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3 years ago

I need help with these 2 algebra questions

Mathematics

2 answers:

omeli [17]3 years ago

8 0

Answer:

question.4.

answer . b

question. 5

answer . 7m and -2m

Romashka-Z-Leto [24]3 years ago

3 0

Answer:

Step-by-step explanation:

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-2x + 5 (1 -7x ) =190

luda_lava [24]

The answer is x= -5 ...explained... -2x + 5 -35x= 190...-37x +5 = 190 ... now subtract 190-5 = 185... divide 185 by - 37 and get -5

8 0

3 years ago

Two parallel lines are _____ coplanar.

dimulka [17.4K]

Answer:

1. always

2. sometimes

Step-by-step explanation:

Two lines are coplanar if they lie in the same plane or in parallel planes. There is ALWAYS a plane which contains two parallel lines (see first attached image for details).

Two lines that lie in parallel planes are sometimes parallel. For example, see at second image. Planes $\alpha$ and $\beta$ are parallel. Consider pair of lines $a$ and $a_1$ - they are parallel, but if you consider the pair of lines $a$ and $b_1$ , you can see they are not parallel.

8 0

3 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

Please help, will give brainliest! :)

Nonamiya [84]

Answer:

the answer is 2>am sorry if wrong

6 0

2 years ago

Can anyone help me out ?

AnnZ [28]

no se pero de seguro alguien más te ayudará

8 0

3 years ago

I need help with these 2 algebra questions​

I need help with these 2 algebra questions