All categories

3 years ago

The length of a rectangle is 5 cm longer than its width.

Mathematics

2 answers:

blondinia [14]3 years ago

4 0

Equation To write :2 {x+ ( x+5) }=30

Let's solve your equation step-by-step.

2(x+x+5)=30

Step 1: Simplify both sides of the equation.

2(x+x+5)=30

(2)(x)+(2)(x)+(2)(5)=30(Distribute)

2x+2x+10=30

(2x+2x)+(10)=30(Combine Like Terms)

4x+10=30

Step 2: Subtract 10 from both sides.

4x+10−10=30−10

4x=20

Step 3: Divide both sides by 4.

x=5

So... x+5=10.

Area =5x10=50 cm squared.

Semenov [28]3 years ago

4 0

Answer:

Step-by-step explanation:

Okay, so P = perimeter, and the problem says P = 30

Let's call x the width, and y is the length.

In the perimeter formula, 2X + 2Y = P

so 2X + 2Y = 30

The problem also tells us that that x = y+5, so now let's do some substituting and replace X with Y+5

2(y+5) + 2Y = 30

2y+10 +2Y = 30

4y+10=30

-10. -10

4y=20

Y=5

TRIED MY BEST PRETTY SURE ITS RIGHT

You might be interested in

What is the length of segment CD question mark

Yuri [45]

Answer 8 this very easy

3 0

4 years ago

What multiply what equals 26 using numbers between 2 and 9?

nignag [31]

There are only 1*26 and 2*13 so there is no answer unless you allow decimals.

8 0

4 years ago

Find the area of the trapezoid.

kakasveta [241]

I think the answer is 47.6 also

6 0

3 years ago

Which rules define the function whose graph is shown? on x < −2 on −2 ≤ x < 0 on 0 ≤ x < 2 on 2 ≤ x

jeyben [28]

Answer:

-4, -x^2, x^2, 4

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

A 12-cm-long thin rod has the nonuniform charge density λ(x)=(2.0 nC/cm)e−|x|/(6.0 cm), where x is measured from the center of t

Inessa [10]

Answer:

the total charge is

$Q=24(1-\exp(-1))nC\approx15.171nC$

Step-by-step explanation:

Since x is measured from the center, that means that x=0 is the center so the edges of the rod correspond to x=-6 and x=6. that meas that the total charge can be calculated as

$Q=\int^{6}_{-6}2\exp\left(\frac{-|x|}{6}\right)dx$

separating the integral from -6 to 0 and from 0 to 6, taking into account that |x|=-x for x<0 and |x|=x for x >=0, we get $Q=\int^{0}_{-6}2\exp\left(\frac{x}{6}\right)dx+\int^{6}_{0}2\exp\left(\frac{-x}{6}\right)dx$

using the substitution x=-u in the first integral we get $\int^{0}_{-6}2\exp\left(\frac{x}{6}\right)dx=-\int^{0}_{6}2\exp\left(\frac{-u}{6}\right)du=\int^{6}_{0}2\exp\left(\frac{-u}{6}\right)du$

which is the same as the first integral. Thus, the total charge is given by

$Q=2\int^{6}_{0}2\exp\left(\frac{-x}{6}\right)dx$

integrating we get

$Q=4(-6\exp\left(\frac{-x}{6}\right))\big|^{6}_{0}=-24(\exp(-6/6)-\exp(0))=24(1-\exp(-1))$

4 0

3 years ago