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3 years ago

10

10) What is the solution to the system of equations shown below?

1 answer:

kkurt [141]3 years ago

5 0

Answer:

B

X= 7.5

Y= -5

Step-by-step explanation:

In systems of equations, the goal is to add both of the equations together so it cancels one of the variables and you solve for that variable. Before you add they together, you multiply one of the equations by a negative number so when you add them together one of the numbers with x or y cancel.

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45 is 0.9% of what number

RUDIKE [14]

Answer:

0.405

Step-by-step explanation

45 x 0.9%= 0.405

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3 years ago

Read 2 more answers

Find the GCF of 40 and 60 listing factors for each number in order from least to greatest.

Cloud [144]

40:1,2,4,5,8,10,20,40
60:1,2,3,4,5,6,10,12,15,20,30,60
The GCF between 40 and 60 is 20

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4 years ago

9 divided by 4 and 4/5

liraira [26]

Answer:

$\frac{45}{24} = 1.875$

Step-by-step explanation:

$\frac{9}{4+\frac{4}{5}} = \frac{9}{\frac{24}{5}} = \frac{45}{24} = 1.875$

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3 years ago

a table was sold at 20%loss but had it sold for RS 160 more these would have been a profit 20%. find cp

jekas [21]

Answer:

Rs 400

Step-by-step explanation:

Let us assume that the cost price of the article = x

Selling price = S.P

Cost price = C.P

Loss = %Loss × C.P(percentage loss × Cost Price)

S.P = C.P - Loss

In the question, Loss = 20%

S.P = x - 20x/100

Find the Lowest Common Multiple

S.P= 100x - 20x/100

S.P = 80x/100.........Equation 1

In the question, we are told if the article was sold for RS 160 more, it would have made a profit of 20%

Profit = %profit × C.P(percentage profit × Cost Price)

S.P = C.P + Profit

In the question, profit = 20%

S.P = x + 20x/100

Find the Lowest Common Multiple

S.P= 100x + 20x/100

S.P = 120x/100.........Equation 2

Hence,

Equation 2 - Equation 1 = Rs 160

= 120x/100 - 80x/100 = 160

= 40x/100 = 160

Cross multiply

40x = 100 × 160

40x = 16000

x = 16000/40

x = Rs 400

Since x = Cost price, the cost price of the table = Rs 400

4 0

3 years ago

Which ordered pair is a solution to the system of equations y=x+3 and y=x^2-x

Elenna [48]

For this case we have the following system of equations:

$y = x + 3\\y = x ^ 2-x$

Equating the equations we have:

$x ^ 2-x = x + 3\\x ^ 2-x-x-3 = 0\\x ^ 2-2x-3 = 0$

To find the solutions we factor the equation. To do this, we look for two numbers that, when multiplied, result in -3 and when added, result in -2.

These numbers are: -3 and +1

$-3 + 1 = -2\\-3 * (+ 1) = - 3$

Thus, the factored equation is:

$(x-3) (x + 1) = 0$

Then, the solutions for the variable "x" are:

$x_ {1} = 3\\x_ {2} = - 1$

We find the solutions for the variable "y":

$y_ {1} = x_ {1} + 3 = 3 + 3 = 6\\y_ {2} = x_ {2} + 3 = -1 + 3 = 2$

Thus, the solutions are:

$(x_ {1}, y_ {1}): (3,6)\\(x_ {2}, y_ {2}): (-1,2)$

ANswer:

$(x_ {1}, y_ {1}): (3,6)\\(x_ {2}, y_ {2}): (-1,2)$

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4 years ago