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vichka [17]
3 years ago
9

Simplify -4/5 divided by 3/-2

Mathematics
2 answers:
mixer [17]3 years ago
8 0

Answer:

\frac{8}{15}

Step-by-step explanation:

mojhsa [17]3 years ago
7 0

Answer:6/36

Step-by-step explanation:

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Calculate s5 for the sequence defined by
pantera1 [17]
ANSWER

S_5= 42.5

EXPLANATION

The general term of the arithmetic sequence is

a_n=1+\frac{5}{2}n

For the first term,we substitute n=1 to get,

a_1=1+\frac{5}{2}\times 1

a_1=1+2.5=3.5

Similarly

a_5=1+\frac{5}{2}\times 5

a_5=13.5

The sum of the first n-terms is given by the formula,

S_n=\frac{n}{2}(a_1+l)

where l is the last term.
In this case

l=a_5=13.5

We substitute these values to get,

S_5= \frac{5}{2} (3.5 + 13.5)

S_5= \frac{5}{2} (17)
S_5=42.5



We could have also used the formula,

S_n=\frac{n}{2}(2a_1+(n-1)d)

S_5=\frac{5}{2}(2(3.5)+(5-1)2.5)

S_5=42.5
3 0
4 years ago
Read 2 more answers
35 is 28% of what number? Please explain how you got it.
iren2701 [21]

Answer:

125

Step-by-step explanation:

we have

x*.28= 35

we divide both sides by .28 and get

x= 125

and 28% of 125 = 35

Hope it helps !

5 0
3 years ago
Read 2 more answers
"Verify the identity of (sinx cosx)^2/sinx cosx = 2 + secx cscx"
nata0808 [166]

(sinx + cosx)^2/((sinx)(cosx)) = 2 + (secx)(cscx) 
<span>(sinx + cosx)^2/((sinx)(cosx)) = 2 + 1/(sinxcosx); subtract 1/sinxcosx both sides </span>
<span>(sinx + cosx)^2/((sinx)(cosx)) - 1/(sinxcosx)= 2; multiply through by sinxcosx </span>
<span>(sinx + cosx)^2 -1 = 2(sinxcosx) </span>
<span>sin^2 + 2sinxcosx + cos^2 - 1 = 2(sinxcosx); since sin^x + cos^2x = 1 </span>
<span>1 + 2sinxcosx -1 = 2sinxcosx </span>
<span>2sinxcosx = 2sinxcosx</span>
5 0
4 years ago
Read 2 more answers
The sum of two numbers is 54 and difference is 10. what are the numbers?
Firdavs [7]
x+y=54\\&#10;\underline{x-y=10}\\&#10;2x=64\\&#10;x=32\\\\&#10;32+y=54\\&#10;y=22&#10;
7 0
4 years ago
Write your answer in simplest form. 1 over 6 - 1 over 8
Alex
1/6 - 1/8

Find the least common denominator, which is 24.

4/24 - 3/24 = 1/24 
5 0
4 years ago
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