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3 years ago

If h(x)=9+x², solve for x when h(x)=13

Mathematics

2 answers:

AlexFokin [52]3 years ago

6 0

Answer:

I hope you got it. I have presented it clearly and stepwise.

vazorg [7]3 years ago

4 0

Answer:

$\boxed{ \bold{ \huge{ \boxed{ \bold{ \sf{x = 2}}}}}}$

Step-by-step explanation:

Given, h(x) = 13

To find : value of x

$\sf{h(x) = 9 + {x}^{2} }$

$\longrightarrow{ \sf{13 = 9 + {x}^{2} }}$

Swap the sides of equation

$\longrightarrow{ \sf{9 + {x}^{2} = 13}}$

Move 9 to right hand side and change it's sign

$\longrightarrow{ \sf{ {x}^{2} = 13 - 9}}$

Subtract 9 from 13

$\longrightarrow{ \sf{ {x}^{2} = 4}}$

$\longrightarrow{ \sf{ \sqrt{ {x}^{2} } = \sqrt{4} }}$

$\longrightarrow{ \sf{x = \: \sqrt{ {2}^{{2}} }}}$

$\longrightarrow{ \sf{x = 2}}$

Hope I helped!

Best regards!

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20 POINTS

RUDIKE [14]

<em>The correct expressions are as follows:</em>

$7^{\frac{1}{5}} \cdot 49^{\frac{7}{5}}$ Equivalent $343$

$7^{\frac{1}{5}} \cdot 49^{\frac{7}{5}}$ Not Equivalent $49$

$7^{\frac{1}{5}} \cdot 49^{\frac{7}{5}}$ Equivalent $7^{\frac{1}{5}} \cdot 7^{\frac{14}{5}}$

$7^{\frac{1}{5}} \cdot 49^{\frac{7}{5}}$ Not Equivalent $49^{\frac{2}{10}} \cdot 7^{\frac{1}{5}}$

$\texttt{ }$

<h3>Further explanation</h3>

Let's recall following formula about Exponents and Surds:

$\boxed { \sqrt { x } = x ^ { \frac{1}{2} } }$

$\boxed { (a ^ b) ^ c = a ^ { b . c } }$

$\boxed {a ^ b \div a ^ c = a ^ { b - c } }$

$\boxed {\log a + \log b = \log (a \times b) }$

$\boxed {\log a - \log b = \log (a \div b) }$

<em>Let us tackle the problem!</em>

$\texttt{ }$

$7^{\frac{1}{5}} \cdot 49^{\frac{7}{5}} = 7^{\frac{1}{5}} \cdot (7^2)^{\frac{7}{5}}$

$7^{\frac{1}{5}} \cdot 49^{\frac{7}{5}} = 7^{\frac{1}{5}} \cdot (7)^{2\times \frac{7}{5}}$

$7^{\frac{1}{5}} \cdot 49^{\frac{7}{5}} = \boxed{7^{\frac{1}{5}} \cdot 7^{\frac{14}{5}}}$

$7^{\frac{1}{5}} \cdot 49^{\frac{7}{5}} = 7^{\frac{1}{5} + \frac{14}{5}}$

$7^{\frac{1}{5}} \cdot 49^{\frac{7}{5}} = 7^{\frac{15}{5}}$

$7^{\frac{1}{5}} \cdot 49^{\frac{7}{5}} = 7^{3}$

$7^{\frac{1}{5}} \cdot 49^{\frac{7}{5}} = \boxed{343}$

$\texttt{ }$

<em>From the results above, it can be concluded that the correct statements are:</em>

$7^{\frac{1}{5}} \cdot 49^{\frac{7}{5}}$ Equivalent $343$

$7^{\frac{1}{5}} \cdot 49^{\frac{7}{5}}$ Not Equivalent $49$

$7^{\frac{1}{5}} \cdot 49^{\frac{7}{5}}$ Equivalent $7^{\frac{1}{5}} \cdot 7^{\frac{14}{5}}$

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$\texttt{ }$

<h3>Learn more</h3>

Coefficient of A Square Root : brainly.com/question/11337634
The Order of Operations : brainly.com/question/10821615
Write 100,000 Using Exponents : brainly.com/question/2032116

<h3>Answer details</h3>

Grade: High School

Subject: Mathematics

Chapter: Exponents and Surds

Keywords: Power , Multiplication , Division , Exponent , Surd , Negative , Postive , Value , Equivalent , Perfect , Square , Factor.

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Klio2033 [76]

Answer:

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Mean ± (Zcritical * σ/sqrt(n))

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(27.954, 30.046)

b. Does this result lend support to the car's company's claim?

Yes, it does, 28 falls in between the the calculated confidence interval.

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