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Naily [24]
3 years ago
6

The Temperature at 12 noon was 14° C above zero. If it decreases at the

Mathematics
1 answer:
iogann1982 [59]3 years ago
5 0

Answer:

-16

Step-by-step explanation:

you just subtract 2 each hour

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X+ 2 / X + 3 =5 plz give me and<br><br><br><br>​
Anna35 [415]

Answer:

The answer is x = -13/4

Step-by-step explanation:

-13/4 + 2  /  -13/4 +3 = 5

5 0
3 years ago
Which expressions are equivalent to 2^2/7⋅2^2/7? Select all that apply.
sladkih [1.3K]
The answer is E. 2^2/7 + 2/7
5 0
3 years ago
A gym memebership costs a flat fee of $68 and an additional $19.99 per month.Since she became a member Ellen has paid the gym $2
fomenos

Answer:

She has been a member of the gym for seven months

Step-by-step explanation:

In this question, we are asked to calculate the number of months a customer has been a member of a gym, given the pricing template of the gym.

Now, as we can see from the question, there are two kinds of fees to be paid. The flat fee and the subscript class.

First, let’s represent the number of months she has been a member of the gym by m. This means the total amount of money she has paid for m months would be m * 19.99m

Now, recall, we have a flat fee of $68, adding this to the total monthly payment for m months bring the wages to 19.99m + 68

This is said to equal $207.93

This means;

19.99m + 68 = 207.93

19.99m = 207.93 -68

19.99m = 139.93

m = 139.93/19.99 = 7

3 0
3 years ago
Being a good friend is part of developing into a mature person.<br> True<br> False
anygoal [31]

Answer:

That's true

Step-by-step explanation:

8 0
4 years ago
Read 2 more answers
Does the function satisfy the hypotheses of the Mean Value Theorem on the given interval? f(x) = e−5x, [0, 1] Yes, it does not m
Temka [501]

f(x)=e^{-5x} is continuous on [0, 1] and differentiable on (0, 1), so yes, the MVT is satisfied.

By the MVT, there is some c\in(0,1) such that

f'(c)=\dfrac{f(1)-f(0)}{1-0}

The derivative is

f'(x)=-5e^{-5x}

so we get

-5e^{-5c}=e^{-5}-1\implies e^{-5c}=\dfrac{1-e^{-5}}5\implies-5c=\ln\dfrac{1-e^{-5}}5

\implies\boxed{c=-\dfrac15\ln\dfrac{1-e^{-5}}5}

7 0
4 years ago
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