Solve each system of equations by elimination.
y= -x+1 y=4x-14



By ELIMINATION

Answers

Answer 1
The answer is:
x=3, y= -2
(3,-2)
Answer 2

Answer:

X=3, y=-2

Step-by-step explanation:

Solve Each System Of Equations By Elimination. Y= -x+1 Y=4x-14By ELIMINATION

Related Questions

Write numbers to make each line have the same sum

Answers

Answer:

There is nothing here?

Step-by-step explanation:

The price of a 7 ​-minute phone call is ​$1.75. What is the price of a 14 ​-minute phone​ call?

Answers

Answer:

$3.50

Step-by-step explanation:

14 is double 7

So the price should be double

1.75 x 2 = 3.5

(Assuming there's no base fee and the charge is purely for minutes)

This is finding exact values of sin theta/2 and tan theta/2. I’m really confused and now don’t have a clue on how to do this, please help

Answers

First,

tan(θ) = sin(θ) / cos(θ)

and given that 90° < θ < 180°, meaning θ lies in the second quadrant, we know that cos(θ) < 0. (We also then know the sign of sin(θ), but that won't be important.)

Dividing each part of the inequality by 2 tells us that 45° < θ/2 < 90°, so the half-angle falls in the first quadrant, which means both cos(θ/2) > 0 and sin(θ/2) > 0.

Now recall the half-angle identities,

cos²(θ/2) = (1 + cos(θ)) / 2

sin²(θ/2) = (1 - cos(θ)) / 2

and taking the positive square roots, we have

cos(θ/2) = √[(1 + cos(θ)) / 2]

sin(θ/2) = √[(1 - cos(θ)) / 2]

Then

tan(θ/2) = sin(θ/2) / cos(θ/2) = √[(1 - cos(θ)) / (1 + cos(θ))]

Notice how we don't need sin(θ) ?

Now, recall the Pythagorean identity:

cos²(θ) + sin²(θ) = 1

Dividing both sides by cos²(θ) gives

1 + tan²(θ) = 1/cos²(θ)

We know cos(θ) is negative, so solve for cos²(θ) and take the negative square root.

cos²(θ) = 1/(1 + tan²(θ))

cos(θ) = - 1/√[1 + tan²(θ)]

Plug in tan(θ) = - 12/5 and solve for cos(θ) :

cos(θ) = - 1/√[1 + (-12/5)²] = - 5/13

Finally, solve for sin(θ/2) and tan(θ/2) :

sin(θ/2) = √[(1 - (- 5/13)) / 2] = 3/√(13)

tan(θ/2) = √[(1 - (- 5/13)) / (1 + (- 5/13))] = 3/2

Watch help video
In an all boys school, the heights of the student body are normally distributed with a
mean of 69 inches and a standard deviation of 2.5 inches. Using the empirical rule,
what percentage of the boys are between 61.5 and 76.5 inches tall?

Answers

Answer:

99.7%

Step-by-step explanation:

99.7% of boys fall between 61.5 and 76.5

The percentage of the boys that are between 61.5 and 76.5 inches tall is 99.73%

How to determine the percentage between the range?

The given parameters are:

Mean = 69Standard deviation = 2.5

Start by calculating the z score for x = 61.5 and 76.5 using:

[tex]z = \frac{x - \mu}{\sigma}[/tex]

So, we have:

[tex]z_1 = \frac{61.5 - 69}{2.5} = -3[/tex]

[tex]z_2 = \frac{76.5 - 69}{2.5} = 3[/tex]

The percentage is then represented as:

Percentage = P(-3 < x < 3)

Using the z table of probabilities, we have:

Percentage = 0.9973


Express as percentage

Percentage = 99.73%

Hence, 99.73% of the boys are between 61.5 and 76.5 inches tall

Read more about normal distribution at:

https://brainly.com/question/4079902

#SPJ9

3r + 6y - 2z = -6 2x + y + 4z = 19 -5x - 2y+8z = 62
solve each system of equations​

Answers

For the first equation
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