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1 year ago

Find the surface area

Mathematics

2 answers:

steposvetlana [31]1 year ago

6 0

Answer:

should be 872 in^2

Step-by-step explanation:

Arada [10]1 year ago

6 0

Answer:

Area = 872

Step-by-step explanation:

Formula for the surface area of a rectangular prism is

A=2(wl+hl+hw)

Plug in

A =2·(8·17+12·17+12·8)=872

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Select from the following choices those graphs that represent only a dilation applied to this polygon with a scale factor great

romanna [79]

Answer:

B and C

Step-by-step explanation:

Required

Select graphs that are dilated by a scale factor greater than 1

For graph A:

Graph A is smaller than the original graph. This indicates dilation with a scale factor less than 1

For graph B:

Graph B is bigger than the original graph and is dilated over (0,0). This indicates dilation with a scale factor greater than 1

For graph C:

Graph C is bigger than the original graph; however, it is not dilated over (0,0). This indicates dilation with a scale factor greater than 1

For graph D:

Graph D is bigger than the original graph; however, it is not only dilated but also flipped over (i.e. rotated).

<em>Hence, b and c is true</em>

5 0

3 years ago

Which statements accurately describe the function f(x) = 3(StartRoot 18 EndRoot) Superscript x? Select three options. The domain

Studentka2010 [4]

Answer:

The domain is all real numbers.

The initial value is 3

The simplified base is $9\sqrt{2}$

Step-by-step explanation:

The given function is $f(x)=3(\sqrt{18})^x$ .

To find the initial value, we put x=0 into the function:

$f(0)=3(\sqrt{18})^0$ .

$f(0)=3(1)=3$ .

The initial value is actually 3.

The given function is an exponential function, therefore the domain is all real numbers.

The range of this function refers to all values of y for which the function is defined.

The line y=0, is the horizontal asymptote.

The range is $y\:>\:0$

The simplified base is $3\sqrt{18}=3\sqrt{9\times2}$ .

$3\sqrt{18}=3\sqrt{9}\times\sqrt{2}$

$3\sqrt{18}=3\times3\times\sqrt{2}$

$3\sqrt{18}=9\sqrt{2}$

4 0

2 years ago

Read 2 more answers

Multiply the following polynomials, then place the answer in the proper location on the grid. Write the answer in descending pow

shutvik [7]

Answer:

a^3-125

Step-by-step explanation:

(a-5)(a^2+5a+25)

a^3+5a^2+25a-5a^2-25a-125

a^3+5a^2-5a^2+25a-25a-125

a^3-125

5 0

2 years ago

Read 2 more answers

which of the following is equivalent to 3 sqrt 32x^3y^6 / 3 sqrt 2x^9y^2 where x is greater than or equal to 0 and y is greater

Nutka1998 [239]

Answer:

$\frac{\sqrt[3]{16y^4}}{x^2}$

Step-by-step explanation:

The options are missing; However, I'll simplify the given expression.

Given

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$

Required

Write Equivalent Expression

To solve this expression, we'll make use of laws of indices throughout.

From laws of indices $\sqrt[n]{a} = a^{\frac{1}{n}}$

So,

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$ gives

$\frac{(32x^3y^6)^{\frac{1}{3}}}{(2x^9y^2)^\frac{1}{3}}$

Also from laws of indices

$(ab)^n = a^nb^n$

So, the above expression can be further simplified to

$\frac{(32^\frac{1}{3}x^{3*\frac{1}{3}}y^{6*\frac{1}{3}})}{(2^\frac{1}{3}x^{9*\frac{1}{3}}y^{2*\frac{1}{3}})}$

Multiply the exponents gives

$\frac{(32^\frac{1}{3}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

Substitute $2^5$ for 32

$\frac{(2^{5*\frac{1}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

$\frac{(2^{\frac{5}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

From laws of indices

$\frac{a^m}{a^n} = a^{m-n}$

This law can be applied to the expression above;

$\frac{(2^{\frac{5}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$ becomes

$2^{\frac{5}{3}-\frac{1}{3}}x^{1-3}*y^{2-\frac{2}{3}}$

Solve exponents

$2^{\frac{5-1}{3}}*x^{-2}*y^{\frac{6-2}{3}}$

$2^{\frac{4}{3}}*x^{-2}*y^{\frac{4}{3}}$

From laws of indices,

$a^{-n} = \frac{1}{a^n}$ ; So,

$2^{\frac{4}{3}}*x^{-2}*y^{\frac{4}{3}}$ gives

$\frac{2^{\frac{4}{3}}*y^{\frac{4}{3}}}{x^2}$

The expression at the numerator can be combined to give

$\frac{(2y)^{\frac{4}{3}}}{x^2}$

Lastly, From laws of indices,

$a^{\frac{m}{n} = \sqrt[n]{a^m}$ ; So,

$\frac{(2y)^{\frac{4}{3}}}{x^2}$ becomes

$\frac{\sqrt[3]{(2y)}^{4}}{x^2}$

$\frac{\sqrt[3]{16y^4}}{x^2}$

Hence,

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$ is equivalent to $\frac{\sqrt[3]{16y^4}}{x^2}$

8 0

3 years ago

Can you guys help me solve this problem

sleet_krkn [62]

Answer:6

Step-by-step explanation:

5 0

2 years ago

Read 2 more answers