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Tpy6a [65]
3 years ago
8

Look at the table representing the distance covered by a car accelerating from 0 miles per hour. The table shows a relationship.

The differences are all . The function represented by the table is written as
f(x) = x2.

Mathematics
1 answer:
vfiekz [6]3 years ago
8 0

quadratic

second

16

8

✌️

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(Sugg<br> eous equations.<br> 3x + 2y = 12<br> 4x – y = 5
Ostrovityanka [42]

Answer:

1 × (3x + 2y = 12)

2 × (4x - y = 5)

3x + 2y = 12

+ (8x - 2y = 10)

= 11x = 22

divide both sides by the coefficient of 'x' (11)

Therefore x = 2

put (x = 2) in equation 1

3(2) + 2y = 12

6 + 2y = 12

2y = 12 - 6

2y = 6

divide both sides by the coefficient of 'y'(2)

Therefore y = 3

7 0
3 years ago
Help on question c please
ycow [4]
11 meteres per second could’ve the fastest he could travel
3 0
3 years ago
a. Determine whether the Mean Value Theorem applies to the function f (x )equals ln 15 xf(x)=ln15x on the given interval [1 comm
ivolga24 [154]

Answer:

(a)

The function f is continuous at [1,e] and differentiable at (1,e), therefore

the mean value theorem applies to the function.

(b)

c = e-1 \\ = 1.71828

Step-by-step explanation:

(a)

The function f is continuous at [1,e] and differentiable at (1,e), therefore

the mean value theorem applies to the function.

(b)

You are looking for a point c   such that

\frac{1}{c} = \frac{\ln(15e)-\ln(15*1)}{e-1} = \frac{\ln(15e/15)}{e-1} = \frac{\ln(e)}{e-1} = \frac{1}{e-1}

You have to solve for c  and get that

c = e-1 \\ = 1.71828

5 0
3 years ago
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