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RideAnS [48]
3 years ago
10

EASY POINTS

Mathematics
2 answers:
lisov135 [29]3 years ago
8 0

Answer:

1 check

2 check

3 no check

4 no check

5 check

Step-by-step explanation:

weeeeeb [17]3 years ago
4 0

Answer:

1 is true

2 is true

3 is true

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Help please I NEED TO PASS :(((((
adoni [48]

Answer:

The answer should be C

Step-by-step explanation:

I just got done calculating it and according to my calculations it should be C

5 0
3 years ago
Pleaseeee helpppp !!!!!!!! Will mark Brianliest !!!!!!!!!!!!!
Goryan [66]

Answer:

I think you're supposed to add all the fives together. That is what MY teacher taught me to do :/

Step-by-step explanation:

The shape is a square that is tilted and looks like a rhombus. Just add the 5's together.

4 0
3 years ago
HOW DO YOU CONVERT SURFACE AREA TO VOLUME! <br> HELP!!!!<br> WILL MARK BRAINLIST!!
slamgirl [31]

Answer:

Formulas for a rectangular prism:

Volume of Rectangular Prism: V = lwh.

Surface Area of Rectangular Prism: S = 2(lw + lh + wh)

Space Diagonal of Rectangular Prism: (similar to the distance between 2 points) d = √(l2 + w2 + h2)

Step-by-step explanation:

Hope it helps :))))

3 0
3 years ago
If: AB= 2x + 6, AC = 51, and BC=6x +5, Find BC.
Marina86 [1]

Answer:

<h3>Assuming A, B, and C are collinear and B is between A and C then BC = 35 </h3>

Step-by-step explanation:

Assuming A, B, and C are collinear and B is beetween A and C

AB + BC  = AC

2x + 6 + 6x + 5 = 51

 8x + 11 = 51

    8x = 40

     x = 5

BC = 6x+5 = 6×5 + 5 = 35

3 0
3 years ago
For what values of b are the vectors [−18, b, 9] and [b, b2, b] orthogonal? (Enter your answers as a comma-separated list. If an
bearhunter [10]

Answer:

Therefore the given vectors are orthogonal for b = 0,±3.

Step-by-step explanation:

If  \vec a and  \vec b are two vectors orthogonal, then the dot product of \vec a and \vec b will be zero.

i.e \vec a. \vec b =0

If  \vec a = x_1\hat i+y_1\hat j +z_1\hat k  and \vec b = x_2\hat i+y_2\hat j +z_2\hat k

\vec a. \vec b=( x_1\hat i+y_1\hat j +z_1\hat k).( x_2\hat i+y_2\hat j +z_2\hat k)

     =x_1 x_2+y_1y_2+z_1z_2

Given two vectors are (-18,b,9) and (b,b²,b)

Let

\vec P= -18 \hat i+b\hat j +9 \hat k

and

\vec Q = b \hat i+b^2 \hat j +b\hat k

Therefore,

\vec P.\vec Q

=( -18 \hat i+b\hat j +9 \hat k).( b \hat i+b^2 \hat j +b\hat k)

=(-18).b+b.b²+9.b

= -18b+b³+9b

= b³-9b

Since \vec P and \vec Q are orthogonal. Then \vec P.\vec Q = 0.

Therefore,

b³-9b= 0

⇒b(b²-9)=0

⇒b =0 or b²=9

⇒b=0 or b =±3

Therefore the given vectors are orthogonal for b = 0,±3.

8 0
3 years ago
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