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Irina-Kira [14]
2 years ago
11

I don’t understand can someone help me

Mathematics
1 answer:
Elden [556K]2 years ago
6 0
You set 49 equal to 8x - 15

So

8x - 15 = 49
+15 + 15

8x = 64

Then divide by 8

X= 8
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I need help i don’t understand
ch4aika [34]

9514 1404 393

Answer:

  A) J(-3, 1), ...

Step-by-step explanation:

All of the answer choices list point J' first, so it is convenient to use that as an example.

Point J on the given graph has coordinates (x, y) = (3, 2). The x-coordinate is 3 because the point is 3 units to the right of the y-axis.

The problem statement tells you to translate this point 6 units to the left. When you move it 6 units left, it will move left 3 units to the y-axis, then left 3 more units to have an x-coordinate of 3 -6 = -3. That is, each unit of movement to the left subtracts 1 from the x-coordinate. The x-coordinate of J is 3, so the final point J' will have an x-coordinate of 3 - 6 = -3.

At this point, you have enough information to make the correct answer selection. Only one answer choice has the x-coordinate of J' as -3.

__

The other coordinates are translated using similar logic. The y-coordinate of J is 2. Translating it down 1 unit subtracts 1 from the y-coordinate to make it be 2 -1 = 1. Then the coordinates of J' are (-3, 1).

We write the translation rule as ...

  (x, y) ⇒ (x -6, y -1)

This means the coordinates of each translated point have 6 subtracted from the original x-coordinate, and 1 subtracted from the original y-coordinate. The other coordinates of the figure are ...

  I(2, 4) ⇒ I'(2 -6, 4 -1) = I'(-4, 3)

  H(5, 5) ⇒ H'(5 -6, 5 -1) = H'(-1, 4)

  G(4, 1) ⇒ G'(4 -6, 1 -1) = G'(-2, 0)

3 0
3 years ago
Find the circumference and area of a circle with a diameter of 14 inches. Leave your answers in terms of pi. C = 14π; A = 49π. C
Olenka [21]
Given:
diameter = 14 inches
radius = 14 inches / 2 = 7 inches

Circumference of a circle = 2 π r = 2 * π * 7inches = 14 inches * π

Area of a circle = π r² = π * (7in)² = π * 49in² 

C = 14π    and   A = 49π
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Choose the points that are solutions to this:
Scorpion4ik [409]

Answer:

b. (3,3)

Step-by-step explanation:

6 0
3 years ago
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mart [117]

Answer:

\large\boxed{x=1\ \vee\ x=-3i\ \vee\ x=3i}

Step-by-step explanation:

\text{The zeros:}\\\\x^3-x^2+9x-9=0\\\\x^2(x-1)+9(x-1)=0\\\\(x-1)(x^2+9)=0\iff x-1=0\ \vee\ x^2+9=0\\\\x-1=0\qquad\text{add 1 to both sides}\\x-1+1=0+1\\\boxed{x=1}\\\\x^2+9=0\qquad\text{subtract 9 from both sides}\\x^2=-9

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