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AnnyKZ [126]
3 years ago
14

ALGEBRA EQUATION PLEASE HELP ASAP -3=12y-5(2y-7)

Mathematics
2 answers:
Sauron [17]3 years ago
5 0
-3 = 12y -5(2y-7)

-3 = 12y +(-5)*2y + (-5)*(-7)

-3 = 12y -10y + 35

-3 = 2y +35

2y= -3 -35

2y = -38

Therefore:

y=-19

As 2*(-19) = -38
Dovator [93]3 years ago
3 0

Simplify:

−3=12y−5(2y−7)

−3=12y+(−5)(2y)+(−5)(−7)(Distribute)

−3=12y+−10y+35

−3=(12y+−10y)+(35)(Combine Like Terms)

−3=2y+35

Flip the equation.

2y+35=−3

Subtract 35 from both sides.

2y+35−35=−3−35

2y=−38

Divide both sides by 2.

2y/2=−38/2

y=−19


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Answer:

h = \sqrt{72}in

Step-by-step explanation:

We have a right triangle with long side of 11in and smaller sides 14in/2=7in and h.

We know that

h^2 + (7in)^2 = (11in)^2\\h^2 = 121in^2 - 49in^2 = 72in^2\\h = \sqrt{72}in

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Evaluate the line integral by the two following methods. xy dx + x2 dy C is counterclockwise around the rectangle with vertices
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Answer:

25/2

Step-by-step explanation:

Recall that for a parametrized differentiable curve C = (x(t), y(t)) with the parameter t varying on some interval [a, b]

\large \displaystyle\int_{C}[P(x,y)dx+Q(x,y)dy]=\displaystyle\int_{a}^{b}[P(x(t),y(t))x'(t)+Q(x(t),y(t))y'(t)]dt

Where P, Q are scalar functions

We want to compute

\large \displaystyle\int_{C}P(x,y)dx+Q(x,y)dy=\displaystyle\int_{C}xydx+x^2dy

Where C is the rectangle with vertices (0, 0), (5, 0), (5, 1), (0, 1) going counterclockwise.

a) Directly

Let us break down C into 4 paths \large C_1,C_2,C_3,C_4 which represents the sides of the rectangle.

\large C_1 is the line segment from (0,0) to (5,0)

\large C_2 is the line segment from (5,0) to (5,1)

\large C_3 is the line segment from (5,1) to (0,1)

\large C_4 is the line segment from (0,1) to (0,0)

Then

\large \displaystyle\int_{C}=\displaystyle\int_{C_1}+\displaystyle\int_{C_2}+\displaystyle\int_{C_3}+\displaystyle\int_{C_4}

Given 2 points P, Q we can always parametrize the line segment from P to Q with

r(t) = tQ + (1-t)P for 0≤ t≤ 1

Let us compute the first integral. We parametrize \large C_1 as

r(t) = t(5,0)+(1-t)(0,0) = (5t, 0) for 0≤ t≤ 1 and

r'(t) = (5,0) so

\large \displaystyle\int_{C_1}xydx+x^2dy=0

 Now the second integral. We parametrize \large C_2 as

r(t) = t(5,1)+(1-t)(5,0) = (5 , t) for 0≤ t≤ 1 and

r'(t) = (0,1) so

\large \displaystyle\int_{C_2}xydx+x^2dy=\displaystyle\int_{0}^{1}25dt=25

The third integral. We parametrize \large C_3 as

r(t) = t(0,1)+(1-t)(5,1) = (5-5t, 1) for 0≤ t≤ 1 and

r'(t) = (-5,0) so

\large \displaystyle\int_{C_3}xydx+x^2dy=\displaystyle\int_{0}^{1}(5-5t)(-5)dt=-25\displaystyle\int_{0}^{1}dt+25\displaystyle\int_{0}^{1}tdt=\\\\=-25+25/2=-25/2

The fourth integral. We parametrize \large C_4 as

r(t) = t(0,0)+(1-t)(0,1) = (0, 1-t) for 0≤ t≤ 1 and

r'(t) = (0,-1) so

\large \displaystyle\int_{C_4}xydx+x^2dy=0

So

\large \displaystyle\int_{C}xydx+x^2dy=25-25/2=25/2

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According with this theorem

\large \displaystyle\int_{C}Pdx+Qdy=\displaystyle\iint_{A}(\displaystyle\frac{\partial Q}{\partial x}-\displaystyle\frac{\partial P}{\partial y})dydx

where A is the interior of the rectangle.

so A={(x,y) |  0≤ x≤ 5,  0≤ y≤ 1}

We have

\large \displaystyle\frac{\partial Q}{\partial x}=2x\\\\\displaystyle\frac{\partial P}{\partial y}=x

so

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on Saturday nikki and tyler walked to the local farmers market nikki bout 8 apples for $6 dollars and tyler brought 3 fewer appl
erica [24]

On Saturday Nikki and Tyler walked to the local farmers market Nikki bought 8 apples for $6 dollars and Tyler brought 3 fewer apples than Nikki. How much did Tyler spend on apples?​

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If 8 apples cost $6​

Then 5 apples will cost $X​

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​Answer : Tyler spent on apples $3.75

​Hope this helps!​​​​

\textit{\textbf{Spymore}}​​​​​​​

4 0
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olasank [31]
8 million 3 hundred sixty-four thousand nine hundred eighty-seven
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ICE Princess25 [194]

Answer:

It goes right

Step-by-step explanation:

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6 0
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