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3 years ago

Please help me and please say all the answers to this it has to be more than 1 answer

Mathematics

1 answer:

abruzzese [7]3 years ago

3 0

I believe the correct answers would be A & C

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3x − 2y = −2

Lina20 [59]

Answer: C. -3y+14

Step-by-step explanation:

$3x-2y=-2\\x+3y=14$

Move 3y to ther right side.

$3x-2y=-2\\x=-3y+14$

So the answer is C. Because x = -3y+14.

7 0

3 years ago

Read 2 more answers

Find the value of 7v-6 given that -8v-9=7 Simplify your answer as much as possible 7v-6=

evablogger [386]

Answer:

-20

Step-by-step explanation:

-8v-9=7

Add 9 to each side

-8v-9+9=7+9

-8v = 16

Divide by -8

-8v /-8 = 16/-8

v = -2

Now find

7v -6

7(-2) -6

-14 -6

-20

4 0

4 years ago

Read 2 more answers

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

-5x+10>-15 help please

nexus9112 [7]

Answer:

Step-by-step explanation:

-5x + 10 > -15 is your original equation.

You need to isolate the x variable.

Start by subtracting 10.

-5x > -25

x > 5

3 0

3 years ago

If g(x) = x+1/x-2 and h(x) = 4-x, what is the value of (g•h)(-3)?

Yakvenalex [24]

Answer:

I think my answer is

C. 15/2

4 0

3 years ago