Simplify each expression. (y + 22) + 3

14 lessons costs $210. What would 72 lessons cost

Oksi-84 [34.3K]

Answer:

$1,080

Step-by-step explanation:

To solve, we can make a proportion

14 lessons are $210 and 72 lessons are $x

14/210=72/x

Cross multiply

14*x=210*72

14x=15120

To solve for x, we need to get x by itself. Since x is being multiplied by 14, divide both sides by 14.

14x/14=15120/14

x=1080

So, 72 lessons will cost $1,080

4 0

3 years ago

Read 2 more answers

Given the function f(x) = 4(x+3) − 5, solve for the inverse function when x = 3.

Dimas [21]

Hello,

f(x)=4(x+3)-5=4x+12-5=4x+7

y=4x+7==>c=(y-7)/4

f^(-1)(x)=(x-7)/4
f^(-1)( 3)=(3-7)/4=-4/4=-1

Answer D

4 0

3 years ago

Find the four arithmetic means between -21 and -36.

lara31 [8.8K]

The numbers given in the problem above are part of an arithmetic sequence with first and sixth terms equal to -21 and -36, respectively. Firstly, calculate for the common difference (d).

d = (-36 - -21) / (6 - 1) = -3

The arithmetic mean is calculated by adding -3 to the term prior to it.

a2 = -21 + -3 = -24 a3 = -24 + -3 = -27
a4 = -27 + -3 = -30 a5 = -30 + -3 = -33

Thus the four arithmetic means are -24, -27, -30, and -33.

4 0

3 years ago

A concert hall has 8,000 seats and two categories of ticket prices, $29 and $34. Assume that all seats in each category can be s

Greeley [361]

Using a system of equations, it is found that:

For a profit of $255,000, 3400 tickets of $29 and 4600 tickets of $34 must be sold.
For a profit of $271,000, 200 tickets of $29 and 7800 tickets of $34 must be sold.
For a profit of $235,000, 7400 tickets of $29 and 600 tickets of $34 must be sold.

--------------------

The variables of the system are:

x, which is the number of $29 tickets sold.
y, which is the number of $34 tickets sold.

Total of 8,000 seats, all can be sold, thus:

$x + y = 8000 \rightarrow x = 8000 - y$

--------------------

For a profit of $255,000, we have that:

$29x + 34y = 255000$

$29(8000 - y) + 34y = 255000$

$5y = 23000$

$y = \frac{23000}{5}$

$y = 4600$

$x = 8000 - 4600 = 3400$

For a profit of $255,000, 3400 tickets of $29 and 4600 tickets of $34 must be sold.

--------------------

For a profit of $271,000, we have that:

$29x + 34y = 271000$

$29(8000 - y) + 34y = 271000$

$5y = 39000$

$y = \frac{39000}{5}$

$y = 7800$

$x = 8000 - 7800= 200$

For a profit of $271,000, 200 tickets of $29 and 7800 tickets of $34 must be sold.

--------------------

For a profit of $235,000, we have that:

$29x + 34y = 235000$

$29(8000 - y) + 34y = 235000$

$5y = 3000$

$y = \frac{3000}{5}$

$y = 600$

$x = 8000 - 600 = 7400$

For a profit of $235,000, 7400 tickets of $29 and 600 tickets of $34 must be sold.

A similar problem is given at brainly.com/question/22826010

3 0

3 years ago

What are the last three terms of the binomial expansion 32x³ + 80x'y + 80x³y² +

Gnesinka [82]

The last three terms of the given binomial expansion are 40x²y³ + 10xy⁴ + y⁵

<h3>Binomial expansion</h3>

From the question, we are to determine the last three terms of the binomial expansion 32x⁵ + 80x⁴y + 80x³y² + __ + __ + __

The highest power of the expansion is 5

Using the binomial expansion theorem

(m +n)⁵ = m⁵ + 5m⁴n + 10m³n² + 10m²n³ + 5mn⁴ + n⁵

Now, we will determine the values of m and n by comparison

By comparison,

m⁵ = 32x⁵

m⁵ = (2x)⁵

∴ m = 2x

Also,

5m⁴n = 80x⁴y

Divide both sides by 5, to get

m⁴n = 16x⁴y

Put m = 2x into the equation

(2x)⁴n = 16x⁴y

16x⁴n = 16x⁴y

∴ n = y

Thus, the expression being expanded is (2x + y)⁵

By binomial expansion, the expansion of (2x + y)⁵ is

(2x + y)⁵ = (2x)⁵ + 5(2x)⁴y + 10(2x)³(y)² + 10(2x)²(y)³ + 5(2x)(y)⁴ + y⁵

(2x + y)⁵ = 32x⁵ + (5×16)x⁴y + (10×8)x³(y)² + (10×4)x²(y)³ + (5×2)x(y)⁴ + y⁵

(2x + y)⁵ = 32x⁵ + 80x⁴y + 80x³y² + 40x²y³ + 10xy⁴ + y⁵

Hence, the last three terms of the given binomial expansion are 40x²y³ + 10xy⁴ + y⁵

Learn more on Binomial expansion here: brainly.com/question/13602562

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8 0

2 years ago