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3 years ago

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The Robinson family and the Moore family are going on a day trip to visit the Square Caverns on Monday. The Robinson family has

2 parents and 4 children and paid $75 for admission. The Moore family has 2 parents and 3 children and paid $60. If x represents the admission price of each parent and y represents the admission price of each child, which is a system of equations that could be used to determine the admission price of an adult ?

1 answer:

sladkih [1.3K]3 years ago

6 0

Answer:

admission price of each parent = $7.5

admission price of each child = $15

Step-by-step explanation:

x = admission price of each parent

y = admission price of each child

Robinson family:

2x + 4y = 75

Moore family:

2x + 3y = 60

2x + 4y = 75 (1)

2x + 3y = 60 (2)

Subtract (2) from (1)

4y - 3y = 75 - 60

y = 15

Substitute y = 15 into (1)

2x + 4y = 75

2x + 4(15) = 75

2x + 60 = 75

2x = 75 - 60

2x = 15

x = 15/2

x = 7.5

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Y = -(x+5)2 + 9 into standard form

mote1985 [20]

Answer:

y = -2x - 1

Step-by-step explanation:

y = -(x+5)2 + 9

Foil method to get rid of parenthesis

y= -2x - 10 + 9

y = -2x - 1

Have a great day!

(Please mark this question as brainiest if it worked or ask me a question if you do not understand)

4 0

2 years ago

A cereal box has a length of 8 inches, a width of 2.5 inches and a height of 12 inches. The cereal box is made completely of car

ozzi

We know that

[the surface area]=2*[area of the base]+[perimeter of the base]*height
[the surface area]=2*[8*2.5]+[2*(8+2.5)]*12
[the surface area]=40+252-----> 292 in²

the answer is
<span>the least amount of cardboard needed to make the cereal box is 292 in</span>²

5 0

3 years ago

Solve the initial value problem

tigry1 [53]

$9(t+1)\dfrac{\mathrm dy}{\mathrm dt}-7y=14t\implies\dfrac{\mathrm dy}{\mathrm dt}-\dfrac7{9(t+1)}y=\dfrac{14t}{9(t+1)}$

Look for an integrating factor $\mu(t)$ :

$\ln\mu=\displaystyle-\frac79\int\frac{\mathrm dt}{t+1}=-\frac79\ln(t+1)\implies\mu=(t+1)^{-7/9}$

Multiply both sides by $\mu$ :

$(t+1)^{-7/9}\dfrac{\mathrm dy}{\mathrm dt}-\dfrac79(t+1)^{-16/9}y=\dfrac{14}9t(t+1)^{-16/9}$

Condense the left side as the derivative of a product:

$\dfrac{\mathrm d}{\mathrm dt}\left[(t+1)^{-7/9}y\right]=\dfrac{14}9t(t+1)^{-16/9}$

Integrate both sides:

$(t+1)^{-7/9}y=\displaystyle\frac{14}9\int t(t+1)^{-16/9}\,\mathrm dt$

For the integral on the right, substitute

$u=t+1\implies t=u-1\implies\mathrm dt=\mathrm du$

$\displaystyle\int t(t+1)^{-16/9}\,\mathrm dt=\int(u-1)u^{-16/9}\,\mathrm du$

$\displaystyle=\int\left(u^{-7/9}-u^{-16/9}\right)\,\mathrm du=\frac92u^{2/9}+\frac97u^{-7/9}+C$

$\implies(t+1)^{-7/9}y=\dfrac{14}9\left(\dfrac92(t+1)^{2/9}+\dfrac97(t+1)^{-7/9}+C\right)$

$\implies(t+1)^{-7/9}y=7(t+1)^{2/9}+2(t+1)^{-7/9}+C$

$\implies y=7(t+1)+2+C(t+1)^{7/9}=7t+9+C(t+1)^{7/9}$

Given that $y(0)=12$ , we get

$12=9+C\implies C=3$

$\implies\boxed{y(t)=7t+9+3(t+1)^{7/9}}$

6 0

3 years ago

Find the volume of the pyramid below.

olganol [36]

Answer:

Option B. $192\ units^{3}$

Step-by-step explanation:

we know that

The volume of the square pyramid is equal to

$V=\frac{1}{3}b^{2}H$

where

b is the length side of the square base

H is the height of the pyramid

we have

$b=8\ units$

$H=9\ units$

substitute

$V=\frac{1}{3}(8)^{2}(9)$

$V=192\ units^{3}$

5 0

3 years ago

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A cell phone company recently noticed a 26% sales decrease in the month of July compared to June. Let n represent June’s sales.

maw [93]

N x 26% = Y
Hopes this helps

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3 years ago

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