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2 years ago

Determine the value of the variable that makes the equation true. 12-b=12.5

Mathematics

1 answer:

dusya [7]2 years ago

6 0

Answer:

B = -.5

Step-by-step explanation:

If B is negative is cancels out the prexisting negative making it positive.

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What is the perimeter of this rectangle?

garik1379 [7]

Answer:

B.) (X2 -x1)(y2 - y1

Step-by-step explanation:

I learned the formula

7 0

3 years ago

What is 10 to the 3rd power times 10 to the 2nd power

blagie [28]

Answer:100,000

Step-by-step explanation:

4 0

2 years ago

400 people are asked if they drink tea

Korvikt [17]

Answer:

56 people drink 3 cups daily

Step-by-step explanation:

What's the total number of people that drink 3 cups?

Total number of people = 400

7 / 10 were asked said yes

Number of people that said yes = (7 / 10) × 400 = 280

280 people drink tea.

20% of 280 people drink 3 cups daily, whats the number = ?

20% of 280 = ?

20 / 100 = x / 280

(20 × 280) = (100 × x)

5600 = 100x

X = 5600 / 100

X = 56 people

56 people drink 3 cups of tea daily.

5 0

3 years ago

two years of local internet service cost $685 including the installation fee of $85 what is the monthly fee?

zimovet [89]

2 years is the same as 24 months, so the equation would be:
685 = 24x + 85, where x is the monthly fee.
600 = 24x
600/24 = x
x = 25
The monthly fee is $25.

7 0

2 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