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cupoosta [38]
2 years ago
7

Last year, Chau had $10,000 to invest. He invested some of it in an account that paid

Mathematics
1 answer:
viva [34]2 years ago
3 0

Answer:

Account with 8% interest per year: $3000

Account with 5% interest per year: $7000

8% of 3000 = 240

5% of 7000 = 350

240+ 350 = 590

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Which pair of expressions are equivalent?
Zielflug [23.3K]

Answer:

second one

Step-by-step explanation:

Because when you try to remove the brackets then you should multiply coefficient of x and coefficient of y by 2

So it will be 2x+18y

2*9=18

4 0
2 years ago
Adrian has a points card for a movie theater.
ANEK [815]

Answer: v ≥ 6

This means that Adrian needs to do at least 6 visits.

Step-by-step explanation:

First, we know that he gets 20 points just for signing up, so he starts with 20 points.

Now, if he makes v visits, knowing that he gets 2.5 points per visit, he will have a total of:

20 + 2.5*v

points.

And he needs to get at least 35 points, then the total number of points must be such that:

points ≥ 35

and we know that:

points = 20 + 2.5*v

then we have the inequality:

20 + 2.5*v  ≥ 35

Now we can solve this for v, so we need to isolate v in one side of the equation:

2.5*v ≥ 35 - 20 = 15

2.5*v ≥ 15

v ≥ 15/2.5 = 6

v ≥ 6

So he needs to make at least 6 visits.

6 0
2 years ago
517 37/50 + 312 3/100
fgiga [73]

Answer: 829 3/4 or 829 75/100

Step-by-step explanation:

6 0
2 years ago
(x+3)^2=4
GalinKa [24]

Answer:

x = -1, -5

Step-by-step explanation:

{(x + 3)}^{2}  = 4 \\  =  >  {x}^{2}  + 2 \times x \times 3 +  {3}^{2}  = 4 \\  =  >  {x}^{2}  + 6x + 9 = 4 \\  =  >  {x}^{2}  + 6x = 4 - 9 \\  =  >  {x}^{2}  + 6x =  - 5 \\  =  >  {x}^{2}  + 6x  + 5 = 0 \\  =  >  {x}^{2}  + 5x + x + 5 = 0 \\  =  > x(x + 5) + 1(x + 5) = 0 \\  =  > (x + 1)(x + 5) \\  =  > x =  - 1 \:  \: or \:  \:  - 5

Hope it helps.

Don't forget to click the "Brainliest" button if you like my answer.

6 0
2 years ago
Read 2 more answers
Please prove this........​
Crazy boy [7]

Answer:  see proof below

<u>Step-by-step explanation:</u>

Given: A + B + C = π    →     C = π - (A + B)

                                    → sin C = sin(π - (A + B))       cos C = sin(π - (A + B))

                                    → sin C = sin (A + B)              cos C = - cos(A + B)

Use the following Sum to Product Identity:

sin A + sin B = 2 cos[(A + B)/2] · sin [(A - B)/2]

cos A + cos B = 2 cos[(A + B)/2] · cos [(A - B)/2]

Use the following Double Angle Identity:

sin 2A = 2 sin A · cos A

<u>Proof LHS → RHS</u>

LHS:                        (sin 2A + sin 2B) + sin 2C

\text{Sum to Product:}\qquad 2\sin\bigg(\dfrac{2A+2B}{2}\bigg)\cdot \cos \bigg(\dfrac{2A - 2B}{2}\bigg)-\sin 2C

\text{Double Angle:}\qquad 2\sin\bigg(\dfrac{2A+2B}{2}\bigg)\cdot \cos \bigg(\dfrac{2A - 2B}{2}\bigg)-2\sin C\cdot \cos C

\text{Simplify:}\qquad \qquad 2\sin (A + B)\cdot \cos (A - B)-2\sin C\cdot \cos C

\text{Given:}\qquad \qquad \quad 2\sin C\cdot \cos (A - B)+2\sin C\cdot \cos (A+B)

\text{Factor:}\qquad \qquad \qquad 2\sin C\cdot [\cos (A-B)+\cos (A+B)]

\text{Sum to Product:}\qquad 2\sin C\cdot 2\cos A\cdot \cos B

\text{Simplify:}\qquad \qquad 4\cos A\cdot \cos B \cdot \sin C

LHS = RHS: 4 cos A · cos B · sin C = 4 cos A · cos B · sin C    \checkmark

7 0
3 years ago
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