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3 years ago

Solve using substitution. x + 2y = 4 and –4x − 7y = –9

Mathematics

2 answers:

xxMikexx [17]3 years ago

4 0

Answer:

X=-10 y=7

Step-by-step explanation:

1) in first equation subtract 2y on both sides

X=4-2y

Substitute as "x" in the second equation:

-4(4-2y)-7y=-9

-16+8y-7y=-9

-16+y=-9

Y=7

Substitute y as 7 to get x)

X=4-2(7)

X=4-14

X=-10

Hope this helps

RSB [31]3 years ago

3 0

Step-by-step explanation:

Given

x + 2y = 4

x = 4 - 2y.......... equation 1

-4x -7y = - 9.......equation 2

Putting equation 1 in equation 2 

- 4 ( 4 - 2y) - 7y = - 9

-16 + 8y - 7y = -9

y = - 9 + 16

y = 7

Now 

x = 4 - 2* 7 = 8 - 7 = 1 

hope this helps. 

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Find the area. Show your work.

Maksim231197 [3]

9*10(90) - 7*3(21) = 69

5 0

3 years ago

Find the value of c that makes x^2–6x + c a perfect square trinomial and write as a binomial squared.

denpristay [2]

Answer: 9

Step-by-step explanation:

Given

The function is $x^2-6x+c$

To make it a perfect square, compare it with $(a-b)^2=a^2+b^2-2ab$

Here,

$x=a,\\-6x=-2ab\\\\\therefore b=3$

So, c must be square of b i.e. 9

$x^2-6x+c\Rightarrow x^2-6x+9\\\Rightarrow (x-3)^2$

Therefore, value of c is 9.

3 0

3 years ago

The line width used for semiconductor manufacturing is assumed to be normally distributed with a mean of 0.5 micrometer and a st

Alinara [238K]

Answer:

There is a 0.82% probability that a line width is greater than 0.62 micrometer.

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by

$Z = \frac{X - \mu}{\sigma}$

After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X. The sum of the probabilities is decimal 1. So 1-pvalue is the probability that the value of the measure is larger than X.

In this problem

The line width used for semiconductor manufacturing is assumed to be normally distributed with a mean of 0.5 micrometer and a standard deviation of 0.05 micrometer, so $\mu = 0.5, \sigma = 0.05$ .