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3 years ago

Solve for X 7(X-3)+3(4-X)=-8

1 answer:

7 0

$7(x-3)+3(4-x)=-8\\\\7x-21+12-3x=-8\\\\4x-9=-8\ \ \ \ |add\ 9\ to\ both\ sides\\\\4x=1\ \ \ \ |divide\ both\ sides\ by\ 4\\\\\boxed{x=0.25}$

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it's from Khan Academy and there are answer choices it's called interpet negative number addition and subtraction expression

kolbaska11 [484]

Answer: Hi!

Answers B and D are correct.

For B, since negative numbers are on the left side of the number line, a positive number would have to be to the right of a negative number. Since we are adding a negative and a positive number of the same absolute value (9), they would cancel out to 0, which makes D true.

Hope this helps!

4 0

3 years ago

Read 2 more answers

A conical container can hold 120π cubic centimeters of water. The diameter of the base of the container is 12 centimeters.

nignag [31]

Answer:

A. 10cm

B. 8 times

Step-by-step explanation:

The question is on volume of a conical container

Volume of a cone= $\pi r^{2} h/3$

where r is the radius of base and h is the height of the cone

Given diameter= 12 cm, thus radius r=12/2 =6 cm

$v=\pi r^2h/3 \\120\pi =\pi *6*6*h/3\\120\pi =12\pi h\\10=h$

h=10 cm

If height and diameter were doubled

New height = 2×10 =20 cm

New diameter = 2×12 = 24, r=12 cm

volume = $v=\pi r^2h/3\\v=\pi *12*12*20/3\\v=960\pi$

To find the number of times we divide new volume with the old volume

$N= 960\pi /120\pi \\\\N= 8$

8 0

3 years ago

Read 2 more answers

Create a factorable polynomial with a GCF of 6. Rewrite that polynomial in two other equivalent forms. Explain how each form was

lubasha [3.4K]

OK well take out the 6 for the first form

= 6(2x^2 + x + 4)

Also we can factor the first 2 terms to give the second form:-

= 6x(2x + 1) + 24

5 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:

$5^{\left(-x\right)}+7=2x+4\\\\\mathrm{Prepare}\:5^{\left(-x\right)}+7=2x+4\:\mathrm{for\:Lambert\:form}:\quad 1=\left(2x-3\right)e^{\ln \left(5\right)x}\\\\\mathrm{Rewrite\:the\:equation\:with\:}\\\left(x-\frac{3}{2}\right)\ln \left(5\right)=u\mathrm{\:and\:}x=\frac{u}{\ln \left(5\right)}+\frac{3}{2}\\\\1=\left(2\left(\frac{u}{\ln \left(5\right)}+\frac{3}{2}\right)-3\right)e^{\ln \left(5\right)\left(\frac{u}{\ln \left(5\right)}+\frac{3}{2}\right)}$

$Simplify\\\\\mathrm{Rewrite}\:1=\frac{2e^{u+\frac{3}{2}\ln \left(5\right)}u}{\ln \left(5\right)}\:\\\\\mathrm{in\:Lambert\:form}:\quad \frac{e^{\frac{2u+3\ln \left(5\right)}{2}}u}{e^{\frac{3\ln \left(5\right)}{2}}}=\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}$

$\mathrm{Solve\:}\:\frac{e^{\frac{2u+3\ln \left(5\right)}{2}}u}{e^{\frac{3\ln \left(5\right)}{2}}}=\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}:\quad u=\text{W}_0\left(\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}\right)\\\\\mathrm{Substitute\:back}\:u=\left(x-\frac{3}{2}\right)\ln \left(5\right),\:\mathrm{solve\:for}\:x$

$\mathrm{Solve\:}\:\left(x-\frac{3}{2}\right)\ln \left(5\right)=\text{W}_0\left(\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}\right):\\\quad x=\frac{\text{W}_0\left(\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}\right)}{\ln \left(5\right)}+\frac{3}{2}$