How do I do this? I’m not very good at geometry.

Mathematics

1 answer:

garri49 [273]2 years ago

6 0

You need to study more

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Express as a single power of 2

den301095 [7]

When you multiply powers, they add, and when you divide them they subtract.
So I would first add the 2^a + 2^b +2^c to get 2^(a+b+c)
Then, divide by 2^(a+b). Because when you divide powers they subtract, you will be taking away the (a+b) from the (a+b+c) and you will be left with c on its own.
The answer is 2^c

6 0

3 years ago

What is the solution for −40 = 10c? <br> A. c = 1 <br> B. C = -1 <br> C. c = -4 <br> D. c = 4

Aleks [24]

Answer:

C. c = -4

Step-by-step explanation:

you must get c by itself. you have to divide both sides by 10. this gets rid of the 10c and keeps both sides equal

-40/10 = c

c = -4

5 0

2 years ago

David bought a poster for an art project. The poster is 2.7 ft wide and 3.9 ft tall. What is the area of the poster? Enter your

algol13

10.53 ft is the Area. :D:D

7 0

2 years ago

Why 1 hundred 4 tabs and 14 tens name the same amount

horrorfan [7]

1414 I believe is the answer

5 0

2 years ago

Find the sum of the first four terms of the geometric sequence shown below. 4, 4/ 3, 4/9, ...

user100 [1]

It's evident that the first four terms are 4, 4/3, 4/9, and 4/27. So the fourth partial sum of the series is

$S_4=4+\dfrac43+\dfrac49+\dfrac4{27}$

It's as easy as adding up the fractions, but I bet this is supposed to be an exercise in taking advantage of the fact that the series is geometric and use the well-known formula for computing such a sum.

Multiply the sum by 1/3 and you have

$\dfrac13S_4=\dfrac43+\dfrac49+\dfrac4{27}+\dfrac4{81}$

Now subtracting this from $S_4$ gives

$S_4-\dfrac13S_4=4-\dfrac4{81}$

That is, all the matching terms will cancel. Now solving for $S_4$ , you
have

$\dfrac23S_4=4\left(1-\dfrac1{81}\right)$
$S_4=6\left(1-\dfrac1{81}\right)$
$S_4=\dfrac{480}{81}=\dfrac{160}{27}$

3 0

3 years ago