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kvasek [131]
3 years ago
12

QUICKLY PLEASE!! 18POINTS! 6(3p – 2) = 24

Mathematics
1 answer:
Gekata [30.6K]3 years ago
6 0

Answer:

p= 2

Step-by-step explanation:

So first you want to distribute the 6 to 3p-2 which basically means multiply

6(3p – 2) = 24

6 times 3p is 18 and 6 time -2 is -12

* remember a negative and a positive equal a negative*

18p-12=24

you want to get 18p by itself, so in order to do that you can add 12 to both sides because if you add 12 on the left side it cancels out -12 and whatever you do to one side you have to do it to the other so add 12 to 24 which is 36

18p = 36

since you want to get p by itself you divide p on both sides so it can cancel out 18 and 36/18 is 2

so p = 2

if you are wondering why i divide it and not subtract or add it is because 18 and p are stuck together so you have to divide but if it  was like 18 + p you can subtract because 18 and p arent together

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3 years ago
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Answer:

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The first step is to multiply 6 and .43 to get how much it was for all of the apples.

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What is the best approximation of the projection of (5,-1) onto (2,6)?
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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.

                                                               

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Step-by-step explanation:

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