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Alex787 [66]
3 years ago
15

FWML is a parallelogram. Find the values of x and y. Solve for the value of z, if z=x−y

Mathematics
1 answer:
Dima020 [189]3 years ago
4 0

Answer:

x=5

y=8

z=-3

Step-by-step explanation:

We have been given a parallelogram. We are asked to solve for the values of x and y.

We know that opposite sides of parallelogram are equal, so we can set equation as:

3x-3=x+7

3x-x-3=x-x+7

2x-3=7

2x-3+3=7+3

2x=10

\frac{2x}{2}=\frac{10}{2}

x=5

Similarly, we will solve for y.

2y-6=y+2

2y-y-6=y-y+2

y-6=2

y-6+6=2+6

y=8

To solve for z, we will subtract y from x as:

z=x-y\\z=5-8\\z=-3

Therefore, the value of z is negative 3.

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Step-by-step explanation:

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<em>Additional comment</em>

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6 0
3 years ago
Tìm một đoạn nghiệm của phương trình<br> 3/x+1 +4/(x+1)^2+5/(x+1)^3+6/(x+1)^4=10
SVETLANKA909090 [29]

Answer:

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Step-by-step explanation:

6 0
3 years ago
Y = (1/4)x^2 - (1/2)lnx..over the interval (1, 7e) ...what is the arc length ?
xenn [34]
So, f[x] = 1/4x^2 - 1/2Ln(x) 
<span>thus f'[x] = 1/4*2x - 1/2*(1/x) = x/2 - 1/2x </span>
<span>thus f'[x]^2 = (x^2)/4 - 2*(x/2)*(1/2x) + 1/(4x^2) = (x^2)/4 - 1/2 + 1/(4x^2) </span>
<span>thus f'[x]^2 + 1 = (x^2)/4 + 1/2 + 1/(4x^2) = (x/2 + 1/2x)^2 </span>
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6 0
3 years ago
-7+(-3) SHOW WORK PLEASE HURRY
DENIUS [597]

Answer:

-10

Step-by-step explanation:

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= -(7+3)

= -10

Side note: Do you not have a caculator?

7 0
3 years ago
Find the volume of the solid generated when R​ (shaded region) is revolved about the given line. x=6−3sec y​, x=6​, y= π 3​, and
Dmitrij [34]

Answer:

V=9\pi\sqrt{3}

Step-by-step explanation:

In order to solve this problem we must start by graphing the given function and finding the differential area we will use to set our integral up. (See attached picture).

The formula we will use for this problem is the following:

V=\int\limits^b_a {\pi r^{2}} \, dy

where:

r=6-(6-3 sec(y))

r=3 sec(y)

a=0

b=\frac{\pi}{3}

so the volume becomes:

V=\int\limits^\frac{\pi}{3}_0 {\pi (3 sec(y))^{2}} \, dy

This can be simplified to:

V=\int\limits^\frac{\pi}{3}_0 {9\pi sec^{2}(y)} \, dy

and the integral can be rewritten like this:

V=9\pi\int\limits^\frac{\pi}{3}_0 {sec^{2}(y)} \, dy

which is a standard integral so we solve it to:

V=9\pi[tan y]\limits^\frac{\pi}{3}_0

so we get:

V=9\pi[tan \frac{\pi}{3} - tan 0]

which yields:

V=9\pi\sqrt{3}]

6 0
3 years ago
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