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2 years ago

Determine if the following system of equations has no solutions ......................

Mathematics

1 answer:

topjm [15]2 years ago

8 0

Answer:

No solutions

Step-by-step explanation:

3x+4y=-4

15x+20y=-22

——————————

-5(3x+4y)=-5(-4) multiply 1st equation by -5 so you can eliminate the variables

-15x-20y=20. Modified 1st equation

15x+20y=-22. 2nd equation

0+0=-2. Added two equations together

since 0 does not equal -2, there is no solution.

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Secθ =__________________ 1/sinθ +-√cos^2 +1 +-√tan^2 +1

Zielflug [23.3K]

Answer:

The answer is last option

As we know 1 + tan^2 = sec^2

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3 years ago

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is (3,-3) a solution to the system 5x+4y=3, 2x-5y=19 need to be taught step by step please and thank you.

8090 [49]

Basically all you’re doing is plugging in the inputs to the equations and see if they’re correct.

5(3)+4(-3)=3 this proves to be true
2(3)-5(-3)=21 not 19

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3 years ago

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How many solutions are in this inequality? -2x+7<-41

geniusboy [140]

Answer:

There is only one solution. Hoped I helped.

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2 years ago

Assume that adults have IQ scores that are normally distributed with a mean of mu equals 105 and a standard deviation sigma equa

kondor19780726 [428]

Answer: 0.6827

Step-by-step explanation:

Given : Mean IQ score : $\mu=105$

Standard deviation : $\sigma=15$

We assume that adults have IQ scores that are normally distributed .

Let x be the random variable that represents the IQ score of adults .

z-score : $z=\dfrac{x-\mu}{\sigma}$

For x= 90

$z=\dfrac{90-105}{15}\approx-1$

For x= 120

$z=\dfrac{120-105}{15}\approx1$

By using the standard normal distribution table , we have

The p-value : $P(90$

$P(z$

Hence, the probability that a randomly selected adult has an IQ between 90 and 120 =0.6827

5 0

3 years ago

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Question 9 - (3 marks available)

finlep [7]

Answer: a) 1,600

b) 35,000

c) 120,000

Step-by-step explanation:

First, round each number to one significant digit (the first digit).

Then multiply them.

a) 17 --> rounds to 20

82 --> rounds to 80

20 x 80 = 2 x 8 with two zeros

= 1 6 0 0

b) 54 --> rounds to 50

671 --> rounds to 700

50 x 700 = 5 x 7 with three zeros

= 3 5 0 0 0

c) 207 --> rounds to 200

643 --> rounds to 600

200 x 600 = 2 x 6 with four zeros

= 1 2 0 0 0 0

6 0

3 years ago