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3 years ago

Select all sets that the number is a part of. You may have to select more than one.

Mathematics

1 answer:

maria [59]3 years ago

6 0

Whole numbers. Rational numbers natural numbers real numbers

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Which is a solution for 4 x + 8 = 3 x + 14?

Liula [17]

Answer:

x=6

Step-by-step explanation:

Step 1:

Subtract 3x from both sides:

4x + 8 - 3x = 3x + 14 - 3x

x + 8 = 14

Step 2:

Subtract 8 from both sides:

x + 8 - 8 = 14 - 8

x = 6

4 0

2 years ago

Read 2 more answers

Y=x+3 and y=x-y help me

Nuetrik [128]

Step-by-step explanation:

Given

y = x + 3.....equation 1

y = x - y

or y + y = x

or 2y = x

or y = x / 2......equation 2

Now

Putting equation 2 in equation 1

x / 2 = x + 3

x - x/2 = - 3

2x - x  = - 3

x / 2 = -3

x = - 6

Now

y = x/ 2

= - 6/2

= - 3

Hope this helps

4 0

3 years ago

Please help!!!!

koban [17]

Answer:

Correct choice is A

Step-by-step explanation:

If a function has an inverse, then there is at most one x-value for each y-value.

The tangent function is periodic with period $\pi.$ Hence, at each value for which $f(x)=\tan x$ is defined, $f(x+n\pi )=\tan x$ for each integer n. Therefore, the function $f(x)=\tan x$ does not have an inverse. Since tangent is not a one-to-one function, the domain must be limited. From examining the graph of the tangent function, we see that in each interval of the form

$\left((2k−1)\dfrac{\pi}{2},(2k+1)\dfrac{\pi}{2}\right)$

where k is an integer, the tangent function assumes every value in its range. Moreover, in each such interval, each y-value is achieved exactly once. Hence, we can create an invertible function by restricting the domain tangent function to one such interval. Such interval is an interval between two consecutive vertical asymptotes $x=(2k−1)\dfrac{\pi}{2}$ and $x=(2k+1)\dfrac{\pi}{2}.$