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tatyana61 [14]
3 years ago
15

√√√√√√√√√√√√√√√√√√√√√√

Mathematics
2 answers:
kondor19780726 [428]3 years ago
7 0
Stan txt (tomorrow by together) 4th gen kings and leaders and BETTER stream freeze on YT Hybe labels.

iogann1982 [59]3 years ago
3 0

Answer:

0

Step-by-step explanation:

This question is impossible to answer

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One student ate 3/20 of all candies and another 1.2 lb. The second student ate 3/5 of the candies and the remaining 0.3 lb. What
Roman55 [17]
The students ate 6 lbs of candies.
8 0
4 years ago
In ΔWXY, the measure of ∠Y=90°, XW = 85, WY = 84, and YX = 13. What is the value of the cosine of ∠W to the nearest hundredth?
IgorLugansk [536]

Answer: 0.99

Step-by-step explanation:

SOH CAH TOA

CAH: cosine = adjacent/hypotenuse

Cosine of angle w = 84/85 = 0.98823 ≈ 0.99

Angle Y is 90 degrees, meaning it's a right triangle. Therefore angle W is the top/tip of a triangle, while X is the angle next to Y. If you plug in the information afterwards and follow the trigonometry rule (SOH CAH TOA) you will find your answer.

4 0
4 years ago
A mathematical sentence which contains an inequality symbol and one variable raised to the first power is called a _ _ _ _ _ _ _
Stella [2.4K]

A mathematical sentence which contains an inequality symbol and one variable raised to the first power is called a linear inequality.

5 0
3 years ago
Read 2 more answers
Identify the exponential function as a growth or decay function.
marusya05 [52]

Answer:

A

Step-by-step explanation:

A. growth

7 0
3 years ago
Find a power series representation for the function. (Give your power series representation centered at x = 0.)f(x) = x3x2 + 1f(
makkiz [27]

I suppose you mean

f(x)=\dfrac{x^3}{x^2+1}

Recall that for |x|, we have

\dfrac1{1-x}=\displaystyle\sum_{n=0}^\infty x^n

Then

\dfrac1{1+x^2}=\dfrac1{1-(-x^2)}=\displaystyle\sum_{n=0}^\infty(-x^2)^n=\sum_{n=0}^\infty(-1)^nx^{2n}

which is valid for |-x^2|=|x|^2, or more simply |x|.

Finally,

f(x)=\displaystyle\frac{x^3}{x^2+1}=\sum_{n=0}^\infty(-1)^nx^{2n+3}

8 0
3 years ago
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