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2 years ago

C(x-2)-5=b(x-3), solve for x

Mathematics

1 answer:

xeze [42]2 years ago

6 0

Answer:

x = (2c - 3b + 5) / (c - b).

Step-by-step explanation:

c(x - 2) - 5 = b(x - 3)

cx - 2c - 5 = bx - 3b Bring the terms in x to left side:-

cx - bx - 2c - 5 = -3b Bring other terms to right side:-

cx - bx = 2c - 3b + 5

x(c - b) = 2c - 3b + 5 Divide both sides by c - b:-

x = (2c - 3b + 5) / (c - b).

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jolli1 [7]

Answer:

Third choice

Step-by-step explanation:

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9 18

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3 0

3 years ago

What fraction of 60 is 45?

JulijaS [17]

This is the answer 45/60 = 3/4 = 0.75 <span>

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8 0

3 years ago

Read 2 more answers

PLS HELP DUE SOON WILL MARK BRAINLIEST

Eva8 [605]

Answer:

you have 9 yes and 3 no. so 3/9=1/3=.333

6 0

3 years ago

Evaluate the double integral.

Fynjy0 [20]

Answer:

$\iint_D 8y^2 \ dA = \dfrac{88}{3}$

Step-by-step explanation:

The equation of the line through the point $(x_o,y_o)$ & $(x_1,y_1)$ can be represented by:

$y-y_o = m(x - x_o)$

Making m the subject;

$m = \dfrac{y_1 - y_0}{x_1-x_0}$

∴

we need to carry out the equation of the line through (0,1) and (1,2)

i.e

y - 1 = m(x - 0)

y - 1 = mx

where;

$m= \dfrac{2-1}{1-0}$

m = 1

Thus;

y - 1 = (1)x

y - 1 = x ---- (1)

The equation of the line through (1,2) & (4,1) is:

y -2 = m (x - 1)

where;

$m = \dfrac{1-2}{4-1}$

$m = \dfrac{-1}{3}$

∴

$y-2 = -\dfrac{1}{3}(x-1)$

-3(y-2) = x - 1

-3y + 6 = x - 1

x = -3y + 7

Thus: for equation of two lines

x = y - 1

x = -3y + 7

i.e.

y - 1 = -3y + 7

y + 3y = 1 + 7

4y = 8

y = 2

Now, y ranges from 1 → 2 & x ranges from y - 1 to -3y + 7

∴

$\iint_D 8y^2 \ dA = \int^2_1 \int ^{-3y+7}_{y-1} \ 8y^2 \ dxdy$

$\iint_D 8y^2 \ dA =8 \int^2_1 \int ^{-3y+7}_{y-1} \ y^2 \ dxdy$

$\iint_D 8y^2 \ dA =8 \int^2_1 \bigg ( \int^{-3y+7}_{y-1} \ dx \bigg) dy$

$\iint_D 8y^2 \ dA =8 \int^2_1 \bigg ( [xy^2]^{-3y+7}_{y-1} \bigg ) \ dy$

$\iint_D 8y^2 \ dA =8 \int^2_1 \bigg ( [y^2(-3y+7-y+1)]\bigg ) \ dy$

$\iint_D 8y^2 \ dA =8 \int^2_1 \bigg ([y^2(-4y+8)] \bigg ) \ dy$

$\iint_D 8y^2 \ dA =8 \int^2_1 \bigg ( -4y^3+8y^2 \bigg ) \ dy$

$\iint_D 8y^2 \ dA =8 \bigg [\dfrac{ -4y^4}{4}+\dfrac{8y^3}{3} \bigg ]^2_1$

$\iint_D 8y^2 \ dA =8 \bigg [ -y^4+\dfrac{8y^3}{3} \bigg ]^2_1$

$\iint_D 8y^2 \ dA =8 \bigg [ -2^4+\dfrac{8(2)^3}{3} + 1^4- \dfrac{8\times (1)^3}{3}\bigg]$

$\iint_D 8y^2 \ dA =8 \bigg [ -16+\dfrac{64}{3} + 1- \dfrac{8}{3}\bigg]$

$\iint_D 8y^2 \ dA =8 \bigg [ -15+ \dfrac{64-8}{3}\bigg]$

$\iint_D 8y^2 \ dA =8 \bigg [ -15+ \dfrac{56}{3}\bigg]$

$\iint_D 8y^2 \ dA =8 \bigg [ \dfrac{-45+56}{3}\bigg]$

$\iint_D 8y^2 \ dA =8 \bigg [ \dfrac{11}{3}\bigg]$

$\iint_D 8y^2 \ dA = \dfrac{88}{3}$

4 0

2 years ago

Lou wants to make a coffee mixture to sell. He is going to mix Sumatra coffee which costs $2.50 per pound with Columbian coffee

Scilla [17]

To answer this question you need to create 2 equations:
S+C=50
2.50S+3.75C=3.35*(50)
S=50-C -so now you can plug in S into the second equation
2.50*(50-C)+3.75C=167.5
Now you just need to solve this equation, the final answer will be 34 pounds of colombian coffee and 16 pounds of sumatra coffee

5 0

3 years ago