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3 years ago

Find the lengths of the diagonals of rectangle QRST when QS = 20x - 5 and RT = 14x + 7 .

Mathematics

1 answer:

givi [52]3 years ago

6 0

Answer:

Simplifying

20x + -5 = 14x + 7

Reorder the terms:

-5 + 20x = 14x + 7

Reorder the terms:

-5 + 20x = 7 + 14x

Solving

-5 + 20x = 7 + 14x

Solving for variable 'x'.

Move all terms containing x to the left, all other terms to the right.

Add '-14x' to each side of the equation.

-5 + 20x + -14x = 7 + 14x + -14x

Combine like terms: 20x + -14x = 6x

-5 + 6x = 7 + 14x + -14x

Combine like terms: 14x + -14x = 0

-5 + 6x = 7 + 0

-5 + 6x = 7

Add '5' to each side of the equation.

-5 + 5 + 6x = 7 + 5

Combine like terms: -5 + 5 = 0

0 + 6x = 7 + 5

6x = 7 + 5

Combine like terms: 7 + 5 = 12

6x = 12

Divide each side by '6'.

x = 2

Simplifying

x = 2

Step-by-step explanation: I'm so sorry if wrong im only in 6th grade :(

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The tip of a pendulum moves through an angle of 0.38 radians. If the radius of the pendulum is 25

AVprozaik [17]

Answer:

The length of arc 'l' = 9.5

Step-by-step explanation:

Explanation:-

Given radius of the pendulum r = 25

The tip of a pendulum moves through an angle of 0.38 radians

 θ = 0.38 radians

The length of arc 'l' = rθ 

 'l' = (25)(0.38)

 The length of arc 'l' = 9.5 

5 0

3 years ago

X is equal 3. What's c? 6x-2(x-c)=10

Lisa [10]

Answer

C = -1

Step-by-step explanation:

substitute

6(3)-2(3-c)=10

18-6+2c=10

12+2c=10

12-12+2c=10-12

2c=-2

c=-1

6 0

3 years ago

27 divided by 53 minus 6 plus 53 times -2 squared

mamaluj [8]

Answer:

$\left(\left(\frac{27}{53}\right)-6\right)+\left(53\left(-\left(2^2\right)\right)\right)=-\frac{11527}{53}\quad \left(\mathrm{Decimal:\quad }\:-217.49056\dots \right)$

Step-by-step explanation:

Considering the expression

27 divided by 53 minus 6 plus 53 times -2 squared

Which can be written as

$\left(\left(\frac{27}{53}\right)-6\right)+\left(53\cdot \left(-\left(2^2\right)\right)\right)$

So, solving the expression

$\left(\left(\frac{27}{53}\right)-6\right)+\left(53\cdot \left(-\left(2^2\right)\right)\right)$

$\mathrm{Remove\:parentheses}:\quad \left(a\right)=a$

$=\frac{27}{53}-6-53\cdot \:2^2$

$\mathrm{Convert\:element\:to\:fraction}:\quad \:6=\frac{6\cdot \:53}{53}$

$=-\frac{6\cdot \:53}{53}+\frac{27}{53}$

$\mathrm{Since\:the\:denominators\:are\:equal,\:combine\:the\:fractions}:\quad \frac{a}{c}\pm \frac{b}{c}=\frac{a\pm \:b}{c}$

$=\frac{-6\cdot \:53+27}{53}$

$=\frac{-291}{53}$ ∵ $-6\cdot \:53+27=-291$

$\mathrm{Apply\:the\:fraction\:rule}:\quad \frac{-a}{b}=-\frac{a}{b}$

$=-2^2\cdot \:53-\frac{291}{53}$

$=-212-\frac{291}{53}$ ∵ $53\cdot \:2^2=212$

$\mathrm{Convert\:element\:to\:fraction}:\quad \:212=\frac{212\cdot \:53}{53}$

$=-\frac{212\cdot \:53}{53}-\frac{291}{53}$

$\mathrm{Since\:the\:denominators\:are\:equal,\:combine\:the\:fractions}:\quad \frac{a}{c}\pm \frac{b}{c}=\frac{a\pm \:b}{c}$

$=\frac{-212\cdot \:53-291}{53}$

$=\frac{-11527}{53}$ ∵ $-212\cdot \:53-291=-11527$

$\mathrm{Apply\:the\:fraction\:rule}:\quad \frac{-a}{b}=-\frac{a}{b}$

$=-\frac{11527}{53}$

Therefore,

$\left(\left(\frac{27}{53}\right)-6\right)+\left(53\left(-\left(2^2\right)\right)\right)=-\frac{11527}{53}\quad \left(\mathrm{Decimal:\quad }\:-217.49056\dots \right)$

Keywords: algebraic expression

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8 0

4 years ago

A person has body fat percentage of 17.2% and weighs 171 pound how many pounds of her weight is made up of fat

Citrus2011 [14]

Answer: 29.41 pounds

Step-by-step explanation:

You have the following information given in the problem:

- The fat percentage that the person has is 17.2%

- The person weighs 171 pounds.

Therefore, to calculate the amount of pounds of her weight is made up of fat (which you can call x), you must multiply the weight of the person by the fat percentage.

Therefore, you obtain the following result:

$x=171lb*0.172\\x=29.41lb$

5 0

4 years ago

What is 3(5x-6)+2(x+8)?

Mashutka [201]

Answer:

I believe it's D. 17x-2

4 0

3 years ago

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