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harkovskaia [24]
3 years ago
9

13/5 as a decimal 2 7/10 as a decimal

Mathematics
2 answers:
irakobra [83]3 years ago
5 0

Answer:

2.6 and 2.7

Step-by-step explanation:

I think

Marta_Voda [28]3 years ago
5 0

Answer:

13/5 is 2.6 and 2 7/10 is 2.7

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PLEASE HELPP
Snezhnost [94]
-6\cdot q=24p
q=-4

24\cdot (-4) =?
-96=?

Let's check it:
-96\cdot (-4)=384
384=384
L=R

The answer: C

<em>:)</em>
3 0
3 years ago
Can someone tell me what to write for part b ?!!!!
Sergeu [11.5K]
No, because if it was 2 gal. per min., then the first coordinates would be (1,2)
7 0
4 years ago
Read 2 more answers
.Find (g o f)(x) for f(x) = 2x + 4 and g(x) = x^2 . Show all of your work, it'll help me for my next one ! (: thanks
adelina 88 [10]
You should know that <span>(g o f)(x) is the same as g(f(x)), or g of f(x). To solve for g(f(x)), substitute whatever f(x) equals as x into g(x).

So, if you have the two functions
g(x) = x^2
f(x) = 2x + 4, 
you substitute what f(x) equals as x into the g(x) equation.
</span>g(x) = x^2 \\ &#10;g(f(x)) = (f(x))^2 \\ &#10;g(f(x)) = (2x + 4)^2 \\ &#10;g(f(x)) = 4x^2 + 16x + 16<span>

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5 0
4 years ago
Step-by-step explanation: Is (2, 3) a solution to the equation y=x ?
Diano4ka-milaya [45]

Answer:

no

Step-by-step explanation:

if you substitute it:

3 = 2

but 3 is not equal to 2 so i dont think its a solution

5 0
3 years ago
Read 2 more answers
In a sequence of ten terms, each term (starting with the third term) is equal to the sum of the two previous terms. The seventh
poizon [28]

Answer:

\boxed{a_1+a_2+a_3+a_4+a_5+a_6+a_7+a_8+a_9+a_{10}=66}

Step-by-step explanation:

We know that in a sequence of ten terms, each term (starting with the third term) is equal to the sum of the two previous terms. The seventh term is equal to 6. Therefore, we have

a_1+a_2=x\\a_3=x\\a_4=2x\\a_5=3x\\a_6=5x\\a_7=8x\\

We know that:

a_7=6\\\implies 8x=6\\x=\frac{3}{4}

We get:

a_1+a_2=x=\frac{3}{4}\\\\a_3=x=\frac{3}{4}\\\\a_4=2x=\frac{3}{2}\\\\a_5=3x=\frac{9}{4}\\\\a_6=5x=\frac{15}{4}\\\\a_7=8x=6\\\\a_8=13x=\frac{39}{4}\\\\a_9=21x=\frac{63}{4}\\\\a_{10}=34x=\frac{102}{4}\\

Therefore, we calculate

\boxed{a_1+a_2+a_3+a_4+a_5+a_6+a_7+a_8+a_9+a_{10}=66}

4 0
3 years ago
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