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3 years ago

Find the equation of the line passing through the points (1, -2) and (-2, 7). Write the equation in slop-intercept from.

Mathematics

1 answer:

alex41 [277]3 years ago

5 0

Y=-3x+1 because in the points the equation will be y+2=-3(x-1) this converted into slope intercept form is y=-3(x-1)

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NEED Math help please?

Tamiku [17]

Answer:

Step-by-step explanation:5,7 3,9

8 0

3 years ago

Function f(x) represents the population of bacteria x hours after 9 a.m. What does F(2)-F(1) represent?

bezimeni [28]

Answer:

f(2)-f(1) => Difference between the count of bacteria between 11AM and 10AM

Step-by-step explanation:

Here x is the number of hours after 9AM

f(2) means the count of bacteria after 2 hours of 9AM i.e. at 11AM

f(1) means the count of bacteria after 1 hour of 9AM i.e. at 10 AM

f(2) - f(1) will give us the difference between the count of bacteria between 11 AM and 10 AM

Hence,

f(2)-f(1) => Difference between the count of bacteria between 11AM and 10AM

8 0

3 years ago

The side of an Equileteral triangle is 12cm. What is its Area?

elena55 [62]

Answer:

A = 62.35 cm²

Step-by-step explanation:

Use the area formula A = $\frac{\sqrt{3}a^2}{4}$ , where a is the side length.

Plug in the values:

A = $\frac{\sqrt{3}(12^2)}{4}$

A = $\frac{\sqrt{3}(144)}{4}$

A = 62.35 cm²

3 0

4 years ago

The expression 100 + 20m100+20m100, plus, 20, m gives the volume of water in Marcel’s pool (in liters) after Marcel spends mmm m

Nady [450]

All you have to do is you only need to evaluate the given equation which is:100 + 20m where m would be the number of minutes = 5 1/4 minutes
100 + 20 (5 1/4)
= 100 + 105
= 205 cubic liters is the volume of water in Marcel’s pool after he fills it.

8 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