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2 years ago

PLEASE HELP ME WITH THIS QUESTION ^^

Mathematics

1 answer:

AURORKA [14]2 years ago

4 0

I’m sorry I don’t get that whole situation

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Can someone help me with this question

Inessa [10]

Answer:

i cant see the pic please send a better one and i reloaded it and it still does not work

Step-by-step explanation:

4 0

3 years ago

Solve the system of equations given . -2x+5y=-3

algol13

For this case we have the following system of equations:

$-2x + 5y = -3\\y-15 = 3x$

We multiply the second equation by -5:

$-5y + 75 = -15x$

Now we add the equations:

$-2x-5y + 5y + 75 = -3-15x\\-2x + 75 = -3-15x\\-2x + 15x = -75-3\\13x = -78\\x = \frac {-78} {13}\\x = -6$

We find the value of the variable "y":

$y = 3x + 15\\y = 3 (-6) +15\\y = -18 + 15\\y = -3$

THE solution is: (-6, -3)

Answer:

(-6, -3)

3 0

3 years ago

1. Mariana took 50 minutes to complete her math assignment. She took 2 minutes to solve each problem and 10 minutes to check her

Anarel [89]

Answer:

Step-by-step explanation:

It's the only thing not listed aka the number of math problems.

6 0

2 years ago

In the following answer, why does 830 convert to 83000?

Mekhanik [1.2K]

Then the money borrowed insurance company is $1,000.

Money borrowed from insurance company = $1000

Let the money borrowed by her friend at 8% interest = x

Then the money borrowed by the bank at 9% interest = 2x

<h3>What is the total money?</h3>

Total money borrowed = $10,000

So, the money borrowed from the insurance company = 10,000-(2x+x)

= 10,000-3x

Total interest for the first year = $830

We have the equation,

8%(x) + 9%(2x) +5% (10,000-3x) = 830

8x+18x+50,000-15x = 83,000

11x = 83000-50000

11x = 33,000

x = $3000

Then the money borrowed insurance company = 10,000-3x

Then the money borrowed insurance company is $1,000.

To learn more about the insurance visit:

brainly.com/question/25855858

#SPJ1

4 0

2 years ago

Find the oth term of the geometric sequence 5,--25, 125,

Genrish500 [490]

Given the geometric progression below

$5,-25,125,\ldots$

The nth term of a geometric progression is given below

$T_n=ar^{n-1},\begin{cases}a=\text{first term} \\ r=\text{common ratio}\end{cases}$

From the geometric progression, we can deduce the following

$\begin{gathered} T_1=a=5 \\ T_2=ar=-25 \\ T_3=ar^2=125 \end{gathered}$

To find the value of r, we will take ratios of two consecutive terms

$\begin{gathered} \frac{T_2}{T_1}=\frac{ar}{a}=\frac{-25}{5} \\ \Rightarrow r=-5 \end{gathered}$

To find the 9th term of the geometric, we will have that;

$\begin{gathered} T_9=ar^8=5\times(-5)^8=5\times390625 \\ =1953125 \end{gathered}$

Hence, the 9th term of the geometric progression is 1953125

8 0

11 months ago