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Vera_Pavlovna [14]
3 years ago
15

Find the area. 7 cm 4 cm 1 1 4 cm 122 cm . 1 8 cm 18 cm

Mathematics
1 answer:
Naddik [55]3 years ago
4 0

Answer: ok i got this. 9.79 cm^2 ???

Its really hard to understand the numbers shown in the picture or even written in the question ≡(▔﹏▔)≡

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What is (f–g)(x)?<br> f(x)=<br> –<br> 2x2+10x<br> g(x)=<br> –<br> 3x2
WINSTONCH [101]

Answer:

What is (f–g)(x)?

f(x)=

–

2x2+10x

g(x)=11

–

3x2

12

Step-by-step explanation:

5 0
2 years ago
Pleaseeee helpp it’s due in 9 minutes helpp
defon
C. $30.00 is the answer !!!!!!
4 0
3 years ago
1. The product of five and a number is 15
dedylja [7]

Answer:

Step-by-step explanation:

(1) 5x = 15

Dividing both sides by 5 will give us the answer:

x = 3

(2) x + 6 = 16

Subtracting both sides by 6 will give us the answer:

x = 10

(3) \frac{27}{x} = 9

First let's multiply both sides by x

27 = 9x

Now, let's divide both sides by 9 to get our answer:

x = 3

(4) 6x = 72

Dividing both sides by 6 will give us the answer:

x = 12

6 0
3 years ago
$ 57,000 at 14% for 2years
olga_2 [115]

Answer:

$288,810.94

Step-by-step explanation:

7 0
3 years ago
Jaclyn plays singles for the varsity tennis team where she won the sudden death tie breaker point with a cross-court passing sho
podryga [215]

Answer:

Incomplete question

Complete question: Jaclyn plays singles for South's varsity tennis team. During the match against North, Jaclyn won the sudden death tiebreaker point with a cross-court passing shot. The 57.5-gram ball hit her racket with a northward velocity of 26.7 m/s. Upon impact with her 331-gram racket, the ball rebounded in the exact opposite direction (and along the same general trajectory) with a speed of 29.5 m/s.

a. Determine the pre-collision momentum of the ball.

b. Determine the post-collision momentum of the ball.

c. Determine the momentum change of the ball.

Answer:

A. 1.5353kgm/s

B. 1.6963kgm/s

C. 0.161kgm/s

Step-by-step explanation:

A. The pre-collision momentum of the ball = mass of ball × velocity of ball

Mass of ball = 57.5g = 0.0575kg

Velocity of ball = 26.7m/s

Pre-collision momentum of ball = 0.0575×26.7

= 1.5353kgm/s

B. Post collision momentum of the ball = mass of ball × velocity of ball after impact

Velocity of ball after impact = 29.5m/s

Post collision momentum of ball after impact = 0.0575×29.5

= 1.6963kgm/s

C. Momentum change of ball = momentum after impact - momentum before imlact

= 1.6963kgm/s - 1.5353kgm/s

= 0.161kgm/s

7 0
3 years ago
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