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3 years ago

What is the sum of the angles of a quadrilateral

Mathematics

2 answers:

tia_tia [17]3 years ago

7 0

The angles of a quadrilateral always add up to 360 degrees.

posledela3 years ago

6 0

A quadrilateral always has interior angles that add up to 360 degrees.

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It is for a grade please help asap

guajiro [1.7K]

Answer:

I believe the answer is 64. I'm not exactly sure but let's hope it's correct.

Step-by-step explanation:

I did 4 times 16 since 1 pound is 16 ounces. So I multiplied 4 by 16 and it gave me 64.

5 0

2 years ago

Read 2 more answers

Soledads regular pay rate is $7.50 per hour. If she works more than 40 hours in one week, she receives 1.5 times

monitta

Answer:

Step-by-step explanation:

pay for 40 hours: 7.50 * 40= 300

pay for 1 hour of overtime pay: 1.5 * 7.5= 11.25 per hour

11.25 * 4= 45

45 + 300= 345

40 + 4= 44

8 0

3 years ago

Given a/b=x+5, solve for B

anygoal [31]

Answer:

Refer the attachment for the answer

Hope it helps

Please mark me as the brainliest

Thank you

7 0

3 years ago

Need help with my homework

Volgvan

Answer:

$\displaystyle y=\frac{16-9x^3}{2x^3 - 3}$

$\displaystyle y=-\frac{56}{13}$

Step-by-step explanation:

<u>Equation Solving</u>

We are given the equation:

$\displaystyle x=\sqrt[3]{\frac{3y+16}{2y+9}}$

To make y as a subject, we need to isolate y, that is, leaving it alone in the left side of the equation, and an expression with no y's to the right side.

We have to make it in steps like follows.

Cube both sides:

$\displaystyle x^3=\left(\sqrt[3]{\frac{3y+16}{2y+9}}\right)^3$

Simplify the radical with the cube:

$\displaystyle x^3=\frac{3y+16}{2y+9}$

Multiply by 2y+9

$\displaystyle x^3(2y+9)=\frac{3y+16}{2y+9}(2y+9)$

Simplify:

$\displaystyle x^3(2y+9)=3y+16$

Operate the parentheses:

$\displaystyle x^3(2y)+x^3(9)=3y+16$

$\displaystyle 2x^3y+9x^3=3y+16$

Subtract 3y and $9x^3$ :

$\displaystyle 2x^3y - 3y=16-9x^3$

Factor y out of the left side:

$\displaystyle y(2x^3 - 3)=16-9x^3$

Divide by $2x^3 - 3$ :

$\mathbf{\displaystyle y=\frac{16-9x^3}{2x^3 - 3}}$

ii) To find y when x=2, substitute:

$\displaystyle y=\frac{16-9\cdot 2^3}{2\cdot 2^3 - 3}$

$\displaystyle y=\frac{16-9\cdot 8}{2\cdot 8 - 3}$

$\displaystyle y=\frac{16-72}{16- 3}$

$\displaystyle y=\frac{-56}{13}$

$\mathbf{\displaystyle y=-\frac{56}{13}}$

8 0

2 years ago

Can I get some help on this question please

mr Goodwill [35]

things were not as complicated when I was in middle school.

3 0

2 years ago