15 Mary needs to work out the size of angle x in this diagram.
DO NOT WRITE IN THIS AREA
63°
X
B
С
She writes
x = 63° because base angles of an isosceles triangle are equal.
Mary is wrong.
(a) Explain why.

Answer:

3+1.20c > 13.50

Step-by-step explanation:

hope this helps

Answer:

nasa pic yung answer

Step-by-step explanation:

hope its help

correct me if im wrong

Answer:

(a) [tex]\sf y=\dfrac{11}{20}x+7[/tex]

(b) $16.90

Step-by-step explanation:

Part (a)

Add a line of best fit (see attached).

Find two points on the line:  (7, 0) and (20, 18)

Use these points to find the slope of the line:

[tex]\sf slope\:(m)=\dfrac{change\:in\:y}{change\:in\:x}=\dfrac{18-7}{20-0}=\dfrac{11}{20}[/tex]

Use the found slope and one of the points in the point-slope form of the linear equation to find an equation for the line of best fit:

[tex]\sf\implies y-y_1=m(x-x_1)[/tex]

[tex]\sf\implies y-7=\dfrac{11}{20}(x-0)[/tex]

[tex]\sf\implies y=\dfrac{11}{20}x+7[/tex]

Part (b)

Substitute x = 18 into the equation and solve for y:

[tex]\sf\implies y=\dfrac{11}{20}(18)+7=\$16.90[/tex]

[tex]\text{Let it be x percent}\\ \\\dfrac 14 \times x\% = \dfrac 15\\\\\\\implies\dfrac 14 \times \dfrac x{100} = \dfrac 15\\ \\\\\implies x = \dfrac{400}{5} = 80\\\\\\\text{Hence the answer is}~ 80\%[/tex]

[tex]\textit{area of a sector of a circle}\\\\ A=\cfrac{\theta \pi r^2}{360}~~ \begin{cases} r=radius\\ \theta =\stackrel{degrees}{angle}\\[-0.5em] \hrulefill\\ A=300\pi \\ \theta =270 \end{cases}\implies 300\pi =\cfrac{(270)\pi r^2}{360}\implies 300\pi =\cfrac{3\pi r^2}{4} \\\\\\ 1200\pi =3\pi r^2\implies \cfrac{1200\pi }{3\pi }=r^2\implies 400=r^2\implies \sqrt{400}=r\implies 20=r[/tex]

Answer:

88 scoops

Step-by-step explanation:

Hello!

If 25% of 352 scoops were strawberry, we find 25% of 352

Solve:

This means 88 scoops of strawberry were sold.

Image attatched for conversion of decimals, percents, and fractions.

Step-by-step explanation:

According to the question,

Let the width of rectangle be x and length of rectangle be 3x

Perimeter of Rectangle :- 2(L+B) = 64 in

Putting the values we get ,

2(3x+x) = 64 in

8x = 64 in

x = 8 in

Putting the value of x ,

Width :- 8 Inch

Length :- 24 inch

[tex]\rightarrow[/tex] Length(l) of the rectangle is three times it's width(w) = 3w.

[tex]\rightarrow[/tex] Width(w) of the rectangle = w.

[tex]\rightarrow[/tex] Perimeter of the rectangle = 64in.

[tex]\rightarrow[/tex]Length and width of the rectangle.

[tex]\rightarrow[/tex] Perimeter of rectangle = [tex]\sf{2(l+w)}[/tex] putting the value of perimeter, l and w from the above given)

[tex]\rightarrow[/tex] 64 = [tex]\sf{2(3w+w)}[/tex]

[tex]\rightarrow[/tex] 64 = [tex]\sf{2(4w)}[/tex]

[tex]\rightarrow[/tex] 64 = [tex]\sf{8w}[/tex]

[tex]\rightarrow[/tex] [tex]\sf{\frac{64}{8}}[/tex]= [tex]\sf{w}[/tex]

[tex]\rightarrow[/tex] [tex]\sf{8}[/tex]= [tex]\sf{w}[/tex]

Therefore, width of the rectangle = 8in.

And Length = 3(8)in. = 24in.

To check whether the answer is correct or not, we can put the value of length and width in the formula = [tex]\sf{2(l+w)}[/tex]

[tex]\rightarrow[/tex] [tex]\sf{=\: 2(24+8)}[/tex]

[tex]\rightarrow[/tex] [tex]\sf{=\: 2(32)}[/tex]

[tex]\rightarrow[/tex] [tex]\sf{=\: 64in.}[/tex]

Since, the perimeter of the rectangle is same as given in the question, therefore the value of length and width are correct.

_______________________________

Hope it helps you:)

Answer:

n = 7

Step-by-step explanation:

simplify the logs according to their bases

Answer:

n = 7

Step-by-step explanation:

[tex]\sf \log_4(64)^{n+1}=log_5(625)^{n-1}[/tex]

Change 64 to an exponent with base 4 and 625 to an exponent with base 5:

[tex]\implies \sf \log_4(4^3)^{n+1}=log_5(5^4)^{n-1}[/tex]

Using exponent rule [tex](a^b)^c=a^{bc}[/tex]

[tex]\implies \sf \log_4(4)^{3(n+1)}=log_5(5)^{4(n-1)}[/tex]

Using log rule:  [tex]\log_a(b^c)=c \log_a(b)[/tex]

[tex]\implies \sf 3(n+1)\log_4(4)}=4(n-1)log_5(5)[/tex]

Using log rule: [tex]\sf \log_a(a)=1[/tex]

[tex]\implies \sf 3(n+1)=4(n-1)[/tex]

[tex]\implies \sf 3n+3=4n-4[/tex]

[tex]\implies \sf 3+4=4n-3n[/tex]

[tex]\implies \sf n=7[/tex]

Answer:

16.50 per week

Step-by-step explanation:

6.60 times 10 equals 66 minus 75% equals 16.50

Answer:

The lines of symmetry in the wheel is infinite.

Explanation:

Since there are an infinite number of lines through the center, a circle has an infinite number of lines of symmetry.

Function: y = 2x²-5

Find y-intercept:

y = 2(0)²-5

y = -5

Find x-intercept:

2x²-5 = 0

2x² = 5

x² = 2.5

x = ±√2.5

x = -1.5811 , 1.5811

Graph plotted:

Answer:

Vertex and y-intercept (0, -5)

x-intercepts (-1.58, 0)  (1.58, 0)

opens upwards

other plot points: (-2, 3) (-1, -3) (1, -3) (2, 3)

Step-by-step explanation:

The graph is not quite correct - it's a little too narrow and doesn't go through the points on the graph.

The y-intercept is when x = 0:

f(0) = 2(0)² - 5

      = - 5

Therefore, the y-intercept is at (0, -5)

We also know that the y-intercept is the vertex since the equation is in the form [tex]f(x)=ax^2+c[/tex]

The x-intercepts are when f(x) = 0:

[tex]\implies 2x^2 - 5 = 0[/tex]

[tex]\implies x^2 =\dfrac52[/tex]

[tex]\implies x=\pm1.58113883...[/tex]

As the leading coefficient is positive, the parabola opens upwards.

Finally, input values -2 ≤ x ≤ 2 to find plot points:

(-2, 3)

(-1, -3)

(0, -5)

(1, -3)

(2, 3)

Answer: 372 sq. un

Step-by-step explanation:

7+24=31

Divided by 2 = 15.5

15.5x24 = 372

D. 40 ft.

Based on the given conditions, formulate:

[tex]30\times20\div15[/tex]

Calculate

[tex]\frac{30\times20}{15}[/tex]

Cross out the common factor

[tex]2\times20[/tex]

Calculate the product or quotient

[tex]40[/tex]

___________________________________________________

C. Yes, Right Triangle Similarity Theorem

[tex]-\ If\ an\ altitude\ is\ drawn\ to\ the\ hypotenuse\ of\ a\\ right\ triangle,\ then\ the\ two\ triangles\ formed\ are\\ similar\ to\ the\ original\ triangle\ and\ to\ each\ other.[/tex]

____________________________________________________

D. 23.40 ft.

[tex]height\ of\ tree:x[/tex]

[tex]\frac{x}{s_0-d_0}=\frac{39}{50}[/tex]

[tex]x=\frac{30.39}{50}[/tex]

[tex]=23.40ft[/tex]

[tex]\left \{ {{calculate\}[/tex]

I hope this helps you

:)

Answer: b = -0.5

Step-by-step explanation:

Given

-3b + 2.5 = 4

Subtract 2.5 on both sides

-3b + 2.5 - 2.5 = 4 - 2.5

-3b = 1.5

Divide -3 on both sides

-3b / -3 = 1.5 / -3

[tex]\boxed{b=-0.5}[/tex]

Hope this helps!! :)

Please let me know if you have any questions

Answer:

[tex]\Longrightarrow: \boxed{\sf{b=-0.5}}[/tex]

Step-by-step explanation:

To solve this problem, all you have to do is isolate the term of b from one side of the equation.

First, you have to subtract by 2.5 from both sides.

→ -3b+2.5-2.5=4-2.5

Solve.

Add or subtract the numbers from left to right.

→ -3b=1.5

Then, you divide by -3 from both sides.

→ -3b/-3=1.5/-3

Solve.

Divide the numbers from left to right.

→ 1.5/-3=-0.5

b=-0.5

I hope this helps! Let me know if you have any questions.

Answer:

390 m (perpendicular to river) x 780 m (parallel to river)

Step-by-step explanation:

Let y be the length of the side parallel to the river, and let x be the length of the sides perpendicular to the river.

The total area and length of fence required are given by:

Rewriting the length of fence as a function of only x:

The value of x for which the derivate of L(x) is zero is the length of x that uses the least amount of fencing:

If x = 390 m, then:

The dimensions that will use the least amount of fencing are 390 m x 780 m

[tex]\left[3+(-7)\times 2\right] - \left[10 \div(-5)\right]\\\\=(3-14) +\left(10\times \dfrac 15\right)\\\\=-11+2\\\\=-9[/tex]

Answer:

1/2

Step-by-step explanation:

The scale factor is 1/2 because each side length of [tex]\triangle{A'B'C'}[/tex] is 1/2 of the length of the side lengths of [tex]\triangle{ABC}[/tex].

Hope this helps :)

Answer:

1/2

Step-by-step explanation:

took the test

[tex] \frac{3}{10 } = 0.3[/tex]

[tex]0.25 < 0.3 < 0.333[/tex]

[tex] \frac{32}{100} = \frac{8}{25} = 0.32[/tex]

[tex]0.25 < 0.32 < 0.333[/tex]

[tex] \frac{26}{100} = \frac{13}{50} = 0.26[/tex]

[tex]0.25 < 0.26 < 0.333[/tex]

[tex] \frac{28}{100} = \frac{7}{25} = 0.28[/tex]

[tex]0.25 < 0.28 < 0.333[/tex]

[tex] \frac{252}{1000} = \frac{63}{250} = 0.252[/tex]

[tex]0.25 < 0.252 < 0.333[/tex]

[tex] \frac{ \frac{ - 2}{5} + \frac{1}{2} }{2} = \frac{ \frac{ - 4 + 5}{10} }{2} = \frac{ \frac{1}{10} }{2} = \frac{1}{20} [/tex]

Answer:

[tex]\sf g(x+5) = \dfrac{x+6}{4x+17}[/tex]

explanation:

g(x + 5) || this function refers to x = x + 5

solving steps:

[tex]\sf g(x) = \dfrac{1+x}{-3+4x}[/tex]

[tex]\rightarrow \sf g(x+5) = \dfrac{1+(x+5)}{-3+4(x+5)}[/tex]

[tex]\rightarrow \sf g(x+5) = \dfrac{6+x}{-3+4x+20}[/tex]

[tex]\rightarrow \sf g(x+5) = \dfrac{x+6}{4x+17}[/tex]

Answer:

Given function:

[tex]g(x)=\dfrac{1+x}{-3+4x}[/tex]

To find  [tex]g(x+5)[/tex], substitute  [tex]x+5[/tex]  into the function in place of [tex]x[/tex]:

[tex]\implies g(x+5)=\dfrac{1+(x+5)}{-3+4(x+5)}[/tex]

                     [tex]=\dfrac{1+x+5}{-3+4x+20}[/tex]

                     [tex]=\dfrac{x+6}{4x+17}[/tex]

The error made by Tnaya in constructing the box plot is the first quartile and third quartile depicited is wrong.

A box plot is used to study the distribution and level of a set of data. The box plot consists of two lines known as whiskers and a box. The first whisker represents the minimum value and the last whisker represents the maximum value.

On the box, the first line to the left represents the first quartile. 25% of the score represents the lower quartile. The next line on the box represents the median. 50% of the score represents the median. The third line on the box represents the third quartile. 75% of the scores represents the third quartile.

For the data given, the:

To learn more about median, please check: https://brainly.com/question/20434777

The  rectangles have the same areas but greater perimeters are 12 ft by 3 ft, 9 ft by 4 ft and 18 ft by 2 ft.

Area is the amount of space occupied by a two dimensional shape or object.

A rectangle is shown with a length and height of 6 ft. Hence:

Area = (6 * 6) = 36 ft²

Perimeter = 2(6 + 6) = 24 ft

For 12 ft by 3 ft:

Area = (12 * 3) = 36 ft²

Perimeter = 2(12 + 3) = 30 ft

For 9 ft by 4 ft:

Area = (9 * 4) = 36 ft²

Perimeter = 2(9 + 4) = 26 ft

For 8 ft by 3 ft:

Area = (8 * 3) = 24 ft²

Perimeter = 2(8 + 3) = 24 ft

For 7 ft by 4 ft:

Area = (7 * 4) = 28 ft²

Perimeter = 2(7 + 4) = 26 ft

For 18 ft by 2 ft:

Area = (18 * 2) = 36 ft²

Perimeter = 2(18 + 2) = 40 ft

The  rectangles have the same areas but greater perimeters are 12 ft by 3 ft, 9 ft by 4 ft and 18 ft by 2 ft.

Find out more on area at: https://brainly.com/question/25292087

The similar circles P and Q can be made equal by dilation and translation

From the figure, we have the centers to be:

P = (-5,4)

Q = (6,8)

The distance is then calculated using:

d = √(x2 - x1)^2 + (y2 - y1)^2

So, we have:

d = √(6 + 5)^2 + (8 - 4)^2

Evaluate the sum

d = √137

Evaluate the root

d = 11.70

Hence, the horizontal distance between the center of circles P and Q is 11.70 units

We have their radius to be:

P = 2

Q = 5

Divide the radius of Q by P to determine the scale factor (k)

k = Q/P

k = 5/2

k = 2.5

Hence, the scale factor of dilation from circle P to Q is 2.5

Read more about dilation at:

https://brainly.com/question/3457976

Answer:

[tex]\frac{x^2}{25} + \frac{y^2}{16} = 1[/tex]

Step-by-step explanation:

Great, so the question already tells you that you're using the standard form of an ellipse. All you gotta do is apply your knowledge of the characteristics of an ellipse graph.

The standard form of an ellipse is:

[tex]1 = \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2}[/tex]

[tex]h[/tex] and [tex]k[/tex] are the offset, or the coordinates of the center of the ellipse. The center of the ellipse will always be on (h,k). Now because the question says the center is (0,0), we can make h and k equal to 0 in the equation giving:

[tex]1 = \frac{x^2}{a^2} + \frac{y^2}{b^2}[/tex].

Now, the height of the ellipse will always be equal to 2a. The width will be equal to 2b.

Since we are told what the height and width should be, we can find the a and b values quite easily using algebra.

So first, height:

[tex]10 = 2a\\5 = a[/tex]

Now width:

[tex]8 = 2b\\4 = b[/tex]

Subbing for a and b in the equation give you:

[tex]1 = \frac{x^2}{5^2} + \frac{y^2}{4^2}\\=\\1 = \frac{x^2}{25} + \frac{y^2}{16}[/tex]

Since [tex]A^2=A[/tex], we have

[tex](I + A)^2 = I^2 + IA + AI + A^2 = I + A + A + A = I + 3A[/tex]

[tex](I + A)^3 = (I + A) (I + 3A) = I^2 + 3IA + AI + 3A^2 = I + 3A + A +  3A = I + 7A[/tex]

Then the target expression is

[tex]7A - (I + A)^3 = 7A - (I + 7A) = \boxed{-I}[/tex]

Answer:

0.5,0.2,0.3

Step-by-step explanation:

divide 60,23,37 by 120

Answer:

A

Step-by-step explanation:

Remember domain is input and range is output.

Input: All real numbers

Output: [0, infinity)

A

Answer:

Two angles whose sides are opposite rays are called vertical angles. Two coplanar angles with a common side, a common vertex, and no common interior points are called adjacent angles.

Further explanation

Adjacent Angles are two angles that share a common vertex, a common side, and no common interior points. They share a vertex and side, but do not overlap. The example is shown in the picture below.  

Where ∠1 and ∠2 are adjacent angles, ∠ABC and ∠1 are NOT adjacent angles because ∠ABC overlaps ∠1.

Vertical Angles are two angles whose sides form two pairs of opposite rays (straight lines). Vertical angles are located across from one another in the corners of the "X" formed by the two straight lines.  The example is shown in the picture below.

Where ∠1 and ∠2 are vertical angles, ∠3 and ∠4 are vertical angles, Vertical angles are not adjacent.  ∠1 and ∠3 are not vertical angles (they are a linear pair) and Vertical angles are always equal in measure.

Step-by-step explanation:

[tex]lim \: f(x) = ( \infty + 4)( \infty - 2) {}^{2} \\ x - > \infty [/tex]

[tex]lim \: f(x) = \infty \times \infty \\ x - > \infty [/tex]

[tex]lim \: f(x)= \infty \\ x - > \infty [/tex]

[tex]lim \: \frac{f(x)}{x} = \frac{x(1 - \frac{4}{x})(x - 2) {}^{2} }{x} \\ x - > \infty [/tex]

[tex]lim \: \frac{f(x)}{x} = (1)( \infty - 2) {}^{2} \\ x - > \infty [/tex]

[tex]lim \: \frac{f(x)}{x} = \infty \\ x - > \infty [/tex]

We can then say that the function f(x)=(x-4)(x-2)² admits an asymptotic direction parallel to the y-axis at +∞ and -∞ as well since we have to follow the same steps.