Locate and label at least five points on the coordinate plane that has an X coordinate of 6

1 answer:

8 0

Answer: you could use any point on the coordinate plane it just has to have the x coordinate for an example ( -1,6 ) is a point you could use .... I dunno maybe i'm wrong another helper you can use is symbolab it really helps me with my homework ... anyways let me know if I helped.

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Solve for x (10 points)

ella [17]

8^2 + 5^2 = c^2
64 + 25 = c^2
89 = c^2
c = 9.43
x = 9.43 - 3
x = 6.43

8 0

3 years ago

1) f(4) = g(4) 2) f(4) = g(-2) 3) f(2) = g(-2) 4) f(-2) = g(-2)

aniked [119]

Answer: 4)

Step-by-step explanation:

f(4) = -14

g(4) = 10

g(-2) = 4

f(2) = -8

f(-2) = 4

The correct statement is 4) f(-2) = g(-2)

Or by inspection, we can see that the graph f(x) intersects g(x) at x = -2 and y = 4. So, f(-2) = g(-2) = 4 is true

6 0

3 years ago

Which of the following goes through (-2,4)

Setler79 [48]

-3x+y=10 is the answer. If you plug the coordinates in to the respective spots (-2 for x and 4 for y), you will get an equal true statement.

8 0

3 years ago

The solution to an any quality is giving in the same building notation as The solution to an inequality is giving in the set bui

Lostsunrise [7]

Options

$(A)\left(-\infty , \dfrac23\right]\\\\(B)\left(-\infty , \dfrac23\right) \\\\(C)(\frac23\right, \infty ) \\\\(D) [\frac23\right, \infty )$

Answer:

$(C)(\frac23\right, \infty )$

Step-by-step explanation:

Given the solution to an inequality

{x|x>2/3}

The solution set does not include $\dfrac23$ , therefore, it must be open at the left. Recall that we use a curvy bracket ( to denote openness at the left.

Since x is greater than $\dfrac23$ , the solution set contains all values of larger than $\dfrac23$ up till infinity. Since infinity is an arbitrarily large value, we also use an open bracket at the right.

Therefore, another way to represent the solution {x|x>2/3} is:

$(\frac23\right, \infty )$