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just olya [345]
3 years ago
12

Please help me !!!!!

Mathematics
2 answers:
Lelu [443]3 years ago
6 0

Answer:

thats hard

Step-by-step explanation:

Airida [17]3 years ago
4 0

Answer:

The third table. I am pretty sure 'bout that;-)

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Please help me with this
Sever21 [200]

Answer:

\begin{array}{c|c}x&y\\-2&13\\-1&9\\0&5\\1&1\\2&-3\end{array}

Step-by-step explanation:

Put x = -2, x = -1, ... , x = 2 to the equation of the function y = -4x + 5 and calculate the values of y:

for x = -2

y = -4(-2) + 5 = 8 + 5 = 13

for x = -1

y = -4(-1) + 5 = 4 + 5 = 9

for x = 0

y = -4(0) + 5 = 0 + 5 = 5

for x = 1

y = -4(1) + 5 = -4 + 5 = 1

for x = 2

y = -4(2) + 5 = -8 + 5 = -3

5 0
3 years ago
Read 2 more answers
29, 19, 9, -1, ... <br> please help me out;(
anyanavicka [17]
The next one is -11.
7 0
3 years ago
Read 2 more answers
If another line is perpendicular to the line shown on the graph, what must be its slope
Shtirlitz [24]

It's B

Slope is negative, the other must be positive due to m1 x m2 = - 1

m1 = -2/5

The diagram shows a gradient of -2/5. So, the

reciprocal of this valve to make - 1 is 5/2 (which is the - 2/5 flipped upside down, but positive)

-2/5 x 5/2 = - 10/10 (or - 1)

Thus, the perpendicular line to this has a gradient of 5/2.

Hope this helps!

4 0
1 year ago
How do I graph this linear function <br> x &lt; -1
wolverine [178]
Hope this helps! Let me know if you need anything else.

6 0
2 years ago
Read 2 more answers
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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