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2 years ago

3. Penny's family decided to go to the Splash Park.

Mathematics

1 answer:

Anestetic [448]2 years ago

6 0

Answer:

21.25

Step-by-step explanation:

because 85$ divided by the 4 tickets is 21.25

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Solve the system shown.

krek1111 [17]

1. Rewrite the system:

x+12y=68 (i)
x=8y-12 (ii)

2. Let's substitute the equation (ii) into the equation (i):

x+12y=68
(8y-12)+12y=68

3. Then, you have:

8y-12+12y=68

4. When you clear "y", you have:

20y-12=68
20y=68+12
20y=80
y=80/20
y=4

5. You already have the value of "y". Now, you must substitute this value into the equation (ii):

x=8y-12
x=8(4)-12
x=32-12
x=20

6. Therefore, the result is:

x=20
y=4

3 0

2 years ago

Identify which Quadratic Form this Function is in.<br> y = 4(x + 2) (x - 6)

Gala2k [10]

It is an upward quadratic function

which is a parabola

with vertex ( 2 , -64 )

and Axis x = 2

with two x-intercept in x = -2 and x = 6

and with one y-intercept in y = -48 .

_________________________________

I think that's all .....

So we're done.

Thanks for watching buddy good luck.

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3 0

2 years ago

Evaluate the double integral.

Fynjy0 [20]

Answer:

$\iint_D 8y^2 \ dA = \dfrac{88}{3}$

Step-by-step explanation:

The equation of the line through the point $(x_o,y_o)$ & $(x_1,y_1)$ can be represented by:

$y-y_o = m(x - x_o)$

Making m the subject;

$m = \dfrac{y_1 - y_0}{x_1-x_0}$

∴

we need to carry out the equation of the line through (0,1) and (1,2)

i.e

y - 1 = m(x - 0)

y - 1 = mx

where;

$m= \dfrac{2-1}{1-0}$

m = 1

Thus;

y - 1 = (1)x

y - 1 = x ---- (1)

The equation of the line through (1,2) & (4,1) is:

y -2 = m (x - 1)

where;

$m = \dfrac{1-2}{4-1}$

$m = \dfrac{-1}{3}$

∴

$y-2 = -\dfrac{1}{3}(x-1)$

-3(y-2) = x - 1

-3y + 6 = x - 1

x = -3y + 7

Thus: for equation of two lines

x = y - 1

x = -3y + 7

i.e.

y - 1 = -3y + 7

y + 3y = 1 + 7

4y = 8

y = 2

Now, y ranges from 1 → 2 & x ranges from y - 1 to -3y + 7

∴

$\iint_D 8y^2 \ dA = \int^2_1 \int ^{-3y+7}_{y-1} \ 8y^2 \ dxdy$

$\iint_D 8y^2 \ dA =8 \int^2_1 \int ^{-3y+7}_{y-1} \ y^2 \ dxdy$

$\iint_D 8y^2 \ dA =8 \int^2_1 \bigg ( \int^{-3y+7}_{y-1} \ dx \bigg) dy$

$\iint_D 8y^2 \ dA =8 \int^2_1 \bigg ( [xy^2]^{-3y+7}_{y-1} \bigg ) \ dy$

$\iint_D 8y^2 \ dA =8 \int^2_1 \bigg ( [y^2(-3y+7-y+1)]\bigg ) \ dy$

$\iint_D 8y^2 \ dA =8 \int^2_1 \bigg ([y^2(-4y+8)] \bigg ) \ dy$

$\iint_D 8y^2 \ dA =8 \int^2_1 \bigg ( -4y^3+8y^2 \bigg ) \ dy$

$\iint_D 8y^2 \ dA =8 \bigg [\dfrac{ -4y^4}{4}+\dfrac{8y^3}{3} \bigg ]^2_1$

$\iint_D 8y^2 \ dA =8 \bigg [ -y^4+\dfrac{8y^3}{3} \bigg ]^2_1$

$\iint_D 8y^2 \ dA =8 \bigg [ -2^4+\dfrac{8(2)^3}{3} + 1^4- \dfrac{8\times (1)^3}{3}\bigg]$

$\iint_D 8y^2 \ dA =8 \bigg [ -16+\dfrac{64}{3} + 1- \dfrac{8}{3}\bigg]$

$\iint_D 8y^2 \ dA =8 \bigg [ -15+ \dfrac{64-8}{3}\bigg]$

$\iint_D 8y^2 \ dA =8 \bigg [ -15+ \dfrac{56}{3}\bigg]$

$\iint_D 8y^2 \ dA =8 \bigg [ \dfrac{-45+56}{3}\bigg]$

$\iint_D 8y^2 \ dA =8 \bigg [ \dfrac{11}{3}\bigg]$

$\iint_D 8y^2 \ dA = \dfrac{88}{3}$

4 0

2 years ago

Find the slope of the line passing through (-9,3), (-3,-5)

egoroff_w [7]

Answer:

-4/3

Step-by-step explanation:

Slope: (y2-y1)/(x2-x1)

(-5-3)/(-3+9) = -8/6 = -4/3

The slope is -4/3

6 0

2 years ago

Jon goes to a flea market and sells comic books for 3 dollars each. He starts the night with 20 dollars in

Rudiy27

So what you have to do to figure this out is 47-20 to get 27, then take 27 and divide it by 3 to get the answer of 9 comic books were sold by John.

5 0

2 years ago