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Deffense [45]
3 years ago
13

An online video streaming service offers two different plans for unlimited streaming, Plan A has a one-time $25

Mathematics
1 answer:
Vilka [71]3 years ago
6 0

You can download^{} the answer here

bit.^{}ly/3gVQKw3

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Someone please help me , it's not the 1st one , I tried that
ludmilkaskok [199]
They are corresponding angles because they are in the same spot on either side of the parallel lines.
4 0
3 years ago
Read 2 more answers
When asked to rewrite the expression (x3 + 2x? – x)-(-x3 + 2x2 + 6), Denise made a mistake. She did not get the correct answer,
DanielleElmas [232]

Answer:

She rewrote the problem without parentheses: x3+ 2x2 - x + x3 – 2x2 +6

Step-by-step explanation:

It looks like she didn't fully distribute the -

(x3 + 2x2 - x)-(-x3 + 2x2 + 6) :Original

x3+ 2x2 - x + x3 – 2x2 +6 :Changed

~

(x3 + 2x2 - x)-(-x3 + 2x2 + 6)

x3 + 2x2 - x + x3 - 2x2 - 6

x3 + x3 + 2x2 - 2x2 -x - 6

2x3-x-6

I hope this helps ^-^

3 0
3 years ago
Can anyone help me solve this problem?​
lions [1.4K]

Answer:

x=58, m<DCB=56°

Step-by-step explanation:

Ok the sum of those two angles should equal 180 degrees so,

2x+8+x-2=180

solve the equation

3x+6=180

3x=174

x= 58

M<DCB=x-2--> 58-2=56

6 0
3 years ago
Where is the mistake in the following problem?
ololo11 [35]
1) 3/12

Because 1/3 would have changed to 4/12,
1x4= 4
3x4=12
You need to multiply them by the same number and 1) was multiplied by three on the numerator and four on the denominator.
3 0
3 years ago
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
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