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3 years ago

How would you find AC and DC

Mathematics

1 answer:

Harrizon [31]3 years ago

6 0

Because you know that segments AD and DC are congruent, that means that DC=20.

You also know that segments AD+DC=AC.

You can plug in all the information you know and begin to solve for x.

AD+DC=AC
20+20=3x+4
40=3x+4
36=3x
12=x

Now we know that:

x=12
AC=40
DC=20

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Factor 9c³– 6c² Please help this is due soon

Makovka662 [10]

The answer would be 3c^2

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3 years ago

Read 2 more answers

If the length of diagonal of a square is 4√2 cm, find it's length, perimeter and area.

Sonbull [250]

Answer:

As Per Provided Information

Length of diagonal of square is 4√2 cm

We have been asked to find the length , perimeter and area of square .

First let's calculate the side of square .

Using Formulae

$\boxed{\bf \:Diagonal_{(Square)} \: = side \sqrt{2}}$

On substituting the value in above formula we obtain

$\qquad\longrightarrow\sf \:4 \sqrt{2} = side \sqrt{2} \\ \\ \\ \qquad\longrightarrow\sf \:4 \cancel{\sqrt{2}} = side \cancel{ \sqrt{2}} \\ \\ \\ \qquad\longrightarrow\sf \:side \: = 4 \: cm$

Therefore,

Length of its side is 4 cm.

Finding the perimeter of square.

$\boxed{\bf \: Perimeter_{(Square)} = 4 \times side}$

Substituting the value we obtain

$\qquad\longrightarrow\sf \:Perimeter_{(Square)} \: = 4 \times 4 \\ \\ \\ \qquad\longrightarrow\sf \:Perimeter_{(Square)} = 16 \: cm$

Therefore,

Perimeter of square is 16 cm .

Finding the area of square .

$\boxed{\bf \: Area_{(Square)} = {side}^{2}}$

Substituting the value we get

$\qquad\longrightarrow\sf \:Area_{(Square)} \: = {4}^{2} \\ \\ \\ \qquad\longrightarrow\sf \:Area_{(Square)} = 16 \: {cm}^{2}$

Therefore,

Area of square is 16 cm².

3 0

2 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 $d$ 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

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