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3 years ago

Solve the quadratic equation using the table

Mathematics

2 answers:

nekit [7.7K]3 years ago

7 0

F.no solution

Is the answer

algol [13]3 years ago

4 0

F) no solution BC it’s not consistent

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What is the midpoint of the segment shown below?

Amanda [17]

Answer:

answer is c (0,2)

hope it helps

7 0

2 years ago

2x-5y=10 ;what is the x intercept, what is the y intercept, what is the slope?

creativ13 [48]

First we put the equation in y = mx + b form where m is ur slope and b is ur y int.

2x - 5y = 10
-5y = -2x + 10
y = 2/5x - 2.......so ur slope is 2/5 and ur y int is (0,-2) <==

to find ur x int, sub in 0 for y and solve for x
2x - 5y = 10
2x - 5(0) = 10
2x = 10
x = 10/2
x = 5......and ur x int is (5,0) <==

another way to find the y int is to sub in 0 for x and solve for y...but u might as well solve it by putting it in y = mx + b form because that way u have ur slope as well.

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3 years ago

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DONT ANSWER IF YOU DO NOT KNOW!

Lynna [10]

Answer:

~2.09

Step-by-step explanation:

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3 years ago

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MsDaily can grade 54 essays in 6 hours. Which of the following rates represent the number of essays per hour ms. Daley can grade

marshall27 [118]

Step-by-step explanation:

Rate = 54/6

= 9 essays per hour

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3 years ago

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Use the graph of △ABC with midsegments DE, EF and DF. Show that EF is parallel to AC and that EF=1/2 AC

Vinvika [58]

According to the midsegment theorem, the midsegments are parallel to

and half the length of the opposite side.

The completed statement are as follows;

Because the slopes of $\overline{EF}$ and $\overline{AC}$ are both -4, $\overline{EF}$ ║ $\overline{AC}$

EF = $\underline{\sqrt{17}}$ and AC = $\underline{2 \cdot \sqrt{17} }$

Because $\underline{\sqrt{17} }$ = $\underline{\frac{1}{2} \cdot 2 \cdot \sqrt{17} }$ , $EF = \frac{1}{2} \cdot AC$

Reasons:

From the given graph of ΔABC, we have;

Coordinates of the points A, B, and C are; A(-5, 2), B(1, -2), and C(-3, -6)

The coordinates of the point D and E on $\mathbf{\overline{DE}}$ are; D(-4, -2), and E(-2, 0)

The coordinates of the point F is; F(-1, -4)

$Slope, \, m =\dfrac{y_{2}-y_{1}}{x_{2}-x_{1}}$

$\displaystyle Slope \ of \ line \ \overline{AC} = \mathbf{\frac{(-6) - 2}{-3 - (-5)} = \frac{-8}{2}} = -4$

$\displaystyle Slope \ of \ line \ \overline{EF} = \frac{0 - (-4)}{-2 - (-1)} = \frac{4}{-1} = -4$

$Length \ of \ segment,\ l = \sqrt{\left (x_{2}-x_{1} \right )^{2}+\left (y_{2}-y_{1} \right )^{2}}$

Length of EF = √((-1 - (-2))² + (-4 - 0)²) = √(17)

Length of AC = √((-3 - (-5))² + (-6 - 2)²) = √(4 × 17) = 2·√(17)

Therefore, we have;

Because the slope of $\mathbf{\overline {EF}}$ and $\mathbf{\overline {AC}}$ are both , -4, $\overline {EF}$ ║ $\overline {AC}$ . EF = $\underline{\sqrt{17}}$ , and AC

= $\underline{\frac{1}{2} \cdot 2 \cdot \sqrt{17}}$ ,. Because $\underline{\sqrt{17}}$ = $\underline{\frac{1}{2} \cdot 2 \cdot \sqrt{17}}$ , $EF =\mathbf{ \frac{1}{2} AC}$

Learn more about midsegment theorem of a triangle here:

brainly.com/question/7423948

8 0

2 years ago