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vlada-n [284]
3 years ago
13

Rashida deposits $2,000 into an account that earns 5% annually.

Mathematics
1 answer:
vitfil [10]3 years ago
7 0

Answer:

16 years because she still too young

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What number is between 0.40 and 0.47
san4es73 [151]
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3 years ago
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5х + y = 16<br> х+ 6у = 38
Ainat [17]

Answer:

(2,6)

Step-by-step explanation:

Solve by substitution. Solve the second equation for x:

x+6y=38\\\\x+6y-6y=38-6y\\\\x=38-6y

Insert the value of x into the first equation:

5(38-6y)+y=16

Solve for y. Simplify multiplication using the distributive property:

5(38)+5(-6y)+y=16\\\\190-30y+y=16\\\\190-29y=16

Subtract 190 from both sides:

190-190-29y=16-190\\\\-29y=-174

Isolate the variable to find its value. Divide both sides by -29:

\frac{-29y}{-29}=\frac{-174}{-29}  \\\\y=6

Now take the value of y and insert back into either equation to solve for x:

5x+6=16

Subtract 6 from both sides:

5x+6-6=16-6\\\\5x=10

Divide both sides by 5:

\frac{5x}{5}=\frac{10}{5}\\\\  x=2

Therefore, the solution to the system is (2,6).

:Done

8 0
3 years ago
Find y when x=4 and z=15; if y varies jointly as x and z y=5 when z=8 and x=10
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I think the answer to y when x=4 and z=15; if y varies jointly as x and z y=5 when z=8 and x=10 is B. Y=1/2. Y•15; 12/4.
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3 years ago
Use the equation and type the ordered-pairs. y=3^x
aleksandrvk [35]

The ordered pairs of the equation y = 3^x are (0,1), (1,3) and (2,9)

<h3>How to type the ordered pairs?</h3>

The equation is given as:

y = 3^x

Let x = 0, 1 and 2

y = 3^0 = 1

y = 3^1 = 3

y = 3^2 = 9

So, we have the following ordered pairs (0,1), (1,3) and (2,9)

Hence, the ordered pairs of the equation y = 3^x are (0,1), (1,3) and (2,9)

Read more about exponential functions at:

brainly.com/question/2456547

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How do you write 309,099,990 in expanded form
kolbaska11 [484]
Three hundred million, ninety nine thousand, nine hundred ninety <span />
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