3 years ago

I have to do Decompose fractions and my problem is 3 over 7 what do I do?

Mathematics

1 answer:

mart [117]3 years ago

4 0

Just simplify them to a decimal by dividing the numerator by the denominator.

You might be interested in

What is 175% of 14.2

LUCKY_DIMON [66]

175% of 14.2 is saying that

1.75(14.2)=24.9

Any questions please just ask. Thank you.

7 0

3 years ago

Read 2 more answers

What are the solutions to the quadratic equation x^2-16=0

frutty [35]

Simplifying this, we get:
(x+4)(x-4) = 0
Now, to get 0, we can replace x for one of these things to make 0, so:
x can either be 4, or -4

8 0

3 years ago

Read 2 more answers

Which of the following is most likely the next step in the series?<br>

Andrej [43]

Answer:

Step-by-step explanation:

They are increasing by 1 vertically. Hope this helps!! :)

5 0

3 years ago

Read 2 more answers

Find the critical numbers of the function. (Enter your answers as a comma-separated list. If an answer does not exist, enter DNE

EleoNora [17]

Answer:

0, 10

Step-by-step explanation:

The given function is:

$g(y) = \frac{y-5}{y^2-3y+15}$

According to the quotient rule:

$d(\frac{f(y)}{h(y)}) = \frac{f(y)*h'(y)-h(y)*f'(y)}{h^2(y)}$

Applying the quotient rule:

$g(y) = \frac{y-5}{y^2-3y+15}\\g'(y)=\frac{(y-5)*(2y-3)-(y^2-3y+15)*(1)}{(y^2-3y+15)^2}$

The values for which g'(y) are zero are the critical points:

$g'(y)=0=\frac{(y-5)*(2y-3)-(y^2-3y+15)*(1)}{(y^2-3y+15)^2}\\(y-5)*(2y-3)-(y^2-3y+15)=0\\2y^2-3y-10y+15-y^2+3y-15\\y^2-10y=0\\y=\frac{10\pm \sqrt 100}{2}\\y_1=\frac{10-10}{2}= 0\\y_2=\frac{10+10}{2}=10$

The critical values are y = 0 and y = 10.

5 0

3 years ago

What kind of quadrilaterals is this?

DanielleElmas [232]

Answer:

Trapezoid

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers