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Arlecino [84]
3 years ago
5

What does 2b+9 when b=3 like the hell

Mathematics
2 answers:
BabaBlast [244]3 years ago
5 0
You plug 3 into B so 2(3) + 9 = 15
beks73 [17]3 years ago
3 0
So let’s start from the beginning. 2 is by B, so that means we have to MULTIPLY 2 by whatever B equals. B equals to 3, so we have to multiply 2 by 3, which equals 6. Then, we add 6 and 9, which is 15. Your final answer is 15. Hope that helps!
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15<br> G<br> F<br> H<br> 4) What is the value of x?<br> A) 10<br> B) 12<br> C) 15<br> D) 17
coldgirl [10]

Answer:

C) 15

Step-by-step explanation:

This is an isosceles triangle (two of the three angles are equal) so two of the sides are also equal. x is the same length as 15.

5 0
2 years ago
The sum of three times a number, and 3, is equal to two times the number.
ale4655 [162]

3x + 3 = 2x

three times a number = 3x ,and 3 = 3x + 3, is equal to two times the number =

3x + 3 = 2x

6 0
3 years ago
Police Chase: A speeder traveling 40 miles per hour (in a 25 mph zone) passes a stopped police car which immediately takes off a
zubka84 [21]

Answer:

a. 18.34 s b. 327.92 m

Step-by-step explanation:

a. How long before the police car catches the speeder who continued traveling at 40 miles/hour

The acceleration of the car a in 10 s from 0 to 55 mi/h is a = (v - u)/t where u = initial velocity = 0 m/s, v = final velocity = 55 mi/h = 55 × 1609 m/3600 s = 24.58 m/s and t = time = 10 s.

So, a =  (v - u)/t =  (24.58 m/s - 0 m/s)/10 s = 24.58 m/s ÷ 10 s = 2.458 m/s².

The distance moved by the police car in 10 s is gotten from

s = ut + 1/2at² where u = initial velocity of police car = 0 m/s, a = acceleration = 2.458 m/s² and t = time = 10 s.

s = 0 m/s × 10 s + 1/2 × 2.458 m/s² (10)²

s = 0 m + 1/2 × 2.458 m/s² × 100 s²

s = 122.9 m

The distance moved when the police car is driving at 55 mi/h is s' = 24.58 t where t = driving time after attaining 55 mi/h

The total distance moved by the police car is thus S = s + s' = 122.9 + 24.58t

The total distance moved by the speeder is S' = 40t' mi = (40 × 1609 m/3600 s)t' =  17.88t' m where t' = time taken for police to catch up with speeder.

Since both distances are the same,

S' = S

17.88t' = 122.9 + 24.58t

Also, the time  taken for the police car to catch up with the speeder, t' = time taken for car to accelerate to 55 mi/h + rest of time taken for police car to catch up with speed, t

t' = 10 + t

So, substituting t' into the equation, we have

17.88t' = 122.9 + 24.58t

17.88(10 + t) = 122.9 + 24.58t

178.8 + 17.88t = 122.9 + 24.58t

17.88t - 24.58t = 122.9 - 178.8

-6.7t = -55.9

t = -55.9/-6.7

t = 8.34 s

So, t' = 10 + t

t' = 10 + 8.34

t' = 18.34 s

So, it will take 18.34 s before the police car catches the speeder who continued traveling at 40 miles/hour

b. how far before the police car catches the speeder who continued traveling at 40 miles/hour

Since the distance moved by the police car also equals the distance moved by the speeder, how far the police car will move before he catches the speeder is given by S' = 17.88t' = 17.88 × 18.34 s = 327.92 m

4 0
2 years ago
Can someone help me with theeseeeeeeee
Whitepunk [10]
Wait but that is a test don't you need to look at a book or something insteadt of waiting for the answers
8 0
3 years ago
Read 2 more answers
How to do this question
Hoochie [10]
312.5 x 13. Which is <span>4062.5. Then add 1562.5 to get 5625. So by 1998 there would be 5625 bald eagle pairs</span>
5 0
3 years ago
Read 2 more answers
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