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3 years ago

How to find area of a triangle

Mathematics

2 answers:

aliya0001 [1]3 years ago

8 0

Answer:

Base × Height ÷ 2

Otrada [13]3 years ago

5 0

Answer:

Hope this helps! :)

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Find the surface area<br> Will give brainliest to first answer!!!!

user100 [1]

120 I think I’m not sure

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2 years ago

Find the work done when a crane lifts a -pound boulder through a vertical distance of feet. Round to the nearest foot-pound.

goldenfox [79]

Answer:

72000ft-1b

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3 years ago

How do I do this problem <br> -6(x+1)-18=26+4

pogonyaev

Answer:

x=-9

Step-by-step explanation:

Step 1: Simplify both sides of the equation.

−6(x+1)−18=26+4

(−6)(x)+(−6)(1)+−18=26+4(Distribute)

−6x+−6+−18=26+4

(−6x)+(−6+−18)=(26+4)(Combine Like Terms)

−6x+−24=30

−6x−24=30

Step 2: Add 24 to both sides.

−6x−24+24=30+24

−6x=54

Step 3: Divide both sides by -6.

−6x

−6

x=−9

Answer:

x=−9

5 0

3 years ago

ILL GIVE EVERYTHING IF U GET IT RIGHTTT

Veronika [31]

Answer:

Step-by-step explanation:

Because if you did the math it's A

I'm 100% Sure

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

6 0

3 years ago

How to find area of a triangle ​

How to find area of a triangle