If RS = 3x + 1, ST = 2x – 2, and RT = 64, find the value of x.

Point N is on line segment MO. Given NO=2x−3, MO=3x+5, and MN=2x+3, determine the numerical length of MO.

ELEN [110]

Answer:-19

Step-by-step explanation:

3x+5-(2x-3)=0

3x+5-2x+3=0

x+8=0

x=-8

substitute x into 3x+5 to find the length of MO

3x+5

3(-8)+5

-24+5

-19

6 0

3 years ago

A rectangle has a length 6 more than it's width if the width is decreased by 2 and the length decreased by 4 the resulting has a

Rashid [163]

Answer:

Length of original rectangle: 11 units.

$\frac{\text{Area of original rectangle}}{\text{Area of new rectangle}}=\frac{55}{21}$

$\text{Perimeter of new rectangle}=20$

Step-by-step explanation:

Let x represent width of the original rectangle.

We have been given that a rectangle has a length 6 more than it's width. S the length of the original rectangle would be $x+6$ .

We have been given that when the width is decreased by 2 and the length decreased by 4 the resulting has an area of 21 square units.

The width of new rectangle would be $x-2$ .

The length of new rectangle would be $x+6-4=x+2$ .

The area of new rectangle would be $(x+2)(x-2)$ .

Now we will equate area of new rectangle with 21 and solve for x as:

$(x+2)(x-2)=21$

Applying difference of squares, we will get:

$x^2-2^2=21$

$x^2-4=21$

$x^2-4+4=21+4$

$x^2=25$

Since width cannot be negative, so we will take positive square root of both sides.

$\sqrt{x^2}=\sqrt{25}$

$x=5$

Therefore, the width of original rectangle is 5 units.

Length of the original rectangle would be $x+6\Rightarrow x+5=11$ .

Therefore, the length of original rectangle is 11 units.

$\text{Area of original rectangle}=5\times 11$

$\text{Area of original rectangle}=55$

Therefore, area of the original rectangle is 55 square units.

Now we will find ratio of the original rectangle area to the new rectangle area as:

$\frac{\text{Area of original rectangle}}{\text{Area of new rectangle}}=\frac{55}{21}$

We know that perimeter of rectangle is two times the sum of length and width.

$\text{Perimeter of new rectangle}=2((x+2)+(x-2))$

$\text{Perimeter of new rectangle}=2((5+2)+(5-2))$

$\text{Perimeter of new rectangle}=2(7+3)$

$\text{Perimeter of new rectangle}=2(10)$

$\text{Perimeter of new rectangle}=20$

Therefore, the perimeter of the new rectangle is 20 units.

7 0

4 years ago

Hi can anybody tell me the answer

MArishka [77]

<u>Answer</u>:

8x^5y^7

<u>Step by step explanation</u>

$4 {x}^{2} {y}^{3} \times 2 {x}^{3} {y}^{4} \\ 4 \times 2 \times {x}^{2} \times {x}^{3} \times {y}^{3} \times y ^{4} \\ 8 \times {x}^{2 + 3} \times {y}^{3 + 4} \\ =8 {x}^{5} {y}^{7}$

6 0

3 years ago

Read 2 more answers

Forty pounds of candy is distributed in 5/7 lb bags. How many bags can be filled?

Morgarella [4.7K]

Answer:56 bags

Step-by-step explanation:

you are dividing 40 into 5/7 so its technically 40*7 divided by 5, which equals 56. double check by doing 56*5 divided by 7 and you get 40 pounds,

3 0

2 years ago

Read 2 more answers

Honon spent 3 1/4 hours on homework while Sequoia spent 2 5/6 hours on homework. How much more time did Honon spend on homework

Rzqust [24]

Answer:

5/12 hours

Step-by-step explanation:

Honon = 3 1/4 hours

Sequoia = 2 5/6 hours

How much more time did Honon spend on homework than Sequoia

= number of hours Honon spent - number of hours Sequoia spent

= 3 1/4 hours - 2 5/6 hours

= 13/4 - 17/6

= (39-34) /12

= 5/12 hours

6 0

3 years ago