All categories

3 years ago

What's the formula to find the width

Mathematics

2 answers:

vlabodo [156]3 years ago

5 0

2l + 2w = p

If you already have perimeter and length, you can use this formula to solve for width.

Pavel [41]3 years ago

4 0

Answer:

Area= length x width

Step-by-step explanation:

width of a rectangle, use the formula: area = length × width. Just use the area and length of the rectangle and the formula and answer for the width. If you don't have the area, you could use the rectangle's perimeter .

You might be interested in

Find an equation for the line that passes through the point (2, −5) and is parallel to the line 2x − 4y = 7.

viva [34]

X=2y+3.5 is the awnser too this one

6 0

3 years ago

Find the volume of the cylinder in terms of pi.

Marizza181 [45]

Answer is c as the equation for the volume of a cylinder is pi x r^2 x h which equated to pi x 25 x 11 which equals 275 and your units have to be cubed as it is volume.

3 0

3 years ago

Read 2 more answers

Solve the system by substitution.<br> -2x + 7y = 15<br> бу = 3

Salsk061 [2.6K]

Answer:

Step-by-step explanation:

-2x + 7(1/2) =15

-4x +7 =30

-4x =23

x=-23/4

6 0

2 years ago

Helppp, graph of function!!

Debora [2.8K]

Yes I will help you but give me point then I will help you

3 0

2 years ago

16) Please help with question. WILL MARK BRAINLIEST + 10 POINTS.

Katyanochek1 [597]

We will use the sine and cosine of the sum of two angles, the sine and consine of $\frac{\pi}{2}$ , and the relation of the tangent with the sine and cosine:

$\sin (\alpha+\beta)=\sin \alpha\cdot\cos\beta + \cos\alpha\cdot\sin\beta

\cos(\alpha+\beta)=\cos\alpha\cdot\cos\beta-\sin\alpha\cdot\sin\beta$

$\sin\dfrac{\pi}{2}=1,\ \cos\dfrac{\pi}{2}=0$

$\tan\alpha = \dfrac{\sin\alpha}{\cos\alpha}$

If you use those identities, for $\alpha=x,\ \beta=\dfrac{\pi}{2}$ , you get:

$\sin\left(x+\dfrac{\pi}{2}\right) = \sin x\cdot\cos\dfrac{\pi}{2} + \cos x\cdot\sin\dfrac{\pi}{2} = \sin x\cdot0 + \cos x \cdot 1 = \cos x$

$\cos\left(x+\dfrac{\pi}{2}\right) = \cos x \cdot \cos\dfrac{\pi}{2} - \sin x\cdot\sin\dfrac{\pi}{2} = \cos x \cdot 0 - \sin x \cdot 1 = -\sin x$

Hence:

$\tan \left(x+\dfrac{\pi}{2}\right) = \dfrac{\sin\left(x+\dfrac{\pi}{2}\right)}{\cos\left(x+\dfrac{\pi}{2}\right)} = \dfrac{\cos x}{-\sin x} = -\cot x$

3 0

3 years ago