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2 years ago

6

Given the equation: (a-c)/(x-a) = m, solve for x.

2 answers:

Iteru [2.4K]2 years ago

6 0

We are given with the equation (a-c)/(x-a) = m. In this case, to isolate x, we can use the division, multiplication, addition adn subtraction properties. In this case, Divide the whole equation by m, to give, (a-c)/(x-a)/m = 1. Then multiply by (x-a). Then to isolate x, add a. this results to
x = [(a-c)/m]

REY [17]2 years ago

4 0

If you would like to solve the equation<span> (a - c) / (x - a) = m for x, you can do this using the following steps:

</span><span>(a - c) / (x - a) = m
</span>a - c = m * (x - a)
a - c = mx - ma
mx = a - c + ma
x = (a - c + ma) / m
x = a/m - c/m + a

The correct result would be <span>x = a/m - c/m + a.</span>

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Consider the following function. f(x) = |x − 9| Find the derivative from the left at x = 9. If it does not exist, enter NONE. -1

nydimaria [60]

Answer with Step-by-step explanation:

We are given that

$f(x)=\mid x-9\mid$

$f(x)=-(x-9)$ when x<9

$f(x)=x-9$ when $x\geq 9$

LHD

$\lim_{x\rightarrow 9-}f'(x)$

= $-\lim_{x\rightarrow 9-}(x-9)=-1$

RHD

$\lim_{x\rightarrow 9+}f'(x)$

$=\lim_{x\rightarrow 9+}(x-9)=1$

$LHD\neq RHD$

Hence, the function is not differentiable at x=9

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3 years ago

Whoever gives me the correct answer to this will get a brainlest <br> I need the answer ASAP!

Amiraneli [1.4K]

Volume of cone= 1/3πr^2h

8842.24 m^3= 1/3πr^2h

1/3*3.14*33*r^2= 8842.24m^3

r^2 = 8842.24*3/3.14*33

r^2= 26526.72/103.62

r^2 = 256

r= √256

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2 years ago

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a_sh-v [17]

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3 years ago

1. Find the Least Common Multiple of these two monomials:<br> See picture

gtnhenbr [62]

Answer:

<em>The last choice is correct</em>

<em /> $LCM=120a^4b^7c^5$ <em />

Step-by-step explanation:

<u>Least Common Multiple (LCM)</u>

To find the LCM we can follow this procedure:

List the prime factors of each monomial.

Multiply each factor the greatest number of times it occurs in either factor.

We have two monomials:

$12a^4b^2c^5$

$40a^3b^7c^1$

The prime factors of the first monomial are:

$2^2,3,a^4,b^2,c^5$

The prime factors of the second monomial are:

$2^3,5,a^3b^7c^1$

LCM = Multiply $2^3*3*5*a^4*b^7*c^5$

These are all the factors the greatest number of times they occur.

Operating:

$LCM=8*15*a^4*b^7*c^5$

$\boxed{LCM=120a^4b^7c^5}$

The last choice is correct

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Step-by-step explanation:

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