All categories

2 years ago

Y=2x and x+3y=-14 substitution to solve each system of equations

1 answer:

7 0

You can use the substitution method.
You substitute the first equation to the second equation to get the value of x.
x + 3y = -14
x + 3(2x) = -14
x + 6x = -14
7x = -14
x = -14/7 = -2

Substitute the value of x to the first equation.
y = 2(-2) = 4

You might be interested in

Sin2A+sin2B+sin2C÷4cosA/2.cosB/2.cosC/2

Mrrafil [7]

The answer is c2 have a good day !!!! Bc you add all the numbers up and you will get it

3 0

2 years ago

1. On which road would there be a larger standard deviation in speed: a backroad or an interstate? Justify your answer

White raven [17]

Answer:

Your answers may vary slightly. 5.2 Normal Distributions: Finding Probabilities If you are given that a random variable Xhas a normal distribution, nding probabilities corresponds to nding the area between the standard normal curve and the x-axis, using the table of z-scores. The mean (expected value) and standard deviation ˙should be given

Step-by-step explanation:

7 0

3 years ago

Let x1 = 12, y1 = 15, and y2 = 3. Let y vary inversely with x. Find x2.

vladimir1956 [14]

Answer:

x2 = 60

Step-by-step explanation:

If the variables x and y are inversely proportional, the product x * y is a constant.

So using x1 and y1 we can find the value of this constant:

$x1 * y1 = k$

$12 * 15 = k$

$k = 180$

Now, we can use the same constant to find x2:

$x2 * y2 = k$

$x2 * 3 = 180$

$x2 = 180 / 3 = 60$

So the value of x2 is 60.

3 0

2 years ago

2+2+4= 8 x 12=<br> .........

Sveta_85 [38]

Answer:

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

How to solve this :') please help

blsea [12.9K]

Answer:

3/2

Step-by-step explanation:

sin(3π/4 - β) = sin(3π/4)cosβ - cos(3π/4)sinπ =

$\sin(\frac{3\pi}{4}-\beta)=\sin(\frac{3\pi}{4})\cos\beta-\cos\frac{3\pi}{4}\sin\beta\\=\frac{1}{\sqrt{2}}\cos\beta-(-\frac{1}{\sqrt{2}})\sin\beta\\\\=\frac{1}{\sqrt{2}}(\cos\beta +\sin\beta)\\\\\sin^2(\frac{3\pi}{4}-\beta)=\frac{1}{2}\cos\beta +\sin\beta)^2=\frac{1}{2}(\cos^2\beta +\sin^2\beta+2\sin\beta\cos\beta\\=\frac{1}{2}(1+\sin2\beta)=\frac{1}{2}(1-\frac{1}{5}) = \frac{2}{5}\\$

Use $\cot^2x=\csc^2x-1=\frac{1}{\sin^2x}-1$

$\cot^2(\frac{3\pi}{4}-\beta)=\frac{1}{\sin^2(\frac{3\pi}{4}-\beta)}-1 = \frac{1}{\frac{2}{5}}-1=\frac{5}{2}-1=\frac{3}{2}$

7 0

2 years ago