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2 years ago

7

The diagram below shows the measurements of two identical rectangular prisms joined together. Austin says the combined volume of

the prisms is 225 cubic meters. Explain Austin’s error and give the correct volume.

1 answer:

Ksivusya [100]2 years ago

7 0

Answer:

450 cubic meters

Step-by-step explanation:

( 15 X 5 X 3) + ( 15 x 5 x 3)
225 + 225
450 cubic meters

Austin didn't add the 2 prisms, he multiplied the 1st prism and kept it as the answer.

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goldfiish [28.3K]

$26-2\left(10-1\right)+\left(3+4\cdot 11\right)$

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$55$

"55 is correct answer".

Hope this helps!

Thanks!

-Charlie

5 0

2 years ago

Put the equation in standard form.<br><br> y - 1 = -1/3 (x - 6)

Drupady [299]

Answer:

x + 3y = 9

Step-by-step explanation:

The equation of a line in standard form is

Ax + By = C ( A is a positive integer and B, C are integers )

Given

y - 1 = - $\frac{1}{3}$ (x - 6)

Multiply through by 3 to clear the fraction

3y - 3 = - (x - 6) ← distribute parenthesis by - 1

3y - 3 = - x + 6 ( add x to both sides )

x + 3y - 3 = 6 ( add 3 to both sides )

x + 3y = 9 ← in standard form

3 0

2 years ago

On a coordinate plane, 2 lines intersect at (0, 1).

choli [55]

Answer:

(0,1)

Step-by-step explanation:

i just did it

4 0

2 years ago

Read 2 more answers

Solve<br> -3x - y = 8<br> - 7x + 3y = -24

Nookie1986 [14]

The solution is (0,-8)

7 0

2 years ago

Read 2 more answers

Find f. f ″(x) = x^−2, x > 0, f(1) = 0, f(6) = 0

marin [14]

If you do in fact mean $f(1)=f(6)=0$ (as opposed to one of these being the derivative of $f$ at some point), then integrating twice gives

$f''(x) = -\dfrac1{x^2}$

$f'(x) = \displaystyle -\int \frac{dx}{x^2} = \frac1x + C_1$

$f(x) = \displaystyle \int \left(\frac1x + C_1\right) \, dx = \ln|x| + C_1x + C_2$

From the initial conditions, we find

$f(1) = \ln|1| + C_1 + C_2 = 0 \implies C_1 + C_2 = 0$

$f(6) = \ln|6| + 6C_1 + C_2 = 0 \implies 6C_1 + C_2 = -\ln(6)$

Eliminating $C_2$ , we get

$(C_1 + C_2) - (6C_1 + C_2) = 0 - (-\ln(6))$

$-5C_1 = \ln(6)$

$C_1 = -\dfrac{\ln(6)}5 = -\ln\left(\sqrt[5]{6}\right) \implies C_2 = \ln\left(\sqrt[5]{6}\right)$

Then

$\boxed{f(x) = \ln|x| - \ln\left(\sqrt[5]{6}\right)\,x + \ln\left(\sqrt[5]{6}\right)}$

3 0

1 year ago