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guajiro [1.7K]
4 years ago
12

Find only the x-coordinate of either point of intersection. Y=-x^2+8x-7 Y=x^2-8x+17

Mathematics
1 answer:
postnew [5]4 years ago
6 0
The point of intersections should be (2,5) and (6,5) so the answers are 2 and 6.
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Help I need you right nowwwwwwwwwwww
earnstyle [38]

Answer:

6/8

just add the fractions together after making the bottom the same

4 0
3 years ago
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3+×/2=10 <br> Is ×/2=7<br> And does ×=14
adoni [48]

3+\dfrac{x}{2}=10\qquad|\text{subtract 3 from both sides}\\\\\dfrac{x}{2}=7\qquad|\text{multiply both sides by 2}\\\\\boxed{x=14}

7 0
3 years ago
Abby used the quadratic formula to find the zeros of the
MArishka [77]

Step 2: She did not simplify under the square root correctly

under the square root:

(-6)^2 =36

and -4(5)(-8) =160

therefore under the square root it needs to be square root of 36+160 which is square root of 196.

That answer is 14 when square rooted

5 0
3 years ago
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Find the area of a triangle ABC when measure B = 56 degrees, a = 10 inches, and c = 7 inches.
Alexxandr [17]
The answer is around 29
8 0
3 years ago
Please answer me question 3 solving part​
Illusion [34]

Answer:

1. D

2. B

3. A

Step-by-step explanation:

Question 1:

The pair of <JKL and <LKM can be referred to as linear pairs. They are two adjacent angles that are formed from the intersecting of two lines.

Question 2:

Given that <KLM = x°

<KML = 50°

<JKL = (2x - 15)°

According to the exterior angle theorem, exterior ∠ JKL = <KLM + KML.

2x - 15 = x + 50

Solve for x

2x - x = 15 + 50

x = 65

Therefore, <KLM = 65°

QUESTION 3:

<JKL = 2x - 15

Plug in the value of x

<JKL = 2(65) - 15

= 130 - 15

<JKL = 115°

7 0
3 years ago
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