Question 1 1.1 Given the quadratic number pattern: 5; 10; 17; 26;.. 1.1.1 Write down the next TWO terms of the pattern. 1.1.2 D Determine the formula for the nth term of the patter 1.1.3 Which term of the pattern will have a value of 1765? 1.2 Given the quadratic pattern:x; 6; 9; y; 24; Calculate the sum of x and y. Question 2 COFF 8(1:2)​

The value of x in the trapezoid given, to the nearest tenth, is: 10.5.

A trapezoid is a quadrilateral that has two sides only that are parallel to each other.

To find the area of a trapezoid, the formula used is expressed as: 1/2(a + b) × h, where:

Given the following:

Area of trapezoid = 174 units²

AP = 6

PB = x + 3

CD = 2x + 6

AC = 10

Using the formula, we variables in the formula are:

a = AP + PB = 6 + x + 3 = x + 9

b = CD = 2x + 3

h = PC = √(AC² - AP²)

h = √(10² - 6²) = 8

Plug in the values into 1/2(a + b) × h:

174 = 1/2(x + 9 + 2x + 3) × 8

174 = 1/2(3x + 12) × 8

174 = (3x + 12) × 4

174 = 12x + 48

174 - 48 = 12x

126 = 12x

126/12 = x

10.5 = x

x = 10.5

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answer : 2 1/2

divide y by x

divide  2 1/2 by 1

its 2 1/2

10 divided by 4 is also

2 1/2

17 1/2 divided by 7 is also

2 1/2

Answer:

Step-by-step explanation:

no idea tbh

x is 40 degrees

y is 40 degrees

Hope this helps!

13) 120 + 1.5 + 0.25 = 121.75 pounds

14) 4 - 1.5 = 2.5 pints

15) (2.25)(32)= $72

13. Mrs. Smith's current weight is 121.75 pounds.

14. The students should consume an additional 2.5 pints of water.

15. The nurse's overtime earnings amount to $72.00.

Given are words problems based on mixed numbers:

13. To find Mrs. Smith's current weight, we need to add the weight gained over the last two months to her weight last month.

Weight gained in the first month: 1 1/2 pounds

Weight gained in the second month: 1/4 pounds

Total weight gained over the last two months: 1 1/2 + 1/4 = 1 3/4 pounds

Now, to find Mrs. Smith's current weight, we add the total weight gained to her weight last month:

Current weight = Weight last month + Total weight gained

Current weight = 120 pounds + 1 3/4 pounds

Hence, Mrs. Smith's current weight is 121.75 pounds.

14. To calculate the additional water the students should consume to meet the school nurse's recommendation of at least 4 pints daily, we need to find the difference between the recommended amount and the average amount the students currently drink.

The recommended amount is 4 pints, and the average amount the students currently drink is 1 1/2 pints, which can be written as 1.5 pints.

Additional water needed = Recommended amount - Current average amount

Additional water needed = 4 pints - 1.5 pints

Additional water needed = 2.5 pints

Therefore, the students should consume an additional 2.5 pints of water daily to meet the school nurse's recommendation.

15. To calculate the nurse's overtime earnings, you can use the formula:

Overtime Earnings = Overtime Hours × Overtime Pay Rate

First, let's convert the mixed number of overtime hours into an improper fraction:

2 1/4 hours = 2 + 1/4 = 9/4 hours

Now, we can calculate her overtime earnings:

Overtime Hours = 9/4 hours

Overtime Pay Rate = $32.00/hour

Overtime Earnings = (9/4) hours × $32.00/hour

To simplify the calculation, you can convert the fraction to a decimal:

Overtime Earnings = (9/4) hours × $32.00/hour

Overtime Earnings = 2.25 hours × $32.00/hour

Overtime Earnings = $72.00

So, the nurse's overtime earnings amount to $72.00.

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Answer:

0.6

Step-by-step explanation:

Given what we know, we can set up a proportion to convert 1,056 yards into miles

[tex]\frac{1760 yd}{1mi} =\frac{1056 yd}{x mi} \\\\1760x=1056\\\\x=0.6[/tex]

Answer:

x is equal to 5

Step-by-step explanation:

Answer:

you have to put the formula of perimeter

Answer:

12a. L+8m + 14m

12b. 10m+8m +14m

12c. 66

12d.420m

12e.252m

quick questions

volume= 288cm cubed

Answer:

-4

Step-by-step explanation:

in order to get from -6 to 24, you need to multiply x by -4 since -6* -4 is 24

Answer: [tex](x,y)=\{(-1,1), (0,4) \}[/tex]

Step-by-step explanation:

[tex]3x^2 +6x+4=-3x^2 +4\\\\6x^2 +6x=0\\\\x^2 +x=0\\\\x(x+1)=0\\\\x=-1, 0\\\\x=-1 \longrightarrow y=1\\\\x=0 \longrightarrow y=4\\\\\therefore (x,y)=\{(-1,1), (0,4) \}[/tex]

The solution of the quadratic equations will be ( 0, 4 ) and ( -1, 1).

A quadratic equation is a polynomial with a degree of 2 or the maximum power of the variable is 2 in quadratic equations. It has two solutions as its maximum power is 2.

Given quadratic equations are given as below:-

y = 3x² + 6x + 4

y = -3x² + 4

Equate the equations and solve for the value of x.

3x² + 6x + 4 = -3x² + 4

6x² + 6x = 0

x² + x = 0

x ( x + 1 ) = 0

x = 0 and x = -1

At x=0 and x=-1 the values of y are calculated as below:-

y = -3x² + 4

y = 0+ 4 = 4

y = -3 + 4 = 1

Therefore, the solution of the quadratic equations will be ( 0, 4 ) and ( -1, 1).

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Answer:

0.7

Step-by-step explanation:

Because 0.4 is $5 and 0.3 is $25. So 0.3+0.4=0.7

1. The amount of the adjusting entry for Uncollectible Accounts is $1,250.

2. The adjusted balances of Accounts Receivable Allowance for Doubtful Accounts Bad Debt Expense are:

Accounts Receivable = $3,545,000

Allowance for Doubtful = $7,250

Accounts Bad Debt Expense = $1,250

3. The net realizable value of the accounts receivable is $3,537.750 ($3,545,000 - $7,250).

                               Accounts Receivable    Allowance for Doubtful Accounts

Beginning balance     $645,000 DR               $6,000 CR

Sales                          2,900,000

Bad Debt Estimate = $7,250 ($2,900,000 x 0.0025)

Ending balance      $3,545,000                     $7,250 CR

Bad Debts Expense = $1,250 ($7,250 - $6,000)

Thus, the net realizable value of the accounts receivable shows the amount that the company expects to recover from customers after allowing for doubtful accounts.

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1. Determine the amount of the adjusting entry for uncollectible accounts.

2. Determine the adjusted balances of Accounts Receivable, Allowance for Doubtful Accounts, and Bad Debt Expenses.

3. Determine the net realizable value of accounts receivable

I assume [tex]f(x)=0[/tex] otherwise. If [tex]f(x)[/tex] is indeed a proper PDF, then its integral over the support of [tex]X[/tex] is 1.

[tex]\displaystyle \int_{-\infty}^\infty f(x) \, dx = k \int_0^2 (2x + 3x^2) \, dx = 1[/tex]

Compute the integral.

[tex]\displaystyle \int_0^2 (2x + 3x^2) \, dx = (x^2 + x^3)\bigg|_{x=0}^{x=2} = (2^2 + 2^3) - (0^2 + 0^3) = 12[/tex]

Then

[tex]12k = 1 \implies \boxed{k=\dfrac1{12}}[/tex]

Answer:

(x - 5) ^2 + (y -2)^2 = 1

Step-by-step explanation:

This is for a circle.

We need to complete the square

(x^2 -10x      ) + ( y^2 -4y       )  = -28

(x^2 - 10x +25) + (y^2 - 4y 2y) = -28 + 25 + 4

(x-5)^2 + (y-2)^2 = 1

The center of this circle would be (5,2)  with a radius of 1.

Answer:

2

Step-by-step explanation:

[tex]\frac{3[2(-2)-6] + (-2)^{2} +4[2(-2)+1]}{3[(-2)-5]+2}[/tex]

[tex]\frac{3(-10) - 4 + 4(-3)}{-16}[/tex]

[tex]\frac{-30-4-12}{-16}[/tex]

[tex]\frac{-46}{-16}[/tex]

okay you know what i dont know what im doing but i know the answer is for sure 2..

Given the length of both trail, Trail B is 7/4 miles longer than Trail B.

Given that;

To determine how many miles longer is Trail B more than Trail A, we subtract the length of tail A from Tail B.

Trail B - Trail A

17/2 - 27/4

17/2 is the same as; 34/4

34/4 - 27/4

We combine the numerators over common denominator

(34 - 27 )/4

7/4 miles

Given the length of both trail, Trail B is 7/4 miles longer than Trail B.

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LHS = (1-sin² cos²)/(cos²)

= (sin² + cos² - sin² cos²)/(cos²)

= (sin/cos)² + 1 - sin²

= tan² + cos²

= RHS

The amount that the insurance company will pay is $1800.

It should be noted that the insurance company will pay John for the expenses of his injuries and repair of the car.

For the injury, he will get:

= $500 + ($1400 - $100)

= $500 + $2300

= $1800

Therefore, the amount is $1800.

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The numerical values of the mean and standard deviation are 4 and 1.55, respectively

The given parameters are:

n = 10

π = 0.4

The probability is then calculated as:

[tex]P(x) = ^nC_x * \pi^x *(1-\pi)^{n-x}[/tex]

So, we have:

[tex]P(6) = ^{10}C_6 * 0.4^6 *(1-0.4)^4[/tex]

Apply the combination formula

[tex]P(6) = \frac{10!}{6!4!} * 0.4^6 *0.6^4[/tex]

So, we have:

[tex]P(6) = 210 * 0.4^6 *0.6^4[/tex]

Evaluate

P(6) = 0.1115

Hence, the numerical value of the probability that x = 6 is 0.1115

The mean value is:

[tex]\bar x = n\pi[/tex]

This gives

[tex]\bar x = 10 * 0.4[/tex]

[tex]\bar x = 4[/tex]

The standard deviation value is:

[tex]\sigma = \sqrt{\bar x(1-\pi)[/tex]

This gives

[tex]\sigma = \sqrt{4(1-0.4)[/tex]

[tex]\sigma = 1.55[/tex]

Hence, the numerical values of the mean and standard deviation are 4 and 1.55, respectively

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Answer:

[tex]f(x) = 20x - 4 \\ substitute \: y \: for \: f(x) \\ y = 20x - 4 \\ interchange \: x \: and \: y \\ x = 20y - 4 \\ swap \: the \: sides \: of \: the \: equation \\ 20y - 4 = x \\ move \: the \: constant \: to \: the \: right \: hand \\ 20y = x + 4 \\ divide \: both \: sides \: by \: 20 \\ y = \frac{1}{20} x + \frac{1}{5} \\ substitute \: f {}^{ - 1} (x) \: for \: y \\ f {}^{ - 1} (x) = \frac{1}{20} x + \frac{1}{5} [/tex]

The domain of the inverse of a relation is the same as the range of the original relation. In other words, the y-values of the relation are the x-values of the inverse.

Thus, domain of f(x): x∈R = range of f¯¹(x)

and range of f(x): x∈R =domain of f¯¹(x)

Based on the angles of a + b and b + c, the value of b can be calculated to be 40°.

As this is a triangle, all the internal angles must add up to 180°.

This means that:

a + b + c = 180°

As we have the value of b + c, this becomes:

a + 100 = 180

a = 80°

b is therefore:

a + b = 120

80 + b = 120

b = 120 - 80

= 40 °

Full question is:

If in triangle abc, angle a + angle b= 120° and angle b + angle c = 100°, then find angle b.

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Using the Poisson distribution, we have that:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by:

[tex]P(X = x) = \frac{e^{-\mu}\mu^{x}}{(x)!}[/tex]

The parameters are:

For this problem, the mean is given as follows:

[tex]\mu = 13[/tex]

The probability of having exactly 10 days of precipitation in the month of April is P(X = 10), hence:

[tex]P(X = x) = \frac{e^{-\mu}\mu^{x}}{(x)!}[/tex]

[tex]P(X = 10) = \frac{e^{-13}13^{10}}{(10)!} = 0.0859[/tex]

There is a 0.0859 = 8.59% probability of having exactly 10 days of precipitation in the month of April.

The probability of having less than three days of precipitation in the month of April is:

P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)

In which:

[tex]P(X = x) = \frac{e^{-\mu}\mu^{x}}{(x)!}[/tex]

[tex]P(X = 0) = \frac{e^{-13}13^{0}}{(0)!} \ approx 0[/tex]

[tex]P(X = 1) = \frac{e^{-13}13^{1}}{(1)!} = 0.00003[/tex]

[tex]P(X = 2) = \frac{e^{-13}13^{2}}{(2)!} = 0.00019[/tex]

Then:

P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0 + 0.00003 + 0.00019 = 0.00022

There is a 0.00022 = 0.022% probability of having less than three days of precipitation in the month of April.

For more than 15 days, the probability is:

P(X > 15) = P(X = 16) + P(X = 17) + ... + P(X = 20)

Applying the formula for each of these values and adding them, we have that P(X > 15) = 0.2364, hence:

There is a 0.2364 = 23.64% probability of having more than 15 days of precipitation in the month of April.

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The equation of the parabola is [tex]y=\frac{1}{24} x^2+3[/tex]

None of the given options is correct

Given:

Focus: (0, 3)

Directrix: y = -3

Note that:

f - k = k - (-3)

f - 3 = 3 + 3

f = 6 + 3

f = 9

The equation of the parabola is of the form:

[tex]y=\frac{1}{4(f-k)} (x-h)^2+k[/tex]

Substitute f = 9, k = 3, h = 0 into the equation

[tex]y=\frac{1}{24} (x-0)^2+3\\\\y=\frac{1}{24}x^2+3[/tex]

The equation of the parabola is [tex]y=\frac{1}{24} x^2+3[/tex]

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Answer:

12u²

Step-by-step explanation:

area of a regular quadrilateral is always base x vertical height.

the base here is clearly 4 (just count the squares)

draw an imaginary line connecting the bottom right corner to the line before the top right corner (the line should be vertical)

count the squares here too and you get 3.

so the area = 3 x 4,

the area is 12u²

Using the binomial distribution, we have that:

a) There is a 0.2111 = 21.11% probability that a six will occur exactly twice.

b) The expected number of sixes is of 3.2.

c) The variance of the number of sixes is of 2.35.

The formula is:

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]

The parameters are:

For this problem, the parameters are given by:

p = 8/30 = 0.2667, n = 12.

Item a:

The probability is P(X = 2), hence;

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]P(X = 2) = C_{12,2}.(0.2667)^{2}.(1-0.2667)^{10} = 0.2111[/tex]

Item b:

The expected number of the binomial distribution is:

E(X) = np.

Hence:

E(X) = 12 x 8/30 = 3.2.

Item c:

The variance of the binomial distribution is:

V(X) = np(1-p).

Hence:

E(X) = 12 x 8/30 x 22/30 = 2.35.

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The centre and the radius of the circle is (-7, -1) and 6 units

The equation of the circle in standard from is expressed as:

x^2+y^2+2gx+2fy+C = 0

where;

(-g, -f) is the centre

r= √g²+f²-C

Given the equation below

x^2+y^2+14x+2y+14 = 0

2g = 14

g = 7

2f = 2

f =1

Hence the centre of the circle is (-7, -1)

Radius = √49+1-14

Radius = √36 = 6 units

Hence the centre and the radius of the circle is (-7, -1) and 6 units

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The volume of the figure to the nearest whole number = 756π m³.

The figure shown exists composed of a cone and a cylinder.

The volume of the figure = volume of cone + volume of the cylinder

Height of cone = 12 - 8 = 4

Radius (r) of cone = 18/2 = 9 m

The volume of the cone V = 1/3hπr²

= 1/3 [tex]*[/tex] 4π [tex]*[/tex] 9² = 108π

The volume of the Cylinder = πr²h

radius (r) = 18/2 = 9 m

height (h) = 8 m

Volume of cylinder = π [tex]*[/tex] 9² [tex]*[/tex] 8 = 648π m³

Volume of the figure = 108π + 648π = 756π

The volume of the figure to the nearest whole number = 756π m³.

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Answer:

top left graph

Step-by-step explanation:

The function is not defined for negative x.

Also, the range should not include positive numbers.

So, the answer is the top left graph.

Answer:

All you need to do is suppose x = 3x :

[tex] \tt \: g(x) = \boxed{ \tt{2^{3x} }} [/tex]

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Answer:

D. 5/72

Step-by-step explanation:

The non-zero digits are 1 through 9.

There are 9 different non-zero digits.

p(A) = 1/9

There are 5 odd digits.

p(B) = 5/8

p(A and B) = p(A) × p(B) = 1/9 × 5/8 = 5/72

I think the answer is

A: 1/18