X=(a+b)/2 solve for a

Mathematics

1 answer:

Usimov [2.4K]3 years ago

5 0

Answer:

a=2x-b

Step-by-step explanation:

$x=\frac{a+b}{2}\\\\\frac{a+b}{2} = x\\ a+b = 2x\\a= 2x-b\\$

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Which of the following best describes the change in the range when 13 is

svetoff [14.1K]

The answer is D). It increases by 13.
If you put the numbers in order from least to greatest.. 26,34,38,49,65,75,81
You will see the 26 is the smallest number and 81 in the highest number. The range between the two numbers is 55. To find the range you take the smallest number and subtract it from the biggest number. But when you add the number 13 to the problem then26 is no longer the smallest Number because 13 replaced it. So then you take 81 and subtract 13 from it and get 68. And then you take 55 and subtract it from 68 and come with 13. Which makes the answer D.

6 0

4 years ago

What is the solution for the equation StartFraction 5 Over 3 b cubed minus 2 b squared minus 5 EndFraction = StartFraction 2 Ove

wolverine [178]

Answer:

The solutions are:

$b=0,\:b=4$

Step-by-step explanation:

Considering the expression

$\frac{5}{3b^3-2b^2-5}=\frac{2}{b^3-2}$

Solving the expression

$\frac{5}{3b^3-2b^2-5}=\frac{2}{b^3-2}$

$\mathrm{Apply\:fraction\:cross\:multiply:\:if\:}\frac{a}{b}=\frac{c}{d}\mathrm{\:then\:}a\cdot \:d=b\cdot \:c$

$5\left(b^3-2\right)=\left(3b^3-2b^2-5\right)\cdot \:2$

$5b^3-10=6b^3-4b^2-10$

$\mathrm{Switch\:sides}$

$6b^3-4b^2-10=5b^3-10$

$6b^3-4b^2-10+10=5b^3-10+10$

$6b^3-4b^2=5b^3$

$\mathrm{Subtract\:}5b^3\mathrm{\:from\:both\:sides}$

$6b^3-4b^2-5b^3=5b^3-5b^3$

$b^3-4b^2=0$

$Using\:the\:Zero\:Factor\:Principle:$ $if\:\mathrm ab=0\:\mathrm{then}\:a=0\:\mathrm{or}\:b=0\:\left(\mathrm{or\:both}\:a=0\:\mathrm{and}\:b=0\right)$