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aev [14]
3 years ago
15

What is the Value of X please help!

Mathematics
1 answer:
Brums [2.3K]3 years ago
4 0

Answer:

x = 7cm

Step-by-step explanation:

\frac{36}{6} = \frac{42}{x}\\\\6 = \frac{42}{x}\\\\x = \frac{42}{6} = 7

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Graph the equation by plotting three
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Answer:

(0,7.5) (1,10) (2,12.5)

Step-by-step explanation:

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3 years ago
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Solve the equation<br> 8.3v = -5.1
Mrrafil [7]

Answer:

v= -0.61445783132

Step-by-step explanation:

Divide both sides by 8.3

-5.1 ÷ 8.3= -0.61445783132

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8.3v ÷ 8.3= v

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3 years ago
Which expression is equivalent to 15 y - 8 y​ ?
siniylev [52]

Answer:

D. The expression 5 y plus 2 y is equivalent to 15 y minus 8 y.

Step-by-step explanation:

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3 years ago
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A company is constructing an open-top, square-based, rectangular metal tank that will have a volume of 34.5 ft3. what dimensions
____ [38]

Answer:

l \approx 3.3\,ft, x = 3.2\,ft

Step-by-step explanation:

The equations of volume and surface area are presented below:

34.5 = l^{2}\cdot x

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The length of the tank is:

x = \frac{34.5}{l^{2}}

The expression fo the surface area is therefore simplified into an univariable form:

A_{s} = 2\cdot l^{2} + 4\cdot \left(\frac{34.5}{l} \right)

A_{s} = 2\cdot l^{2} + \frac{138}{l}

The first and second derivatives of the expression are, respectively:

A_{s}' = 4\cdot l -\frac{138}{l^{2}}

A_{s}'' = 4 +\frac{276}{l^{3}}

The first derivative is equalized to zero and length of the square side is now found:

4\cdot l -\frac{138}{l^{2}} = 0

4\cdot l^{3}-138 = 0

l^{3} = \frac{138}{4}

l = \sqrt[3]{\frac{138}{4} }

l \approx 3.3\,ft

Now, the second derivative offers a criteria to determine if solution leads to an absolute minimum:

A_{s}'' = 4 + \frac{276}{(3.3\,ft)^{3}}

A_{s}'' =  11.7 (Absolute minimum)

The depth of the tank is:

x = \frac{34.5\,ft^{3}}{(3.3\,ft)^{2}}

x = 3.2\,ft

6 0
3 years ago
Geomtry plzzz help 15 points
siniylev [52]

Answer:

19

Step-by-step explanation:

they give you xz, and they want half, so just divide 38 by 2

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