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Ivenika [448]
3 years ago
12

HELP ASAP PLS ASAP I ALSO NEED A STEP BY STEP EXPLANATION

Mathematics
1 answer:
Natali5045456 [20]3 years ago
5 0
K=5
if you add 5 to f(x) it will move up five and become g(x)
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A rectangular site measures 125 feet (frontage) by 256 feet (depth), of which 26 feet is in the public right of way. what is the
alexandr1967 [171]

Answer:

32,000 square feet gross area and 28,750 square feet net area

Step-by-step explanation:

Given a rectangular site measures 125 feet (frontage) by 256 feet (depth), of which 26 feet is in the public right of way. We have to find the  the gross and net site area.

As, 26 feet width is in the public right of way which excluded in the net area but included in gross area.

For gross area,

Length=125 feet

Width=256 feet

Area=Length\times width=125\times 256=32000 square feet

For net area,

Length=125 feet

Width=256-26=230 feet

Area=Length\times width=125\times 230=28750 square feet

Hence, last option is correct.

The answer is:  32,000 square feet gross area and 28,750 square feet net area

6 0
3 years ago
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
Amy, Lee, Tom and Uma are candidates in the school election. The person
Reil [10]

Answer:

Tom/Lee

Step-by-step explanation:

5 0
3 years ago
Is the square root of 225 a rational number?
MArishka [77]

Step-by-step explanation:

\text{The principal square root}:\\\\\sqrt{a}=b\iff a^2=b\ \text{for}\ a\geq0\ \text{and}\ b\geq0.\\\\\text{The square root:}\\\\\sqrt{a}=b\iff b^2=a\ \text{for}\ a\geq0\\===================

\text{If you want the principal square root (arithmetic square root):}\\\\\sqrt{225}=15\ \text{because}\ 15^2=225\\\\\text{If you want square root (algebraic square root):}\\\\\sqrt{225}=\pm15\ \text{because}\ (-15)^2=225\ \text{and}\ 15^2=225

7 0
3 years ago
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Olivia goes to Egypt and saw a model of triangle shaped pyramid. She enlarged it using a scale factor of 1.2. What will be the p
sertanlavr [38]
Olivia applied the scale factor to the measurements of the model she saw. We need to know those in order to calculate the new ones.
6 0
3 years ago
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