If x+y = -10 and x-y = 2 what is the value of x

Mathematics

1 answer:

madam [21]2 years ago

5 0

Answer:

x= -4

Step-by-step explanation:

adding the two equations,

x+x+y-y=-10+2

2x= -8

x=-4

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Will give Brainlest Answer please answer i need help

Rashid [163]

Answer:

C. {-1, 5, 8}

Step-by-step explanation:

Use each of the domain values in the function to see what the corresponding range value is.

f(-1) = 5 -3(-1) = 8

f(0) = 5 -3(0) = 5

f(2) = 5 -3(2) = -1

The range is the set of numbers {-1, 5, 8}.

_____

<em>Additional comment</em>

The values in a set are generally listed lowest to highest. The coefficient of x in the equation for f(x) is negative, meaning the lowest range value will correspond to the highest domain value. If you start by finding f(2) = -1, you immediately eliminate all answer choices except B and C.

Those choices differ only in the middle value, so you can tell which is correct by evaluating f(x) for the middle domain value: f(0) = 5. Only one answer choice has both -1 and 5 in the set.

(There are two answers here: how you work the problem, and how you game a multiple choice question.)

4 0

2 years ago

Fill in Sin, Cos, and tan ratio for angle x. <br> Sin X = 4/5 (28/35 simplified)

Fantom [35]

Answer:

Given: $\sin(x) = (4/5)$ .

Assuming that $0 < x < 90^{\circ}$ , $\cos(x) = (3/5)$ while $\tan(x) = (4/3)$ .

Step-by-step explanation:

By the Pythagorean identity $\sin^{2}(x) + \cos^{2}(x) = 1$ .

Assuming that $0 < x < 90^{\circ}$ , $0 < \cos(x) < 1$ .

Rearrange the Pythagorean identity to find an expression for $\cos(x)$ .

$\cos^{2}(x) = 1 - \sin^{2}(x)$ .

Given that $0 < \cos(x) < 1$ :

$\begin{aligned} &\cos(x) \\ &= \sqrt{1 - \sin^{2}(x)} \\ &= \sqrt{1 - \left(\frac{4}{5}\right)^{2}} \\ &= \sqrt{1 - \frac{16}{25}} \\ &= \frac{3}{5}\end{aligned}$ .