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marissa [1.9K]
2 years ago
9

6a2 + 8a - 25 = -3 Using the Quadratic Formula

Mathematics
1 answer:
drek231 [11]2 years ago
4 0

Answer:

6

2

+

8

−

2

5

=

−

3

6a^{2}+8a-25=-3

6

2

+

8

−

2

5

−

(

−

3

)

=

0

Step-by-step explanation:

0/3

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Segments PR and QT are diameters of Circle A.
Dafna11 [192]

Answer:

C. 320

Step-by-step explanation:

Given:

  • PR and QT are diameters

As we can see in the photo, PR and QR are perpendicular with each other at the angle PAQ => the angle QAR = 90°

=> arc PR = 360°/2 = 180°

=> mTAU = 180° - mRAS -mSAT - mPAU = 180° - 50° + 40° + 40° = 50°

Hence,  mPRU = mRAS + mSAT + mTAU + arc PQ

= 50° + 40° + 50° + 180°

= 320°

6 0
4 years ago
How tooo do thissss?
exis [7]

Step-by-step explanation:

m/s² = m/s / s

this is the slope of the line in the graph.

t = 2, that means it is in the first segment of 0 - 6 seconds

the slope in this segment is

(25-15)m/s / (6-0)s

10m/s / 6s = 10/6 m/s² = 5/4 m/s² = 1.25 m/s² ≈ 1.3 m/s²

the same thing for t = 12 in the third segment.

the acceleration or slope is negative here, but that's called here a "forbidden word", and is therefore also a positive number.

in that this segment the slope is

(0-25)m/s / (16-11)s

-25m/s / 5s = -5 m/s²

and so, the answer is 5 m/s².

for the total distance we need to calculate the area under the lines.

the line functions in the 3 segments are :

s(t) = 5/4 t + 15

s(t) = 25

s(t) = -5t + 25

the third one is simplified by moving it from the 11 to 16 seconds interval to a 0 to 5 seconds interval. this creates the same area under that line.

the area under the first line is the integral of the line function between 0 and 6.

that is

[5/8 t² + 15t] between 0 and 6.

and that is

5/8 × 36 + 90 - 5/8 × 0 - 15×0 = 45/4 + 90

= 45/4 + 360/4 = 405/4 = 101.25m

the area under the second segment is simply the area of the rectangle 5×25 = 125m

and the area under the third segment is the integral of the line function between 0 and 5 (remember, we shifted from 11 to 16 to the interval 0 to 5).

that is

[-2.5t² + 25t] in the interval 0 to 5.

and that is

-2.5×25 + 25×5 - -2.5×0 - 25×0 = -62.5 + 125 = 62.5m

as a control check this area is also the area of a right-angled triangle with legs of 5 and 25

the area is 25×5/2 = 125/2 = 62.5

so, it is correct.

the total area under the lines is therefore

101.25 + 125 + 62.5 = 288.75 m ≈ 288.8 m

the average speed is now simply

288.75 m / 16 s = 288.75/16 m/s = 18.046875 m/s ≈

≈ 18 m/s

but if we used the already rounded result of the total distance of 288.8 m, then we would get

288.8 / 16 = 18.05 m/s ≈ 18.1 m/s

3 0
3 years ago
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