Which statement is true about the graphs of the two lines y =

How are you in the bet nobody’s not going to help me on this question

Nataliya [291]

Answer:

See below.

Step-by-step explanation:

Party A

y = x^2 + 1

For each value of x in the table, substitute x in the equation with that value and evaluate y.

x = -2: y = (-2)^2 + 1 = 4 + 1 = 5

x = -1: y = (-1)^2 + 1 = 1 + 1 = 2

Do the same for x = 0, x = 1, x = 2

x y

-2 5

-1 2

0 1

1 2

2 5

Part B

Look at points (-2, 5) and (-1, 2). The change in x from (-2, 5) to (-1, 2) is 1. The change in y is -3.

Now let's look at two other points which have a change in x of 1. Look at points (0, 1) and (1, 2). The change in x from (0, 1) to (1, 2) is 1. The change in y is 1.

You can see that for the first two points, a change of 1 in x produces a change of -3 in y, but for the second two points, the same change of 1 in x produce a change of 1 in y. Since the same change of x does not always produce the same change in y, the function is nonlinear.

Answer: A

4 0

3 years ago

Ad is tangent to circle o at d. find ab. round to the nearest tenth if necessary

Naddik [55]

Check the picture below.

$\bf 0=AB^2+6AB-121\implies \stackrel{\textit{using the quadratic formula}}{AB=\cfrac{-6 \pm \sqrt{6^2-4(1)(-121)}}{2(1)}}
\\\\\\
AB=\cfrac{-6 \pm \sqrt{520}}{2}\implies AB=\cfrac{-6 \pm \sqrt{4\cdot 130}}{2}
\\\\\\
AB=\cfrac{-6 \pm \sqrt{2^2\cdot 130}}{2}\implies AB=\cfrac{-6 \pm 2\sqrt{130}}{2}
\\\\\\
AB=-3\pm\sqrt{130}\implies AB=
\begin{cases}
\boxed{-3+\sqrt{130}}\\\\
-3-\sqrt{130}
\end{cases}$

since the distance AB cannot be a negative value, thus is not -3-√(130).

3 0

3 years ago

Read 2 more answers

109)

Lubov Fominskaja [6]

I strongly believe the answer is c
Hope this helps in some way:)

7 0

3 years ago

Sin4x.sin5x+sin4x.sin3x-sin2x.sinx=0

andreev551 [17]

Recall the angle sum identity for cosine:

cos(x + y) = cos(x) cos(y) - sin(x) sin(y)

cos(x - y) = cos(x) cos(y) + sin(x) sin(y)

==> sin(x) sin(y) = 1/2 (cos(x - y) - cos(x + y))

Then rewrite the equation as

sin(4x) sin(5x) + sin(4x) sin(3x) - sin(2x) sin(x) = 0

1/2 (cos(-x) - cos(9x)) + 1/2 (cos(x) - cos(7x)) - 1/2 (cos(x) - cos(3x)) = 0

1/2 (cos(9x) - cos(x)) + 1/2 (cos(7x) - cos(3x)) = 0

sin(5x) sin(-4x) + sin(5x) sin(-2x) = 0

-sin(5x) (sin(4x) + sin(2x)) = 0

sin(5x) (sin(4x) + sin(2x)) = 0

Recall the double angle identity for sine:

sin(2x) = 2 sin(x) cos(x)

Rewrite the equation again as

sin(5x) (2 sin(2x) cos(2x) + sin(2x)) = 0

sin(5x) sin(2x) (2 cos(2x) + 1) = 0

sin(5x) = 0 or sin(2x) = 0 or 2 cos(2x) + 1 = 0

sin(5x) = 0 or sin(2x) = 0 or cos(2x) = -1/2

sin(5x) = 0 ==> 5x = arcsin(0) + 2nπ or 5x = arcsin(0) + π + 2nπ

… … … … … ==> 5x = 2nπ or 5x = (2n + 1)π

… … … … … ==> x = 2nπ/5 or x = (2n + 1)π/5

sin(2x) = 0 ==> 2x = arcsin(0) + 2nπ or 2x = arcsin(0) + π + 2nπ

… … … … … ==> 2x = 2nπ or 2x = (2n + 1)π

… … … … … ==> x = nπ or x = (2n + 1)π/2

cos(2x) = -1/2 ==> 2x = arccos(-1/2) + 2nπ or 2x = -arccos(-1/2) + 2nπ

… … … … … … ==> 2x = 2π/3 + 2nπ or 2x = -2π/3 + 2nπ

… … … … … … ==> x = π/3 + nπ or x = -π/3 + nπ

(where n is any integer)

5 0

3 years ago

Meg is 6 years older than Victor. Meg's age is 2 years less than five times Victor's age. The equations below model the relation

Verdich [7]

Since no possible correct method is posted, I will suggest a couple.

Method 1: guess and check

Works well for simple problems involving integers like this one.

Victor's age must be zero or greater than one, say one.

Guess v=1, find m=v+6=7, check m=5v-2=5-2=3 no good.

we need to make v bigger

Guess v=2, find m=v+6=2+6=8, check m=5v-2=5*2-2=8 ✔

So v=2, m=8.

Method 2:

Solve the system of two equations.

since the left-hand sides is m in both equations, and since m=m, we just have to equate the right-hand sides to solve for v.

5v-2=v+6

Solve for v

5v-v = 6+2

4v=8

v=2,

so again, v=2, m=v+6=2+6=8.

6 0

4 years ago

Which statement is true about the graphs of the two lines y = –6 and x = 1/6 ?