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2 years ago

What is the solution of the system of equations graphed below?

Mathematics

1 answer:

Dahasolnce [82]2 years ago

6 0

Answer: B

Step-by-step explanation: Look at where the lines intersect

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If angle symbol1 and angle symbol2 are complementary angles and if the measure of angle symbol1 is 5454degrees° more than the me

Misha Larkins [42]

<span>Complementary angles by definition sum up to 90 degrees.
Let x = measure of angle2.
It's given that angle1 = x + 54

Since complementary, we know (x+54)+x = 90
2x+54=90
2x=36
x=18.
So angle2 measures 18 degrees and angle1 is 18 degrees+54 degrees, or 72 degrees. 72+18 = 90 as expected for complementary angles.</span>

7 0

3 years ago

Please tell me the steps to this problem. i will give brainliest

gayaneshka [121]

Option C: -3 is the average rate of change between $x=-1$ and $x=1$

Explanation:

The formula to determine the average rate of change is given by

Average rate of change = $\frac{y_2-y_1}{x_2-x_1}$

We need to find the average rate of change between $x=-1$ and $x=1$

Thus, from the table, we have,

$x_1=-1$ , $y_1=5$

$x_2=1$ , $y_2=-1$

Thus, substituting these values in the formula, we get,

Average rate of change = $\frac{-1-5}{1+1}$

$=\frac{-6}{2}$

$=-3$

Thus, the average rate of change between $x=-1$ and $x=1$ is -3.

Hence, Option C is the correct answer.

6 0

2 years ago

Stuck in this problem help please

marta [7]

Use trigonometry.

tan30° = perpendicular / base

Here for angle 30° , x is perpendicular while 10 is the base.

tan30° = 1/√3
1/√3 = x/10
x = 10/✓3

By rationalising
x = 10√3/3

Hope This Helps You!

3 0

3 years ago

Suppose a simple random sample of size nequals36 is obtained from a population with mu equals 74 and sigma equals 6. (a) Descri

OlgaM077 [116]

Part a)

The simple random sample of size n=36 is obtained from a population with

$\mu = 74$

and

$\sigma = 6$

The sampling distribution of the sample means has a mean that is equal to mean of the population the sample has been drawn from.

Therefore the sampling distribution has a mean of

$\mu = 74$

The standard error of the means becomes the standard deviation of the sampling distribution.

$\sigma_ { \bar X } = \frac{ \sigma}{ \sqrt{n} } \\ \sigma_ { \bar X } = \frac{ 6}{ \sqrt{36} } = 1$

Part b) We want to find

$P(\bar X \:>\:75.9)$

We need to convert to z-score.

$P(\bar X \:>\:75.9) = P(z \:>\: \frac{75.9 - 74}{1} ) \\ = P(z \:>\: \frac{75.9 - 74}{1} ) \\ = P(z \:>\: 1.9) \\ = 0.0287$

Part c)

We want to find

$P(\bar X \: < \:71.95)$

We convert to z-score and use the normal distribution table to find the corresponding area.

$P(\bar X \: < \:71.95) = P(z \: < \: \frac{71.9 5- 74}{1} ) \\ = P(z \: < \: \frac{71.9 5- 74}{1} ) \\ = P(z \: < \: - 2.05) \\ = 0.0202$

Part d)

We want to find :

$P(73\:$

We convert to z-scores and again use the standard normal distribution table.

$P( \frac{73 - 74}{1} \:< \: z$

5 0

2 years ago

Explain how the energy of a toy car is transformed as it slides down a ramp. Give evidence that the energy of the car remains th

yaroslaw [1]

<span>At the top of a ramp, all of the energy of the car is potential: that is, it has yet to be moved but has the ability to move. As the car is sliding down the ramp, the potential energy is being transformed into kinetic energy, which reaches a maximum at halfway down the ramp. After this point, the kinetic energy begins to lessen and the potential energy begins to increase, until the car comes to a full stop, at which point kinetic energy is at zero and the potential energy is again at a maximum.</span>

5 0

2 years ago

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What is the solution of the system of equations graphed below￼?

What is the solution of the system of equations graphed below?