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3 years ago

Convert 534.3% to a decimal

2 answers:

6 0

If you would like to convert 534.3% to a decimal, you can do this using the following steps:

534.3% = 534.3 / 100 = 5.343

The correct result would be 5.343.

Alexandra [31]3 years ago

3 0

534.3% to a fraction is 5343÷1000 to a decimal is 5.343

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Solve for x:<br> 3x^2-5x=2

Yanka [14]

$3{ x }^{ 2 }-5x=2\\ \\ 3{ x }^{ 2 }-5x-2=0\\ \\ \left( 3x+1 \right) \left( x-2 \right) =0\\ \\ \therefore \quad x=2\\ \\ \therefore \quad x=-\frac { 1 }{ 3 }$

8 0

4 years ago

A friend has a 82% average before the final exam for a course. That score includes everything but the final, which counts for 30

Damm [24]

So you want to set up an equation for a weighted average. You know the final is 30% of the grade, so everything else is 70%. This gives you:

(Final)(.30) + (other grades)(.70) = course grade

The best grade the student can get would be if they get a hundred on the final, since that’s the best score you can make on the final. Then,

(100)(.30) + (82)(.70) = course grade
30 + 57.4 = course grade = 87.4 Which, If you round, the student would get an 87.

For the last part, we use the same equation, just filling in different parts.

(Final)(.30) + (other grades)(.70) = course grade

This time, we don’t know the grade for the final, but we know the course grade.

(Final)(.30) + (82)(.70) = 75
(Final)(.30) + 57.4 = 75
(Final)(.30) = 17.6
Final = (17.6)/(.30)
Final = 58.667 Which is approx a 59.

4 0

3 years ago

use the formula A=LW to find the length of a rectangle when the width is 29 yards and the are is 1,044 square yards. Help please

sukhopar [10]

Answer:

L = 36 yd

Step-by-step explanation:

Given the area of a rectangle of 1044 yd², and a width of 29 yards:

Using the formula for finding the area of a rectangle, A = L × W, algebraically solve for the length (L):

A = L × W

where:

A = area of a rectangle = 1044 yd²

L = length of a rectangle

W = width of a rectangle = 29 yd

<h3><u /></h3><h3><u>Solution:</u></h3>

Divide both sides by W to isolate L:

$\displaystyle\mathsf{\frac{A}{W}\:=\:\frac{L\:\times\:W}{W}}$

$\displaystyle\mathsf{L\:=\frac{A}{W}}$

Substitute the values for the area and the width of a rectangle:

$\displaystyle\mathsf{L\:=\frac{1044\:yd^2}{29\:yd}}$

L = 36 yd

Therefore, the <u>length</u> of a rectangle is 36 yards.

3 0

3 years ago

Suppose that 2 ≤ f '(x) ≤ 4 for all values of x. What are the minimum and maximum possible values of

Vesna [10]

Answer:

Step-by-step explanation:

By the Mean Value Theorem, there is at least one number, c, in the interval (1,6) such that

f'(c) = [f(6) - f(1)]/ (6 - 1)

So, f(6) - f(1) = 5f'(c).

Since 2 ≤ f'(c) ≤ 4, 10 ≤ 5f'(c) ≤ 20

So, f(6) - f(1) is between 10 and 20.

5 0

3 years ago

Plz explain how to do this

LUCKY_DIMON [66]

Here is the definition of a negative exponent.

For any non-zero value of a,

$a^{-n} = \dfrac{1}{a^n}$

Look at your problem and follow the definition.

$4^{-3} = \dfrac{1}{4^3}$

Since 4^3 = 4 * 4 * 4 = 64, we can now write:

$4^{-3} = \dfrac{1}{4^3} = \dfrac{1}{4 \times 4 \times 4} = \dfrac{1}{64}$

4 0

4 years ago