All categories

3 years ago

A rectangle is 5 cm wider than twice its length. If the perimeter of the rectangle is 70 cm, then what is the area?

Mathematics

2 answers:

kifflom [539]3 years ago

6 0

Step-by-step explanation:

The dimensions of the rectangle are two unknowns: The length "l" and the width "w"

The perimeter of a rectangle is found as P = 2*l + 2*w

We also know that the length is 5cm more than twice the width. l = 2*w + 5

These two equations gives us a system of linear equations.

P = 2*l + 2*w

l = 2*w + 5

We can use substitution to replace the "l" in the first equation with 2*w + 5

P = 2*(2*w + 5) + 2*w

P = 4*w + 10 + 2*w

P = 6*w + 10

we know the perimeter is 70

70 = 6* w + 10

subtract 10 from both sides

60 = 6* w

w = 10

Now that we know the width, we can find the length by substituting 10 for "w" in the second equation.

l = 2* 10 + 5

l = 20 +5

l = 25

the dimentions of the rectangle are 25cm x 10 cm

HACTEHA [7]3 years ago

6 0

Answer:

Step-by-step explanation:

You might be interested in

Hi please i need the step bye step solution. thank you ♥

wlad13 [49]

Answer:

Step-by-step explanation:

3, 6, 9...

120 = 3+(n-1)3

120 = 3+3n-3

120 = 3n

120 / 3 = 3n / 3

40 = n

3 0

3 years ago

How many are in pounds?

NARA [144]

The answer is 16 ounces

6 0

3 years ago

Josephine has a rectangular garden with an area of 2x2

sergeinik [125]

Answer:

Fourth example, (2x - 3) and (x + 2), are possible dimensions

Step-by-step explanation:

Let's go through all the responses one by one;

First of all the area of the rectangular garden can be computed through length times width, such that:

(x^2 - 3) * (2) = 2x^2 + x - 6,

(2x + 3) * (x-2) = 2x^2 + x - 6,

(2x + 2) * (x-3) = 2x^2 + x - 6,

(2x - 3) * (x + 2) = 2x^2 + x - 6

Now in each of these examples, it should be that the products are equivalent to the area according to the length by width rule, so let's see which is truly the correct dimensions by justifying whether the product is truly equivalent to the area.

First example:

2 * (x^2 - 3) = 2x^2 - 6

Now 2x^2 - 6 ≠ 2x^2 + x - 6 so the first example is not a possibility,

Second example:

(2x + 3) * (x-2) = 2x^2- x - 6

Now 2x^2- x - 6 ≠ 2x^2 + x - 6 so the second example is not a possibility,

Third example:

(2x + 2) * (x-3) = 2x^2 - 4x - 6

Now 2x^2 - 4x - 6 ≠ 2x^2 + x - 6 so the third example is not a possibility,

Fourth example:

(2x - 3) * (x + 2) = 2x^2 + x - 6

Here 2x^2 + x - 6 = 2x^2 + x - 6 so the fourth example can act as possible dimensions

6 0

3 years ago

Read 2 more answers

Assume Y=1+X+u, where X, Y, and u=v+X are random variables, v is independent of X; E(v)=0, Var(v)=1, E(X)=1, and

kobusy [5.1K]

Answer:

a) $E(u|X=1)= E(v|X=1) + E(X|X=1) = E(v) +1 = 0 +1 =1+$

b) $E(Y| X=1)= E(1|X=1) + E(X|X=1) + E(u|X=1) = E(1) + 1 + E(v) + 0 = 1+1+0=2$

c) $E(u|X=2)= E(v|X=2) + E(X|X=2) = E(v) +2 = 0 +2 =2$

d) $E(Y| X=2)= E(1|X=2) + E(X|X=2) + E(u|X=2) = E(2) + 2 + E(v) + 2 = 2+2+2=6$

e) $E(u|X) = E(v+X |X) = E(v|X) +E(X|X) = E(v) +E(X) = 0+1=1$

f) $E(Y|X) = E(1+X+u |X) = E(1|X) +E(X|X) + E(u|X) = 1+1+1=3$

g) $E(u) = E(v) +E(X) = 0+1=1$

h) E(Y) = E(1+X+u) = E(1) + E(X) +E(v+X) = 1+1 + E(v) +E(X) = 1+1+0+1 = 3[/tex]

Step-by-step explanation:

For this case we know this:

$Y = 1+X +u$

$u = v+X$

with both Y and u random variables, we also know that:

$[tex] E(v) = 0, Var(v) =1, E(X) = 1, Var(X)=2$

And we want to calculate this:

Part a

$E(u|X=1)= E(v+X|X=1)$

Using properties for the conditional expected value we have this:

$E(u|X=1)= E(v|X=1) + E(X|X=1) = E(v) +1 = 0 +1 =1$

Because we assume that v and X are independent

Part b

$E(Y| X=1) = E(1+X+u|X=1)$

If we distribute the expected value we got:

$E(Y| X=1)= E(1|X=1) + E(X|X=1) + E(u|X=1) = E(1) + 1 + E(v) + 0 = 1+1+0=2$

Part c

$E(u|X=2)= E(v+X|X=2)$

Using properties for the conditional expected value we have this:

$E(u|X=2)= E(v|X=2) + E(X|X=2) = E(v) +2 = 0 +2 =2$

Because we assume that v and X are independent

Part d

$E(Y| X=2) = E(1+X+u|X=2)$

If we distribute the expected value we got:

$E(Y| X=2)= E(1|X=2) + E(X|X=2) + E(u|X=2) = E(2) + 2 + E(v) + 2 = 2+2+2=6$

Part e

$E(u|X) = E(v+X |X) = E(v|X) +E(X|X) = E(v) +E(X) = 0+1=1$

Part f

$E(Y|X) = E(1+X+u |X) = E(1|X) +E(X|X) + E(u|X) = 1+1+1=3$

Part g

$E(u) = E(v) +E(X) = 0+1=1$

Part h

E(Y) = E(1+X+u) = E(1) + E(X) +E(v+X) = 1+1 + E(v) +E(X) = 1+1+0+1 = 3[/tex]

8 0

3 years ago

Can yall help with this asap

Gnoma [55]

Answer:

its b

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers