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VashaNatasha [74]
3 years ago
10

Help and explain Don’t use for points or will be reported and I’ll take it back

Mathematics
1 answer:
Fittoniya [83]3 years ago
4 0

Answer:

86+34+x=180

x=180-120

x=60

Hope it helps

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Needed help! Answer when possible.
Lera25 [3.4K]

Answer:

The word "solution" means an action or process of solving a problem

99 is a solution to 1/9x = 11 because  multiplying both side of the equation by 9 makes x = 99

3 0
3 years ago
For , find the value of x for which .
Mice21 [21]
Uh what did you mean to upload a picture or somthing
8 0
3 years ago
Helppppp pls im stuck on thi one problem :(
Ray Of Light [21]

First, find the simplified ratio of the given ratio:

Number of pirates: 22

Number of ships: 2

Divide 22 with 2 to find the ratio: 22/2 = 11

For every ship there is, there will be 11 pirates.

Now, find the ratio for the other questions:

IF there are 5 ships: 5 x 11 = 55

If there are 5 ships, there will be 55 pirates.

If there are 11 ships: 11 x 11 = 121

If there are 11 ships, there will be 121 pirates.

~

5 0
3 years ago
Read 2 more answers
The answer is C that’s why it has a 1 pt in front <br><br> Plz just help me explain how to get that
Ierofanga [76]

First lets get the formula for a triangular prism

v =( \frac{1}{2} bh)h

First lets plug in our numbers

v =( \frac{1}{2} \times 7 \times 4)8.5

It really doesnt matter how you multiply everything, but you end up with 119cubic feet.

3 0
3 years ago
How to solve an expression with variables in the exponents?
gtnhenbr [62]

Answer:

  use logarithms

Step-by-step explanation:

Taking the logarithm of an expression with a variable in the exponent makes the exponent become a coefficient of the logarithm of the base.

__

You will note that this approach works well enough for ...

  a^(x+3) = b^(x-6) . . . . . . . . . . . variables in the exponents

  (x+3)log(a) = (x-6)log(b) . . . . . a linear equation after taking logs

but doesn't do anything to help you solve ...

  x +3 = b^(x -6)

There is no algebraic way to solve equations that are a mix of polynomial and exponential functions.

__

Some functions have been defined to help in certain situations. For example, the "product log" function (or its inverse) can be used to solve a certain class of equations with variables in the exponent. However, these functions and their use are not normally studied in algebra courses.

In any event, I find a graphing calculator to be an extremely useful tool for solving exponential equations.

8 0
3 years ago
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