All categories

4 years ago

X + 2y = 10

Mathematics

1 answer:

weqwewe [10]4 years ago

3 0

Let's multiply the first equation by -2 so that we can eliminate y.

-2x-4y=-20

3x+4y=8

X=-12

Since X is -12, let's plug it in to the equation to check.

-12+2y=10

2y=22; let's divide

y=11

So, D (-12,11)

You might be interested in

The population of a small industrial town was 12 910 in 2000. Each year, the population

balandron [24]

The population in the year 2020 is 4628

<h3>How to determine the population?</h3>

The given parameters are:

Initial, a = 12910

Rate, r = 5%

Since the population decreases, then we make use of an exponential decay function.

This is represented as:

f(n) = a * (1 - r)^n

So, we have:

f(n) = 12910 * (1 - 5%)^n

Evaluate the difference

f(n) = 12910 * 0.95^n

2020 is 20 years from 2000.

So, we have:

f(20) = 12910 * 0.95^20

Evaluate

f(20) = 4628

Hence, the population in the year 2020 is 4628

Read more about exponential functions at:

brainly.com/question/14355665

#SPJ1

6 0

2 years ago

When you solve a problem involving money,what can a negative answer represents

AveGali [126]

Maybe the money you owe someone

4 0

3 years ago

Lacie wants to paint her bedroom walls. She has to cover 1,162.5 square feet. About how many gallons of paint will she need?

Lubov Fominskaja [6]

I dont really know. you dont know how big the walls are and how far 1 gallon goes so figure how far 1 gallon goes and then divide it by the 1,162.5

5 0

3 years ago

Statistics show that a certain soccer player has a 63% chance of missing the goal each time he shoots. If this player shoots twi

Kipish [7]

Answer:

<em>The probability of scoring two goals in both times is</em><em> 0.137 or 13.7%</em>

Step-by-step explanation:

Statistics show that a certain soccer player has a 63% chance of missing the goal each time he shoots.

So $P(A)=0.63$

Hence, the probability of getting success in his shoots will be,

$P(A^c)=1-P(A)=1-0.63=0.37$

The probability of scoring two goals in both times is,

$=P(A^c)\cdot P(A^c)$

$=0.37\times 0.37$

$=0.137$

$=13.7\%$

7 0

3 years ago

Read 2 more answers

A certain large manufacturing facility produces 20,000 parts each week. The manager of the facility estimates that about 1% of t

Slav-nsk [51]

Answer:

198 is the variance for the number of defective parts made each week.

Step-by-step explanation:

We are given the following in the question:

Number of parts produced each week = 20,000

Percentage of defective parts = 1%

We have to calculate the variance for the number of defective parts made each week.

We treat defective part as a success.

P(Defective part) = 1% = 0.01

$p = 0.01$

Then the number defective parts follows a binomial distribution .

Formula for variance =

$\sigma^2 = np(1-p) = 20000(0.01)(1-0.01) = 198$

Thus, 198 is the variance for the number of defective parts made each week.

8 0

4 years ago