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AnnZ [28]
3 years ago
10

3x2(9x)0 = a0x a1 help me people

Mathematics
1 answer:
Temka [501]3 years ago
4 0

X= {0, -3} is the answer hope this helps.

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Simplify the expression<br> 13 (9x)
Ratling [72]

13(9x)

13 * 9x = 117x

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20 POINTS
denis-greek [22]

Answer:

c. 441ft^2

Step-by-step explanation:

Area is height times length and since it's a square, safe to say that the sides are equal in length. So square root both of the given areas of the 2 square:

\sqrt{841} = 29

\sqrt{400} = 20

Now u have side b and c of the triangle, it's a right triangle so use the Pythagorean theorem a^2 + b^2 = c^2

a^2 + b^2 = c^2

a^2 + 20^2 = 29^2

a^2 + 400 = 841

subtract 4 from both sides

a^2 = 441

square root both sides

\sqrt{a^2} = \sqrt{441}

a = 21

Now the side a of the triangle is also a side of the small square therefore

21^2 = 441

8 0
3 years ago
The answer to a system is a(n)<br> A) point<br> B) line<br> C) two points<br> D) equation
love history [14]
I think the answer is D ?
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3 years ago
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5^(-x)+7=2x+4 This was on plato
Setler79 [48]

Answer:

Below

I hope its not too complicated

x=\frac{\text{W}_0\left(\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}\right)}{\ln \left(5\right)}+\frac{3}{2}

Step-by-step explanation:

5^{\left(-x\right)}+7=2x+4\\\\\mathrm{Prepare}\:5^{\left(-x\right)}+7=2x+4\:\mathrm{for\:Lambert\:form}:\quad 1=\left(2x-3\right)e^{\ln \left(5\right)x}\\\\\mathrm{Rewrite\:the\:equation\:with\:}\\\left(x-\frac{3}{2}\right)\ln \left(5\right)=u\mathrm{\:and\:}x=\frac{u}{\ln \left(5\right)}+\frac{3}{2}\\\\1=\left(2\left(\frac{u}{\ln \left(5\right)}+\frac{3}{2}\right)-3\right)e^{\ln \left(5\right)\left(\frac{u}{\ln \left(5\right)}+\frac{3}{2}\right)}

Simplify\\\\\mathrm{Rewrite}\:1=\frac{2e^{u+\frac{3}{2}\ln \left(5\right)}u}{\ln \left(5\right)}\:\\\\\mathrm{in\:Lambert\:form}:\quad \frac{e^{\frac{2u+3\ln \left(5\right)}{2}}u}{e^{\frac{3\ln \left(5\right)}{2}}}=\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}

\mathrm{Solve\:}\:\frac{e^{\frac{2u+3\ln \left(5\right)}{2}}u}{e^{\frac{3\ln \left(5\right)}{2}}}=\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}:\quad u=\text{W}_0\left(\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}\right)\\\\\mathrm{Substitute\:back}\:u=\left(x-\frac{3}{2}\right)\ln \left(5\right),\:\mathrm{solve\:for}\:x

\mathrm{Solve\:}\:\left(x-\frac{3}{2}\right)\ln \left(5\right)=\text{W}_0\left(\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}\right):\\\quad x=\frac{\text{W}_0\left(\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}\right)}{\ln \left(5\right)}+\frac{3}{2}

3 0
3 years ago
If you reflect ABCD across x-axis, what will be the coordinates of A’ and C’
disa [49]

Answer:

A (1, -5), B (2, -5)

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