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3 years ago

Karen deposited $25 in the bank on monday

1 answer:

4 0

Karen deposited $25 in the bank on Monday, $50 on Wednesday and $15 on Friday. On Saturday, she took out $40. Karen’s original balance was $100. What is her balance now?<span>

Given that Karen's original balance was $100.
After the three deposits, her balance is 100 + 25 + 50 + 15 = 190
After her withdrawal, her balance was 190 - 40 = 150

Therefore, Karen's balance now is $150</span>

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<img src="https://tex.z-dn.net/?f=%5Csf%20Solve%5C%3A%20for%5C%3A%20x%3A-" id="TexFormula1" title="\sf Solve\: for\: x:-" alt="\

Mandarinka [93]

Answer:

x = 3

Step-by-step explanation:

=> 4x-5x+9=5x-9

=> -x-5x = -9-9

=> -6x = -18

=> x = 18/6

=> x = 3

6 0

2 years ago

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A photo studio is going to display a portrait in their lobby. The portrait is being re-produced from a photo with dimensions 6in

Burka [1]

Answer:

the form of it will be the same

Step-by-step explanation:

the form will need to be the same or a part will be cropped so it fits the photo/portrait

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3 years ago

The population in Rochester, New York, was 219,773 people in 2000 and 210,565 people in 2010. By about what percent did the popu

garri49 [273]

The population shrank by 23.86762%

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2 years ago

A rectangle has a height 3z and a width of 2x-3<br><br> Express the area of the entire rectangle.

viktelen [127]

Answer:

3z(2x - 3)

Step-by-step explanation:

Area of a rectangle, a = height * width

height, h = 3z

width, w = 2x - 3

a = 3z * (2x -3)

expand

a = 6xz - 9z

a = 3z(2x - 3)

5 0

2 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