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Readme [11.4K]
3 years ago
13

Using a calculator.. I need help please .

Mathematics
1 answer:
marin [14]3 years ago
6 0

Answer:

Just input sin(9), sin(30), sin(81), etc.

For this set, you need a scientific calculator or else, you won't be able to answer that. There is a way but that is already obsolete now.

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G = cst (solve for s)
Harrizon [31]

Step-by-step explanation:

Simplifying

g = cst

Solving

g = cst

Solving for variable 'g'.

Move all terms containing g to the left, all other terms to the right.

Simplifying

g = cst

7 0
3 years ago
A geometric sequence is a sequence of numbers where the next term equals to
exis [7]

Answer:

C

Step-by-step explanation:

Let the first term be a and the common ratio be r.

ATQ, ar^4=24 and ar^6=144, r=sqrt(6) and a=24/(sqrt(6))^2=24/36=2/3

4 0
3 years ago
What is:<br> 1/4 (q-12) + 1/4 (q-4)<br> simplified??
Alexandra [31]

Answer: -b - 9

Step-by-step explanation:

First, since the a cancel eachother out we are left with (4-b) + (-6-7)

Second, you want to add the -6 and -7 to get -13, so (4-b) - 13

Finally, add the 4 to the negative 13 to get (-b - 9)

Hope this helps.

4 0
3 years ago
find the sum of a 9-term geometric sequence when the first term is 4 and the last term is 1,024 and select the correct answer be
olasank [31]
Hello,

u_{1} =4\\&#10; u_{2} =4*q\\&#10; u_{3} =4*q^2\\&#10;...\\&#10;&#10; u_{9} =4*q^8\\\\&#10;==\textgreater\ 4*q^8=1024\\&#10;==\textgreater\ q^8=256\\&#10;==\textgreater\ q=2\mbox{( and there is a problem in the question) } \ or \ q=-2\\&#10;if\ q=2 \ then\  \sum_{i=0}^{8}\ 4*(2)^i= 4*\dfrac{1-2^9}{1-2} =2044\\&#10;&#10;if \ q=-2 \ then\ \sum_{i=0}^{8}\ 4*(-2)^i= 4*\dfrac{1-(-2)^9}{1+2} =684\\&#10;&#10;&#10;&#10;

Answer B (but see the problem)
7 0
3 years ago
Read 2 more answers
Angle 1 and Angle 2 form a linear pair. If the measure of angle is 113°, find the measure of angle 2.​
LuckyWell [14K]

Answer:

<2 =67

Step-by-step explanation:

A linear pair adds to 180 degrees

<1 + <2 =180

113+ <2 = 180

<2 = 180-113

<2 =67

5 0
3 years ago
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