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Komok [63]
2 years ago
5

3(x + 1) - 2x = -6 what is X i need halp

Mathematics
2 answers:
lara [203]2 years ago
5 0
3(x+1)-2x=-6
3x+3-2x=-6
(Like terms) 3x-2x=1x
3+1x=-6
-6-3+-9
Now you’re going to divide -9 by 3 which you will get -3
That will be your answer, hope that help you
Mashutka [201]2 years ago
4 0

Answer:x=−9

Step-by-step explanation:

Apply the distributive property.

3 x + 3 ⋅ 1 − 2 x = − 6

Multiply  3  by  1

.

3 x + 3 − 2 x = − 6

Subtract 2x from 3x.

x + 3 = − 6

Move all terms not containing  x  to the right side of the equation.

Subtract 3 from both sides of the equation.

x = − 6 − 3

Subtract 3 from − 6 .

x = − 9

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42.43 ft

Step-by-step explanation:

Please find attached an image of the softball pitch

h² = p² + f² -2x p×f×cos45

h² = 43² + 60² - 2x (43)(60) x 0.7071

h² = 1849 +  3600 - 3648.636

h² = 1800.364

h = √1800.364

h = 42.43 feet

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2 years ago
A recipe for chocolate cake calls for 3.5 cups of flour the recipe makes seven servings how many cups of flour are needed to mak
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5

Step-by-step explanation:

If 3.5 cups of flour makes 7 servings, then 0.5 cups of flour are needed for each serving if you divide 3.5 by 7, 10 times 0.5 is equal to 5.

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3 years ago
Which of the equations is proportional if y is a function of X?
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2 years ago
The vertices of xyz are x (1,-4)
dexar [7]

Answer:

5. The vertices of ΔX'Y'Z' are (-3, -7), (-6, -4), (-1, -2)

6. The vertices of ΔX'Y'Z' are (6, -7), (3, -4), (8, -2)

Step-by-step explanation:

If the point (x, y) translated by T → (h, k), then its image is (x + h, y + k)

#5

In ΔXYZ

∵ X = (1, -4), Y = (-2, -1), Z = (3, 1)

∵ T → (-4, -3)

∴ h = -4 and k = -3

→ Use the rule above to find the image of the vertices of the Δ

∵ X' = (1 + -4, -4 + -3)

∴ X' = (-3, -7)

∵ Y' = (-2 + -4, -1 + -3)

∴ Y' = (-6, -4)

∵ Z' = (3 + -4, 1 + -3)

∴ Z' = (-1, -2)

∴ The vertices of ΔX'Y'Z' are (-3, -7), (-6, -4), (-1, -2)

#6

In ΔXYZ

∵ X = (1, -4), Y = (-2, -1), Z = (3, 1)

∵ T → (5, -3)

∴ h = 5 and k = -3

→ Use the rule above to find the image of the vertices of the Δ

∵ X' = (1 + 5, -4 + -3)

∴ X' = (6, -7)

∵ Y' = (-2 + 5, -1 + -3)

∴ Y' = (3, -4)

∵ Z' = (3 + 5, 1 + -3)

∴ Z' = (8, -2)

∴ The vertices of ΔX'Y'Z' are (6, -7), (3, -4), (8, -2)

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3 years ago
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the answer to this question is B

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