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Katena32 [7]
4 years ago
7

How do i solve 0.97 on a decimal number line?

Mathematics
1 answer:
e-lub [12.9K]4 years ago
5 0
It's E because it is closer to 1 than the others and look at all the little lines and think of those as the tenths place
Please Mark Brainliest!!!!
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Answer Please!!!!!!!!!!!!!!
Veseljchak [2.6K]
3/10 or 0.3 because when you solve it step by dividing radical of 9 from 100 and then simplifying it you end up with this answer.
5 0
4 years ago
Circle the "best buy."<br> 4 for $1.50<br> 35 cents each<br> 6 for $2.00
denis-greek [22]
Depends on how much you are purchasing
4 0
3 years ago
Does anyone know how to do this ??? <br> -view attachment
cricket20 [7]

As soon as I read this, the words "law of cosines" popped
into my head.  I don't have a good intuitive feeling for the
law of cosines, but I went and looked it up (you probably
could have done that), and I found that it's exactly what
you need for this problem.

The "law of cosines" relates the lengths of the sides of any
triangle to the cosine of one of its angles ... just what we need,
since we know all the sides, and we want to find one of the angles.

To find angle-B, the law of cosines says

       b² = a² + c² - 2 a c cosine(B)

B  =  angle-B
b  =  the side opposite angle-B = 1.4
a, c = the other 2 sides = 1 and 1.9

                  (1.4)² = (1)² + (1.9)² - (2 x 1 x 1.9) cos(B)

                 1.96  =  (1) + (3.61)  -  (3.8) cos(B)

Add  3.8 cos(B)  from each side:

                 1.96 + 3.8 cos(B) = 4.61

Subtract  1.96  from each side:

                             3.8 cos(B) =  2.65

Divide each side by  3.8 :

                                  cos(B)  =  0.69737  (rounded)

Whipping out the
trusty calculator:
                                 B  = the angle whose cosine is 0.69737

                                      =  45.784° .

Now, for the first time, I'll take a deep breath, then hold it
while I look back at the question and see whether this is
anywhere near one of the choices ...

By gosh !  Choice 'B' is  45.8° !                    yay !
I'll bet that's it !

8 0
4 years ago
(2a^x + b^x)^2 i need help ASAP
Finger [1]

Answer:

4a^{2x} + 4a^{x}b^{x} + b^{2x}

Step-by-step explanation:

Using the rule of exponents

a^{m} × a^{n} ⇔ a^{(m+n)}

Given

(2a^{x} + b^{x} ) = (2a^{x} + b^{x} )(2a^{x} + b^{x} )

Each term in the second factor is multiplied by each term in the first factor, that is

2a^{x} (2a^{x} + b^{x} ) + b^{x} (2a^{x} + b^{x} ) ← distribute both parenthesis

= 4a^{2x} +2 a^{x}b^{x} + 2a^{x}b^{x} + b^{2x} ← collect like terms

= 4a^{2x} + 4a^{x}b^{x} + b^{2x}

6 0
3 years ago
What dose equivalent mean
Licemer1 [7]

EQUI = equals

VALent, as in value related


EQUI + VALent, same value

4 0
4 years ago
Read 2 more answers
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