Hurry please Answer ASAP

Mathematics

1 answer:

zepelin [54]2 years ago

4 0

It is non proportional because it doesn’t go through the origin I am pretty sure

You might be interested in

Using the linear combination method what is the solution to the system of linear equations 7x - 2y = -20 and 9x + 4y = -6

Ghella [55]

x should equal -2 and y should equal -3 to solve this i multiplied the first equation by 2 then got rid of the 4y and -4y then i added the 14x to the 9 equaling 23x and added the -40 to the -6 which became 23x=-46 i solved this getting -s for x, then i plugged that into the equations to get y which was -3

8 0

3 years ago

GIVING BRAINLIEST Consider these proportions: 3:7.5 4:10 5:12.5 6:15 Which of the following proportions belongs to the above set

Aleksandr-060686 [28]

Answer:

........ It's C. ............

7 0

3 years ago

Can somone help me withhhhh thisssss pleaseee

Olenka [21]

It is A. 2^12.
The question asks for the value of 8 to the x-power. Divided by 2 to the y-power. the trick to solving this is to rewrite 8^x as 2^3x. so you will have 2^3x divided by 2^y. since the bases are now the same , you can subtract the exponents to get the result of 2^3x-y. since it is given the 3x-y is 12. your final answer is 2^12

5 0

2 years ago

Read 2 more answers

Tonya is 5 years older than Raul. Tonya is 16 years old. Solve the equation r+5=16 to find Raul's age.

AfilCa [17]

Answer:

he is 11 years old

Step-by-step explanation:

7 0

2 years ago

Read 2 more answers

Find all solutions of each equation on the interval 0 ≤ x < 2π.

Korvikt [17]

Answer:

$x = 0$ or $x = \pi$ .

Step-by-step explanation:

How are tangents and secants related to sines and cosines?

$\displaystyle \tan{x} = \frac{\sin{x}}{\cos{x}}$ .

$\displaystyle \sec{x} = \frac{1}{\cos{x}}$ .

Sticking to either cosine or sine might help simplify the calculation. By the Pythagorean Theorem, $\sin^{2}{x} = 1 - \cos^{2}{x}$ . Therefore, for the square of tangents,

$\displaystyle \tan^{2}{x} = \frac{\sin^{2}{x}}{\cos^{2}{x}} = \frac{1 - \cos^{2}{x}}{\cos^{2}{x}}$ .

This equation will thus become:

$\displaystyle \frac{1 - \cos^{2}{x}}{\cos^{2}{x}} \cdot \frac{1}{\cos^{2}{x}} + \frac{2}{\cos^{2}{x}} - \frac{1 - \cos^{2}{x}}{\cos^{2}{x}} = 2$ .

To simplify the calculations, replace all $\cos^{2}{x}$ with another variable. For example, let $u = \cos^{2}{x}$ . Keep in mind that $0 \le \cos^{2}{x} \le 1 \implies 0 \le u \le 1$ .