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Alexxx [7]
3 years ago
10

Jessie had -$68.25 in her bank account. She then spent $72.15 more dollars to

Mathematics
1 answer:
tatiyna3 years ago
5 0

Answer:

-140.4

Explanation:

-68.25 -72.15 = -140.4

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PLEASE HELP!!!
Pavel [41]
The answer would be sas or conguernt angle
7 0
3 years ago
Help on this question ASAP PLEASE!!!
serious [3.7K]
X=4 is the answer if you are looking for x

4 0
3 years ago
12
Slav-nsk [51]

Triangles ABC and DEF may or may not be congruent, because the angle-angle-angle (AAA) postulate is not a criterion for congruency of any two triangles. This is because the angle-angle-angle postulate only incorporates the angles of the triangles, and the side lengths are not specified, so congruency may or may not be true for these triangles.

3 0
2 years ago
How to differentiate y=x^n using the first principle. In this question, I cannot use the rule of differentiation. I have to do t
Zarrin [17]

By first principles, the derivative is

\displaystyle\lim_{h\to0}\frac{(x+h)^n-x^n}h

Use the binomial theorem to expand the numerator:

(x+h)^n=\displaystyle\sum_{i=0}^n\binom nix^{n-i}h^i=\binom n0x^n+\binom n1x^{n-1}h+\cdots+\binom nnh^n

(x+h)^n=x^n+nx^{n-1}h+\dfrac{n(n-1)}2x^{n-2}h^2+\cdots+nxh^{n-1}+h^n

where

\dbinom nk=\dfrac{n!}{k!(n-k)!}

The first term is eliminated, and the limit is

\displaystyle\lim_{h\to0}\frac{nx^{n-1}h+\dfrac{n(n-1)}2x^{n-2}h^2+\cdots+nxh^{n-1}+h^n}h

A power of h in every term of the numerator cancels with h in the denominator:

\displaystyle\lim_{h\to0}\left(nx^{n-1}+\dfrac{n(n-1)}2x^{n-2}h+\cdots+nxh^{n-2}+h^{n-1}\right)

Finally, each term containing h approaches 0 as h\to0, and the derivative is

y=x^n\implies y'=nx^{n-1}

4 0
3 years ago
4<br> For each nth term, find the 50th term.<br> a<br> 4n - 5
alukav5142 [94]
4 x 50 = 200
200 - 5 = 195
195 is the answer hope this helps
5 0
3 years ago
Read 2 more answers
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