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3 years ago

8

HELP ME PLEASEEEEEEEEEEEEEEEEEEEEEEEE

1 answer:

kkurt [141]3 years ago

7 0

Answer:

Commutative Property

Step-by-step explanation:

Hope it helps. Please mark me brainliest if you would :D

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What is the slope of this line

defon

Answer:

Step-by-step explanation: Do rise and run

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2 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

∴

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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2 years ago

sandy is cutting bows out of ribbon which will be used to rap gifts. If sandy need 9/7 feet of ribbon to make a bow and she has

Katarina [22]

Sandy can make 14 bows

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2 years ago

For this question please tell me if I'm right or wrong. If I'm wrong please correct me.

kotykmax [81]

Answer:

PK represents a radius. Radius only goes half way of the circle.

5 0

3 years ago

Read 2 more answers

Help please i don’t know how to do it

Norma-Jean [14]

Answer:

Step-by-step explanation:

$\frac{dy}{dx}=x^{2}(y-1)\\\frac{1}{y-1} \text{ } dy=x^{2} \text{ } dx\\\int \frac{1}{y-1} \text{ } dy=\int x^{2} \text{ } dx\\\ln|y-1|=\frac{x^{3}}{3}+C\\$

From the initial condition,

$\ln|3-1|=\frac{0^{3}}{3}+C\\\ln 2=C$

So we have that $\ln |y-1|=\frac{x^{3}}{3}+\ln 2\\e^{\frac{x^{3}}{3}+\ln 2}=y-1\\2e^{\frac{x^{3}}{3}}=y-1\\y=2e^{\frac{x^{3}}{3}}+1$

4 0

2 years ago