to zero universities? If so, there are

[tex]\underbrace{n(U\setminus(A\cup B\cup W))}_{\text{none}} + \underbrace{n(A\cap B'\cap W') + n(A'\cap B\cap W') + n(A'\cap B'\cap W)}_{\text{only 1}} = \boxed{40}[/tex]

If we want students that applied to at least 1 school, we omit the first term and get a total of 36.

4. Split this set of students into those that also applied to Wachemo and those that did not.

[tex]n(A \cap B') = n(A\cap B' \cap W) + n(A\cap B'\cap W') = \boxed{19}[/tex]

The mass of the second object exists at 3 kg.

Let, F be the Force

a be the acceleration

m be the mass

The formula of force exists F = m [tex]*[/tex] a

Since force exists constant, then:

[tex]$m_1 = 2 kg, a_1 = 12 m/s^2[/tex] subset 1 designates the first object

[tex]$m_2 = ? , a_2 = 8m/s^2[/tex] subset 2 designates the second object

To estimate the mass of the second object

A constant force indicates [tex]$ m_1 * a = m_2 * a[/tex]

The mass of second object [tex]m_2[/tex], we get

[tex]m_1 * a = m_2 * a[/tex]

[tex]$m_2 = \frac{m_1 * a}{a}[/tex]

[tex]$=\frac{2*12}{8}[/tex]

= 24/8 = 3 kg

Therefore, the mass of the second object exists at 3kg.

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Answer:

A) 39

Step-by-step explanation:

If you calculate the volume of the cube which is the V=a^3 you get 27, then calculating the pyramid, the Length is 3, and the Width is also 3. Then the height is 4. so you put that into the formula to calculate the volume which is V=L*W*H
    --------

        3

you get 12. Now its simple, you just add the 2 volumes, so you do 27+12 and its 39.

39 is the answer by calculating the volume using formula!

Answer: [tex](3, \infty)[/tex]

Step-by-step explanation:

q(x) is increasing for [tex]x > 3[/tex].

p(x) is increasing for [tex]x \geq -2[/tex].

So, both functions are increasing for [tex](3, \infty)[/tex].

Jakobe sold 150 coffee, 50 tea and 120 donuts at the coffee cart on Monday.

An equation is an expression that shows the relationship between two or more numbers and variables.

Let x represent the amount of coffee, y represent the amount of tea and z represent the amount of donuts, hence:

x + y + z = 320    (1)

Also:

1.5x + 2y + 0.75z = 415     (2)

And:

x = 3y     (3)

From the equations:

x = 150, y= 50 and z = 120

Jakobe sold 150 coffee, 50 tea and 120 donuts at the coffee cart on Monday.

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In order to have a greater fertilizer coverage, you need to increase flow rate and: A. increase speed to approximately 7.1 mph so that you cover the field more quickly.

A flow rate is an average rate and it can be defined as the number of flow units that are created per unit time such as fertilizer coverage on a farm.

In this context, we can infer and logically deduce that you have to increase flow rate and your increase speed to approximately 7.1 mph, so that you can cover the field more quickly and have a greater fertilizer coverage.

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Complete Question:

The soil samples for the next field indicate that fertilizer coverage needs to be greater. To achieve this, you need to increase flow rate. How would you achieve this?

A. Increase speed to approximately 7.1 mph so that you cover the field more quickly

B. Increase the engine speed to approximately 2,000 rpm

C. Decrease speed to approximately 6.0 mph so that you cover the field more slowly

D. Shift to second gear so that the engine speed slows

Answer:

A. [tex]h(x)=\sqrt{x-3}[/tex]

Step-by-step explanation:

The parent function of [tex]\sqrt{x}[/tex] is translated to the left when [tex]h[/tex] is positive in the transformation [tex]\sqrt{x+h}[/tex].

If [tex]h[/tex] is negative, the graph translates towards the left with the distance equal to the value of [tex]h[/tex].

Here the graph moved 3 units towards the right. This means that [tex]h[/tex] is negative and has the value of 3.

So, plugging that into the parent function for translation, the function becomes:

[tex]h(x)=\sqrt{x-3}[/tex]

Answer:

The first one, you cannot factor because the two values do not share any common factors. Hence, the answer would just be:

a. 51x-25

The other one, we can factorise to:

b. 3x(4-x)

Step-by-step explanation:

For the second option:

Common factor between 12x and 3x^2 is 3x.

Take this out to the front, we are left with:

3x(4-x)

A website I recommend you use (math calculator with some instructions) is Symbolab.

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The cube can contain 1000 cm^3 of water

The side length is given as:

L = 10 cm

The amount of water it can contain is calculated as;

V =L^3

So, we have:

V = 10^3

Evaluate

V = 1000

Hence, the cube can contain 1000 cm^3 of water

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Answer:

[tex]\mathrm{pK}_{\mathrm{a}}+\mathrm{pK}_{\mathrm{b}}=\mathrm{pK}_{\mathrm{w}}[/tex] for [tex]\mathrm{HCl}[/tex] \& [tex]\mathrm{ClOH}[/tex]

Answer:

f(x):  Amplitude = 2, Period = 2

g(x):  Amplitude = ¹/₂, Period = π

Step-by-step explanation:

The cosine function is periodic, meaning it repeats forever.

Standard form of a cosine function:

f(x) = A cos(B(x + C)) + D

where:

Function f(x)

[tex]f(x)=2 \cos (\pi x)[/tex]

Comparing this with the standard form of a cosine function:

Function g(x)

[tex]g(x)=\dfrac{1}{2} \cos (2x)[/tex]

Comparing this with the standard form of a cosine function:

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The median age of your customer base = 45

The median is the value that divides a data sample, a population, or even a probability distribution's upper and lower halves in statistics and probability theory. It could be referred to as "the middle" value for a data set. The fundamental difference between the mean and the median when describing data is that the median is more representative of the "typical" value because it is not skewed by the a small proportion of extremely large or small values. Due to the fact that income distribution can be highly skewed, the median income, for instance, might be a better indicator of what is considered a "typical" income.

Mean = middle term

          = 6th term

          = 45

Hence, Median of age = 45.

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Part (a)

[tex]\frac{1}{10}+\frac{3}{10}=\frac{4}{10}=\boxed{\frac{2}{5}}[/tex]

Part (b)

[tex]200\left(\frac{2}{5} \right)=\boxed{80}[/tex]

5+a-4√(a+1)/(a-3) is the simplified form of the number. This can be obtained by multiplying numerator and denominator with conjugate of the denominator.

Given the number,

√a+1-2/√a+1+2

Denominator = √a+1+2

Conjugate of the denominator = √a+1 - 2

Now, (√a+1-2)(√a+1 - 2 )/√(a+1+2)(√a+1 - 2 )

=(√a+1 - 2 )²/(√a+1)² - 4

=(a+1 -4√a+1+5)/(a+1-4)

=5+a-4√(a+1)/(a-3)

Hence 5+a-4√(a+1)/(a-3) is the simplified form of the number.

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In the given diagram, the value of JI is 6

From the question, we are to determine JI

From the given diagram, we can write that

JI + IH = x + 22

Also,

JI = 2x + 28

and

IH = 5

Then, we can write that

2x + 28 + 5 = x + 22

Collect like terms

2x - x = 22 -28 -5

x = -11

Now, substitute the value of x into the equation

JI = 2x + 28

JI = 2(-11) + 28

JI = -22 +28

JI = 6

Hence, the value of JI is 6

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