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Degger [83]
3 years ago
14

ASNWER THIS AND ILL HUG YOU

Mathematics
1 answer:
alexgriva [62]3 years ago
8 0

She earns 345$ for the whole 3 years.

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Somebody please help meee!!!!
Stels [109]

Answer:

12 m³

Step-by-step explanation:

4 x 2 x 1.5 = x

4 x 2 = 8

1.5 x 8 = 12

5 0
3 years ago
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Plz answer thank you
RoseWind [281]
Jack would need 44 feet(B). This question is just asking for the perimeter, which would be all the sides added up together.
4 0
3 years ago
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What is 1500 biweekly
Damm [24]

Answer:

$1500 income every 2 weeks, I think.

Step-by-step explanation:

6 0
2 years ago
What is the coefficient of the x5y5-term in the binomial expansion of (2x – 3y)10? 10C5(2)5(3)5 10C5(2)5(–3)5 –10C5(2)5(–3)5 10C
butalik [34]

ANSWER

10C_5(2)^{5}( - 3)^5

EXPLANATION

The given binomial expansion is:

{(2x - 3y)}^{10}

Compare this to

{(a + b)}^{n}

we have a=2x , b=-3y and n=10

We want to find the coefficient of the term

{x}^{5}  {y}^{5}

This implies that, r=5.

The terms in the expansion can be obtained using

T_{r+1}=nC_ra^{n-r}b^r

We substitute the given values to obtain;

T_{5+1}=10C_5(2x)^{10-5}(  - 3y)^5

T_{6}=10C_5(2x)^{5}(3y)^5

T_{6}=10C_5(2)^{5}( - 3)^5 {x}^{5}  {y}^{5}

Hence the coefficient is;

10C_5(2)^{5}( - 3)^5

5 0
3 years ago
Read 2 more answers
Invest $6,300 in two different accounts the first account paid 11% the second account paid 6% in interest at the end of the year
MAVERICK [17]

Answer: he invested $5300 at 11% and $1000 at 6%

Step-by-step explanation:

Let x represent the amount which he invested in the first account paying 11% interest.

Let y represent the amount which he invested in the second account paying 6% interest.

He Invest $6,300 in two different accounts the first account paid 11% the second account paid 6% in interest. This means that

x + y = 6300

The formula for determining simple interest is expressed as

I = PRT/100

Considering the first account paying 11% interest,

P = $x

T = 1 year

R = 11℅

I = (x × 11 × 1)/100 = 0.11x

Considering the second account paying 6% interest,

P = $y

T = 1 year

R = 6℅

I = (y × 6 × 1)/100 = 0.06y

At the end of the year, he had earned $643 in interest , it means that

0.11x + 0.06y = 643 - - - - - - - - - -1

Substituting x = 6300 - y into equation 1, it becomes

0.11(6300 - y) + 0.06y = 643

693 - 0.11y + 0.06y = 643

- 0.11y + 0.06y = 643 - 693

- 0.05y = - 50

y = - 50/ - 0.05

y = 1000

x = 6300 - y = 6300 - 1000

x = 5300

7 0
2 years ago
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