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Contact [7]
3 years ago
15

*Urgent* I need help answering this question. Someone please help me, I keep seeing someone type on my questions but they never

get answered. Please help thank you!

Mathematics
2 answers:
devlian [24]3 years ago
6 0

Answer: for B is   (5y - 43) • (y - 7)

Taya2010 [7]3 years ago
5 0

Answer:

5y^2−78y+301   hope you understand it

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Determine the two ordered pair solutions of the equation y = 2x2 + 3x , given (0, ), and (-2, ).
Rudiy27

Answer:

We conclude that the two ordered pairs (0, 0) and (-2, 2) are the solutions of the equation y = 2x² + 3x.

Step-by-step explanation:

Given the expression

y = 2x² + 3x

Substituting x = 0

y = 2(0)² + 3(0)

y = 0+0

y = 0

  • so when x = 0, y = 0

Thus, the ordered pair is: (0, 0)

Now, substituting x = -2

y = 2x² + 3x

y = 2(-2)² + 3(-2)

y = 8 - 6

y = 2

  • so when x = -2, y = 2

Thus, the ordered pair is: (-2, 2)

Therefore, we conclude that the two ordered pairs (0, 0) and (-2, 2) are the solutions of the equation y = 2x² + 3x.

3 0
3 years ago
If u play or know about baseball help me with this
Alenkasestr [34]

Answer:

one hand comes off the bat

3 0
3 years ago
10x=4.1<br>with simplify
hram777 [196]

Solve 10x = 4.1 for x.  Divide both sides of this equation by 10:  x = 4.1 / 10.

x = 0.41

7 0
3 years ago
PQRS is a parallelogram. Answer the questions below.
abruzzese [7]

Answer:

4. SR= 17 (opposite angle of parallogram are equal)

8 0
3 years ago
Helpppppppppppppppppppppppppppppppp
Sonbull [250]
The equation is y=(x-7)^2+7.

We are looking for a function with a vertex above the x-axis and a function that opens upward (has coefficient a > 0).

The first function opens downward and intersects the x-axis. The second function has a vertex below the x-axis. The third function satisfies our requirements. The fourth function has a vertex on the x-axis.

We can solve this algebraically with the knowledge that the real solutions of a quadratic are its x-intercepts. If there are no x-intercepts (because it lies entirely above or below the x-axis), then there are no real solutions. This is true when the discriminant b^2-4ac \ \textless \  0. You can see that from the quadratic formula. This holds true for both answers A and C, so to find the correct one, we remember that when the coefficient a of the x^2 term is positive, the graph opens upwards, so we choose C.
5 0
3 years ago
Read 2 more answers
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