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gayaneshka [121]
3 years ago
8

PLEASE HELP ASAP!! WORDED QUADRATIC QUESTIONS!!!

Mathematics
1 answer:
Roman55 [17]3 years ago
4 0

Answer:

x² - 5x = 14

x = -2   or   x = 7

Step-by-step explanation:

x² - 5x = 14

x² - 5x - 14 = 0

(x+2)(x-7) = 0

x = -2   or   x = 7

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Write the point-slope form of the equation of the line through the given point with the given slope. (-5,0) slope= 2/3
amid [387]

Answer:

Step-by-step explanation:

y - 0 = 2/3(x + 5)

y = 2/3x + 10/3

8 0
4 years ago
Which of the following is most likely the next step in the series?<br>​
tankabanditka [31]

Answer:

D

Step-by-step explanation:

8 0
3 years ago
6 x - 2 [ x + 2 ] &gt; 2 - 3 [ x + 3]<br> could u show me step by step on how to do thiz
Dafna11 [192]

Answer:

x > - 3/7

Step-by-step explanation:

6x - 2x - 4 > 2 - 3 (x + 3)

6x - 2x - 4 > 2 - 3x - 9

4x - 4 > 2 - 3x - 9

4x - 4 > - 7 - 3x

4x - 4 + 3x > - 7

7x > - 7 + 4

7x > - 3

8 0
3 years ago
A brick is thrown upward from the top of a building at an angle of 250 above the horizontal, and with an initial speed of 15 m/s
mojhsa [17]

Answer:

(a) 25.08 m

(b) 2.05 m

Step-by-step explanation:

Let the height of the building be 'h'.

Given:

Angle of projection is, \theta=25°

Initial speed is, u=15\ m/s

Time of flight is, t=3.0\ s

(a)

Consider the vertical motion of the brick.

Vertical component of initial velocity is given as:

u_y=u\sin\theta\\\\u_y=15\sin(25)=6.34\ m/s

Vertical displacement of the brick is equal to the height of the building.

So, vertical displacement = -h (Negative sign implies downward motion)

Acceleration is due to gravity in the downward direction. So,

Acceleration is, g=-9.8\ m/s^2

Now, using the following equation of motion;

-h=u_yt+\frac{1}{2}gt^2\\-h=6.34(3)-\frac{9.8}{2}(3)^2\\\\-h=19.02-44.1\\\\-h=-25.08\\\\h=25.08\ m

Therefore, the building is 25.08 m tall.

(b)

Let the maximum height be 'H'.

At maximum height, the vertical component of velocity is 0 as the brick stops temporarily in the vertical direction.

So, v_y=0\ m/s

Now, using the following equation of motion, we have:

v_y^2=u_y^2+2gH\\\\0=(6.34)^2-2\times 9.8\times H\\\\19.6H=40.2\\\\H=\frac{40.2}{19.6}=2.05\ m

Therefore, the maximum height of the brick is 2.05 m.

5 0
3 years ago
*Can someone please help me out with this one?*
Talja [164]
Either A - the value of r would increase or B- the value of r would not change.
5 0
3 years ago
Read 2 more answers
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