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3 years ago

Write the equation of a line that has a slope of 2 and goes through (3,-4).

1 answer:

8 0

First, start with the basic line equation:
y = mx + b
Then, fill in what you know.
-4 = 2(3) + b
Solve for b
-4 = 6 + b
-6 -6
-10 = b
Now you have your equation:
y = 2x - 10

I hope this helps :)

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Find the coordinates of the midpoint of the segment given its endpoints F(2,-6) and G(-8,5)

Bess [88]

Step-by-step explanation:

see the pic fof the answer

8 0

3 years ago

Solve for XX. Assume XX is a 2×22×2 matrix and II denotes the 2×22×2 identity matrix. Do not use decimal numbers in your answer.

sveticcg [70]

The question is incomplete. The complete question is as follows:

Solve for X. Assume X is a 2x2 matrix and I denotes the 2x2 identity matrix. Do not use decimal numbers in your answer. If there are fractions, leave them unevaluated.

$\left[\begin{array}{cc}2&8\\-6&-9\end{array}\right]$ · X· $\left[\begin{array}{ccc}9&-3\\7&-6\end{array}\right]$ =I.

First, we have to identify the matrix I. As it was said, the matrix is the identiy matrix, which means

I = $\left[\begin{array}{ccc}1&0\\0&1\end{array}\right]$

So, $\left[\begin{array}{cc}2&8\\-6&-9\end{array}\right]$ · X· $\left[\begin{array}{ccc}9&-3\\7&-6\end{array}\right]$ = $\left[\begin{array}{ccc}1&0\\0&1\end{array}\right]$

Isolating the X, we have

X· $\left[\begin{array}{ccc}9&-3\\7&-6\end{array}\right]$ = $\left[\begin{array}{cc}2&8\\-6&-9\end{array}\right]$ - $\left[\begin{array}{ccc}1&0\\0&1\end{array}\right]$

Resolving:

X· $\left[\begin{array}{ccc}9&-3\\7&-6\end{array}\right]$ = $\left[\begin{array}{ccc}2-1&8-0\\-6-0&-9-1\end{array}\right]$

X· $\left[\begin{array}{ccc}9&-3\\7&-6\end{array}\right]$ = $\left[\begin{array}{ccc}1&8\\-6&-10\end{array}\right]$

Now, we have a problem similar to A.X=B. To solve it and because we don't divide matrices, we do X=A⁻¹·B. In this case,

X= $\left[\begin{array}{ccc}9&-3\\7&-6\end{array}\right]$ ⁻¹· $\left[\begin{array}{ccc}1&8\\-6&-10\end{array}\right]$

Now, a matrix with index -1 is called Inverse Matrix and is calculated as: A . A⁻¹ = I.

So,

$\left[\begin{array}{ccc}9&-3\\7&-6\end{array}\right]$ · $\left[\begin{array}{ccc}a&b\\c&d\end{array}\right]$ = $\left[\begin{array}{ccc}1&0\\0&1\end{array}\right]$

9a - 3b = 1

7a - 6b = 0

9c - 3d = 0

7c - 6d = 1

Resolving these equations, we have a= $\frac{2}{11}$ ; b= $\frac{7}{33}$ ; c= $\frac{-1}{11}$ and d= $\frac{-3}{11}$ . Substituting:

X= $\left[\begin{array}{ccc}\frac{2}{11} &\frac{-1}{11} \\\frac{7}{33}&\frac{-3}{11} \end{array}\right]$ · $\left[\begin{array}{ccc}1&8\\-6&-10\end{array}\right]$

Multiplying the matrices, we have

X= $\left[\begin{array}{ccc}\frac{8}{11} &\frac{26}{11} \\\frac{39}{11}&\frac{198}{11} \end{array}\right]$

6 0

3 years ago

Which of these ordered pairs is a solution for this linear inequality 3y-2x<=6? A. (-4,2) B. (1,3) C. (2,-4) D. (2,4)

Usimov [2.4K]

Answer:

Step-by-step explanation:

3y-2x<=6

3(-4) - 2(2)<=6

-12-4<=6

-16<=6

4 0

3 years ago

A person who weighs 100 pounds on Earth weighs 16.6 lb. on the moon.

mario62 [17]

Answer:

See explanation below.

Step-by-step explanation:

Given: 100 lbs on Earth is 16.6 lbs on the moon.

a. The independent variable is weight. The gravity of the Moon and the gravity of the Earth are constant. Weight can change, but gravity is a constant.

b. An equation that relates the weight of someone on the Moon who travels to the Earth:

100 / 16.6 = 6.02. Take the Moon weight and multiply by 6.02:

Moon Weight * 6.02 = Earth Weight.

Proof:

16.6 * 6.024 = 99.99 - approximately 100 lbs Earth weight.

c. A 185 lb astronaut on Earth would weigh:

16.6 / 100 = .166. Take the Earth weight and multiply by .166:

185 * .166 = 30 lbs on the Moon.

d. A person who weighs 50 lbs on the Moon:

50 * 6.024 = 301.2 lbs on Earth.

Hope this helps! Have an Awesome Day! :-)

6 0

3 years ago

Helen plays basketball. For free throws, she makes the shot 78% of the time. Helen must now attempt two free throws. C = the eve

riadik2000 [5.3K]

Answer:

0.6708 or 67.08%

Step-by-step explanation:

Helen can only make both free throws if she makes the first. The probability that she makes the first free throw is P(C) = 0.78, now given that she has already made the first one, the probability that she makes the second is P(D|C) = 0.86. Therefore, the probability of Helen making both free throws is:

$P(C+D) = P(C) *P(D|C) = 0.78*0.86\\P(C+D) = 0.6708$

There is a 0.6708 probability that Helen makes both free throws.

5 0

3 years ago