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2 years ago

Can somebody PLEASE HELP ME??? trying to correct answers and i have no clue how to do this.

Mathematics

1 answer:

jek_recluse [69]2 years ago

4 0

Answer:

613 inches

Step-by-step explanation:

Area right triangle = 10,710

Formula for area of a triangle = 1/2(base x height)

We need to find the base because height and area is given

Reverse divide by 2 by doing multiply by 2

10,710 x 2 = 21420

Revers multiply by 612 by dividing by 612

21420 / 612 = 35

Base = 35

Formula for hypotenuse = a² + b² = c²

In this case a = 35, b = 612, and c = hypotenuse

35² + 612² = c²

1225 + 374544 = c²

375769 = c²

Square root of 375769 = 613

Hypotenuse = 613

If my answer is incorrect, pls correct me!

If you like my answer and explanation, mark me as brainliest!

-Chetan K

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2 years ago

Harry buys 3 granola bars for $5. Sidnee buys

juin [17]

Answer:

The constant of proportionality is 5/3

Step-by-step explanation:

3 * k = 5

k = 5/3

Check

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3 0

2 years ago

Read 2 more answers

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 $\vec b$ = $\frac{1}{5} \hat i+\frac{3}{5} \hat j$ .