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3 years ago

How to find surface area

Mathematics

2 answers:

puteri [66]3 years ago

6 0

Answer:

The way to do this is multiplying all the 3 dimensional sides Then adding them.

Step-by-step explanation:

What i mean by this is by, Multiplying the Length and height together .

Next is Multiplying the Width and Height together

Next would to multiply the Length And Width together.

Lastly would to add all your solutions together.

If you have any questions feel free to ask in the comments.

rjkz [21]3 years ago

5 0

Answer:

Find the area of two sides (Length*Height)*2 sides.

Find the area of adjacent sides (Width*Height)*2 sides.

Find the area of ends (Length*Width)*2 ends.

Add the three areas together to find the surface area.

Example: The surface area of a rectangular prism 5 cm long, 3 cm.

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Which point is an x-intercept of the quadratic equation (x-8)(x+9)

defon

Solve for x in each set of parentheses by setting them to 0:

X -8 = 0

X = 8

X+9 = 0

X =-9

You have two x intercepts (8,0) and (-9,0)

You didn’t list any choices to pick from but it would be either if the ones above.

5 0

3 years ago

Read 2 more answers

Please help If you know the answer!!

prisoha [69]

Answer:

-3 +8a

Step-by-step explanation:

(3-8a) *(-1)

Distribute

3*-1 -8a*-1

-3 +8a

4 0

3 years ago

I fixed the question need help

STatiana [176]

Answer:

Step-by-step explanation:

90-33=57

57-1=56

56/2=28

8 0

3 years ago

let x1,x2, and x3 be linearly independent vectors in R^(n) and let y1=x2+x1; y2=x3+x2; y3=x3+x1. are y1,y2,and y3 linearly indep

Nutka1998 [239]

Answer with Step-by-step explanation:

We are given that

$x_1,x_2$ and $x_3$ are linearly independent.

By definition of linear independent there exits three scalar $a_1,a_2$ and $a_3$ such that

$a_1x_1+a_2x_2+a_3x_3=0$

Where $a_1=a_2=a_3=0$

$y_1=x_2+x_1,y_2=x_3+x_2,y_3=x_3+x_1$

We have to prove that $y_1,y_2$ and $y_3$ are linearly independent.

Let $b_1,b_2$ and $b_3$ such that

$b_1y_1+b_2y_2+b_3y_3=0$

$b_1(x_2+x_1)+b_2(x_3+x_2)+b_3(x_3+x_1)=0$

$b_1x_2+b_1x_1+b_2x_3+b_2x_2+b_3x_3+b_3x_1=0$

$(b_1+b_3)x_1+(b_2+b_1)x_2+(b_2+b_3)x_3=0$

$b_1+b_3=0$

$b_1=-b_3$ ...(1)

$b_1+b_2=0$

$b_1=-b_2$ ..(2)

$b_2+b_3=0$

$b_2=-b_3$ ..(3)

Because $x_1,x_2$ and $x_3$ are linearly independent.

From equation (1) and (3)

$b_1=b_2$ ...(4)

Adding equation (2) and (4)

$2b_1==0$

$b_1=0$

From equation (1) and (2)

$b_3=0,b_2=0,b_3=0$

Hence, $y_1,y_2$ and $y_3$ area linearly independent.

5 0

3 years ago

A line passes through the point (-2,2) and is parallel to the line of y=1/2(x-5/2) l. What is the equation of the line passing t

grin007 [14]

C is your answer.

It contains the proper points and form. A is not in point-slope form and B has the incorrect slope.

3 0

3 years ago