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VARVARA [1.3K]
3 years ago
13

Https://www.hmhco.com/one/assessment/#/formativeLive Assessment/

Mathematics
1 answer:
GuDViN [60]3 years ago
3 0

A. equation:  y = -3x - 2

B. y-intercept:​ -2

<em><u>Solution:</u></em>

Given that,

<em><u>Write the equation 12x = - 4y - 8 in slope-intercept form</u></em>

The slope intercept form is given as:

y = mx + c

Where,

m is the slope and "c" is the y intercept

From given

12x = -4y - 8\\\\Rewrite\\\\-4y-8 = 12x\\\\Move\ the\ constant\ to\ right\ side\\\\-4y = 12x + 8\\\\y = \frac{12x}{-4} + \frac{8}{-4}\\\\y = -3x -2

Thus the slope intercept form is found

Comparing with y = mx + c ,

c = -2

Thus the y intercept is -2

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Answer:

angle B = 78.46°

Step-by-step explanation:

Use the cosine ratio

cosB=\frac{2}{10} =0.2

B=cos^{-1} (0.2)=78.46^{0}

Hope this helps

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1 year ago
5. The combined perimeter of a circle and a square is 16. Find the dimensions of the circle and square
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2 years ago
Gina has 72$ in $1 bills, 5$ bills, and 20$ bills. If she has three times as many 1$ bills as she has 5$ bills, and only half as
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6 0
3 years ago
What is the best approximation of the projection of (5,-1) onto (2,6)?
Hatshy [7]

Answer:

Hence, the scalar projection of \vec a onto \vec b= \frac{\sqrt{10} }{5}, and  the vector projection of \vec a onto \vec b = \frac{1}{5} \hat i+\frac{3}{5} \hat j.

Step-by-step explanation:

We have given two points  (5, -1) and (2, 6).

Let,     \vec a=5\hat {i}-\hat {j}  and  \vec b= 2\hat {i}+6\hat{j} .

and we have calculate the projection of \vec a onto \vec b.

Now,

For the calculation of projection, first we need to calculate the dot product of  \vec a  and \vec b.

\vec a.\vec b=(5\hat {i}-\hat{j}).(2\hat{i}+6\hat{j})

     =10-6

     =4

then, we have to calculate the magnitude of \vec b.

   \mid {\vec {b}}\mid = \sqrt{2^{2}+6^{2}  } = \sqrt{40} = 2\sqrt{10}.

Now, the scalar projection of \vec a onto \vec b = \frac{\vec a.\vec b}{\mid b\mid}

                                                                 = \frac{4}{2\sqrt{10} }\frac{2}{\sqrt{10} } \times\frac{\sqrt{10} }{\sqrt{10} } =\frac{2\sqrt{10} }{10} = \frac{\sqrt{10} }{5}

and the vector projection of \vec a onto \vec b = \frac{\vec a. \vec b}{\mid\vec b \mid^{2} } . \vec b

                                                               = \frac{4}{40} . (2\hat i+ 6\hat j)

                                                                = \frac{1}{5} \hat i+\frac{3}{5} \hat j

Hence, the scalar projection of \vec a onto \vec b= \frac{\sqrt{10} }{5}, and  the vector projection of \vec a onto \vec b = \frac{1}{5} \hat i+\frac{3}{5} \hat j.

                                                               

6 0
3 years ago
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