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3 years ago

Simplify....... will give BRAINLIEST....

Mathematics

2 answers:

Paha777 [63]3 years ago

8 0

Answer:

10^x

Step-by-step explanation:

√2^x×5^2x÷2^-x

√2^x×5^2x×2^x

√10^2x

10^x

boyakko [2]3 years ago

3 0

Answer:

${10}^{x}$

Step-by-step explanation:

$\sqrt{ {2}^{x} \times {5}^{2x} \div {2}^{ - x} } \\ \\ = \sqrt{ {2}^{x} \times {5}^{2x} \: \times {2}^{x} }\\ \\ = \sqrt{ {2}^{x + x} \times {5}^{2x}}\\ \\ = \sqrt{ {2}^{2x} \times {5}^{2x}}\\ \\ = \sqrt{ {(2 \times 5)}^{2x} }\\ \\ = \sqrt{ {(10)}^{2x} }\\ \\ = {10}^{x}$

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Write the quadratic equation in factored form. Be sure to write the entire equation. x^2 + x - 12 = 0

nignag [31]

X^2 + X - 12 = 0

(X + 4)(X - 3) = 0

X = -4 and X = 3.

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3 years ago

Describe how you would find 24+ 36 using mental math

FinnZ [79.3K]

Answer:

First add 6 and 4 together to get 10 now add 2 and 3 and 1 together which gets you 6 and add the zero from the 10 and you get 60

Step-by-step explanation:

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3 years ago

sarah had 28 stickers and put x number of stickers on her binded. haylee had 40 stickers and put three times as many stickers on

Mariana [72]

Answer:

Number of stickers Sarah had = 28

Number of stickers she put on her binder = x.

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=> 3x - x = 40 - 28

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Step-by-step explanation:

7 0

3 years ago

7(x+y)^(2)+13x(x+y)-2^(2) <br> help

artcher [175]

Answer:

https://www.symbolab.com/solver/step-by-step/7%5Cleft(x%2By%5Cright)%5E%7B2%7D%2B13x%5Cleft(x%2By%5Cright)-2%5E%7B2%7D

Step-by-step explanation:

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3 years ago

Help appreciated on question in image!<br> Thanks:)

Verdich [7]

Answer:

$x=-1,\:x=-7,\:x=i,\:x=-i$

Step-by-step explanation:

Considering the equation

$x^4+8x^3+8x^2+8x+7=0$

Solving

$x^4+8x^3+8x^2+8x+7$

$\mathrm{Factor\:}x^4+8x^3+8x^2+8x+7:\quad \left(x+1\right)\left(x+7\right)\left(x^2+1\right)$

$\mathrm{Use\:the\:rational\:root\:theorem}$

$a_0=7,\:\quad a_n=1$

$\mathrm{The\:dividers\:of\:}a_0:\quad 1,\:7,\:\quad \mathrm{The\:dividers\:of\:}a_n:\quad 1$

$\mathrm{Therefore,\:check\:the\:following\:rational\:numbers:\quad }\pm \frac{1,\:7}{1}$

$-\frac{1}{1}\mathrm{\:is\:a\:root\:of\:the\:expression,\:so\:factor\:out\:}x+1$

$=\left(x+1\right)\frac{x^4+8x^3+8x^2+8x+7}{x+1}...[A]$

Solving

$\frac{x^4+8x^3+8x^2+8x+7}{x+1}$

$=x^3+7x^2+x+7$

Putting $\frac{x^4+8x^3+8x^2+8x+7}{x+1}$ = $x^3+7x^2+x+7$ in equation [A]

So,

$\left(x+1\right)\frac{x^4+8x^3+8x^2+8x+7}{x+1}...[A]$

$=\left(x+1\right)x^3+7x^2+x+7$

$x^3+7x^2+x+7=\left(x+7\right)\left(x^2+1\right)$

So,

Equation [A] becomes

$=\left(x+1\right)\left(x+7\right)\left(x^2+1\right)$

So, the polynomial equation becomes

$\left(x+1\right)\left(x+7\right)\left(x^2+1\right)=0$

$\mathrm{Using\:the\:Zero\:Factor\:Principle:\quad \:If}\:ab=0\:\mathrm{then}\:a=0\:\mathrm{or}\:b=0\:\left(\mathrm{or\:both}\:a=0\:\mathrm{and}\:b=0\right)$ $\mathrm{Solve\:}\:x+1=0:\quad x=-1$