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GaryK [48]
3 years ago
14

A square traffic sign has a perimeter of 8 feet. How long is each side?

Mathematics
1 answer:
Lady bird [3.3K]3 years ago
3 0
2 because,8 divided by 4 equals 2
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Can someone answer this fast. I suck at angles and I’ll give brainliest when I can to whoever can answer this with no link. I’m
Kamila [148]

It's the 4th option.

Step-by-step explanation:

Corresponding angles are congruent, so x =43. I hope this helps, have a great day!

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(4x – 1) + (–6x + 3)
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Y = 3x + 13 <br> y = -2x - 22<br><br> substitution method show work <br><br> pls someone help quick
laiz [17]

Answer:

x = -7

Step-by-step explanation:

3x + 13 = -2x - 22

subtract 13 from both sides

3x = -2x - 35

Add 2x to both sides

5x = -35

Divide each side by 5

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3 years ago
A spring has a natural length of 7 m. If a 4-N force is required to keep it stretched to a length of 11 m, how much work W is re
bezimeni [28]

Answer:

18 J is the work required to stretch a spring from 7 m to 13 m.

Step-by-step explanation:

The work done is defined to be the product of the force F and the distance d  that the object moves:

W=Fd

If F is measured in newtons and d<em> </em>in meters, then the unit for is a newton-meter, which is called a joule (J).

This definition work as long as the force is constant, but if the force is variable like in this case, we have that the work done is given by

W=\int\limits^b_a {f(x)} \, dx

Hooke’s Law states that the force required to maintain a spring stretched x    units beyond its natural length is proportional to

f(x)=kx

where k is a positive constant (called the spring constant).

To find how much work W is required to stretch it from 7 m to 13 m you must:

Step 1: Find the spring constant

We know that the spring has a natural length of 7 m and a 4 N force is required to keep it stretched to a length of 11 m. So, applying Hooke’s Law

4=k(11-7)\\\\\frac{k\left(11-7\right)}{4}=\frac{4}{4}\\\\k=1

Thus F=x

Step 2: Find the the work done in stretching the spring from 7 m to 13 m.

Recall that the natural length is 7 m, so when we stretch the spring from 7 m to 13 m, we are stretching it by 6 m beyond its natural length.

Work needed to stretch it by 6 m beyond its natural length

W=\int\limits^6_0 {x} \, dx \\\\\mathrm{Apply\:the\:Power\:Rule}:\quad \int x^adx=\frac{x^{a+1}}{a+1},\:\quad \:a\ne -1\\\\\left[\frac{x^{1+1}}{1+1}\right]^6_0\\\\\left[\frac{x^2}{2}\right]^6_0=18

18 J is the work required to stretch a spring from 7 m to 13 m.

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3 years ago
For the parallelogram find the value of x,y,and z
Nat2105 [25]

Answer:

X = 119, Y = 61, Z = 119

Step-by-step explanation:

These are supplementary angles and equal 180.

180 - 61 = 119

Opposite angles are congruent

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3 years ago
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