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2 years ago

Please please please help me

Mathematics

1 answer:

Gwar [14]2 years ago

8 0

A) 15/4 = 4/4+4/4+4/4+3/4= 3 3/4
B)9/2= 2/2+2/2+2/2+2/2+1/2= 4 1/2
c = 26/12 = 12/12 + 12/12 + 2/12 = 2 2/12
d) 20/6 = 6/6 + 6/6 + 6/6 + 2/6 = 3 2/6

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If hexagon JGHFKI Is congruent to hexagon SRTQUV, which pair of angles must be congruent?

Dima020 [189]

Hey there!

If these hexagons are congruent, then each angle is congruent to the one that is in the same position as it is. In this case, I and V are congruent because they are both last. K and U are congruent because they are both second to last.

While looking over the answer options, you should be able to see that Angle J and Angle S is the only answer with angles that match up, because they are both the first angle in the set.

I hope this helps!

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2 years ago

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1 year ago

Which is greater 9000cm or 2 km?

satela [25.4K]

Convert 2 km into cm

1 km = 100,000 cm

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6 0

2 years ago

Joe receives an average of 780 emails in his personal account and 760 emails in his work account each month. After changing his

Semmy [17]

Answer:

702 emails

Step-by-step explanation:

<h2>This problem bothers on depreciation of value, in this context it is Joe's email that has depreciated by 10%.</h2>

Given data

Average personal emails received monthly = $780 emails$

Average work emails received monthly $= 760 emails$

We are required to solve for the new amount of emails Joe will be receiving after changing his address, to find this value we need to solve for the depreciation of his personal mails.

After solving for the depreciation , we then need to subtract the depreciation from the initial number of mails to get the new number of mails.

let us solve for 10% depreciation.

$depreciation= \frac{10}{100} *780\\depreciation=0.1*780= 78 emails$

The new number of mails

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Joe will be receiving an average of 702 emails in his personal account monthly

7 0

2 years ago

For the binomial expansion of (x + y)^10, the value of k in the term 210x 6y k is a) 6 b) 4 c) 5 d) 7

Sloan [31]

Answer:

a) 6

Step-by-step explanation:

Expanding the polynomial using the formula:

$(x+y)^n=\sum_{k=0}^n \binom{n}{k} x^{n-k} y^k$

Also

$\binom{n}{k}=\frac{n!}{(n-k)!k!}$

I think you mean $210x^6y^4$

We can deduce that this term will be located somewhere in the middle. So I will calculate $k= 5; k=6 \text{ and } k =7$ .

For $k=5$

$\binom{10}{5} (y)^{10-5} (x)^{5}=\frac{10!}{(10-5)! 5!}(y)^{5} (x)^{5}= \frac{10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5! }{5! \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 } \\ =\frac{30240}{120} =252 x^{5} y^{5}$

Note that we actually don't need to do all this process. There's no necessity to calculate the binomial, just $x^{n-k} y^k$

For $k=6$

$\binom{10}{6} \left(y\right)^{10-6} \left(x\right)^{6}=\frac{10!}{(10-6)! 6!}\left(y\right)^{4} \left(x\right)^{6}=210 x^{6} y^{4}$

5 0

2 years ago