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Nadya [2.5K]
3 years ago
15

The answer for this problem

Mathematics
1 answer:
olya-2409 [2.1K]3 years ago
4 0
It would be 12 dollars in US money
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A television that cost $400 is on sale for 20% off do you have a coupon for 20% off the sale price right equation for the end of
vlabodo [156]

Answer:

Step-by-step explanation:

y = 0.8(400)

7 0
3 years ago
13/6 renamed as a mixed number
kozerog [31]

Answer:

2 1/6

Step-by-step explanation:

13/6 is 6/6 + 6/6 + 1/6... each 6/6 = 1 sooo. 2 and 1/6 :)

orrr

6*2 is 12 (closest multiple of 6 to 13, so 2 is whole number w/ 1(/6) as a remainder

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3 years ago
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Help me please asap..<br><br><br><br>​
olga_2 [115]

Answer:

Solution given:

Sin angle is given by opposite side to the hypotenuse

so

Sin E =opposite/hypotenuse=4/5

<u>option</u><u> </u><u>d</u><u>.</u><u>4</u><u>/</u><u>5</u>

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The coordinates of the school is now (5,7).
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First time on Brainly! Please Help!!!!! &gt;^&lt;
lys-0071 [83]
Welcome to Brainly!

Binomial Theorem will follow this pattern:
The powers start at 3 and decrease down to 0 for the first term,
and start at 0 and increase to 3 for the other term.

(2x+4)^3 =

\rm \_\_(2x)^34^0+\_\_(2x)^24^1+\_\_(2x)^14^2+\_\_(2x)^04^3

See how the power counts down on the (2x)
and counts up on the 4?
That's the pattern that our expansion must follow.

I left a little space in front of each term.
The coefficient in front of each term will come from the fourth row of Pascal's Triangle. The one that looks like this:

1 3 3 1

Those are the coefficients we want:

\rm 1(2x)^34^0+3(2x)^24^1+3(2x)^14^2+1(2x)^04^3

We can clean this up a little bit by getting rid of some of the junk. Anything to the 0th power is 1. So let's suppress all of our 1's because multiplying by 1 is not important.

\rm (2x)^3+3(2x)^24+3(2x)4^2+4^3

Now apply exponent rule, distributing the power to both the 2 and the x where applicable.

\rm 2^3x^3+3\cdot2^2x^24+3\cdot2x4^2+4^3

Remember, multiplication is COMMUTATIVE, meaning we can multiply things in any order. So let's bring the numerical portion to the front of each term and multiply it all out.

\rm 2^3x^3+3\cdot2^2\cdot4x^2+3\cdot2\cdot4^2x+4^3
\rm 8x^3+48x^2+96x+64
6 0
3 years ago
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