Ask question

Ask question

All categories

3 years ago

13

A rectangular computer screen is 18 in wide and 11 in high. find the length of its diagonal

2 answers:

Blizzard [7]3 years ago

6 0

Since it is a rectangle, that means that all the corners are 90 degrees.

If we take a rectangle and cut it along the diagonal, that is basically two right triangles.

The diagonal is just the hypotenuse of the triangles

To find the hypotenuse

a^2 + b^2 = c^2
18^2 + 11^2 = c^2
324 + 121 = c^2
445 = c^2
c = sqrt. (445)

Hope this helps!

Reptile [31]3 years ago

5 0

I'm not sure if this answer is correct because I haven't done it since 4th grade but I think it's 14 1/2in

You might be interested in

At this weekend's football game, 12 out of the first 48 people who enter the field were not wearing hats. If this sample is repr

arlik [135]

i think it’s 50 because 48/12 is 4 and 200/4 is 50

3 0

3 years ago

Read 2 more answers

What is the value of x?<br> Enter your answer in the box.

lubasha [3.4K]

Answer:

27

Step-by-step explanation:

X/39=(X+18)/65

65X=39X+702

26X=702

X=27

6 0

3 years ago

Can you help me with answer please

Alexandra [31]

Answer:

Step-by-step explanation:

y is 2,-2,-4,-4,-2,2,8

5 0

3 years ago

The slope of the line passing through the points (-1, 5) and (1, 6) is no slope 1/2 5/4

zloy xaker [14]

Answer:

1/2

Step-by-step explanation:

Here you utilize the formula of slope which is m = y2-y1/x2 -x1

(-1,5) (1,6)

(x1, y1) ( x2, y2)

m = ( 6-5)/ (1-(-1))

= 1/(1+1)

= 1/2

3 0

3 years ago

Read 2 more answers

Pls help me.<br>proove it ^

aivan3 [116]

Answer:

We verified that $a^3+b^3+c^3-3abc=\frac{a+b+c}{2}[(a-b)^2+(b-c)^2+(c-a)^2]$

Hence proved

Step-by-step explanation:

Given equation is $a^3+b^3+c^3-3abc=\frac{a+b+c}{2}[(a-b)^2+(b-c)^2+(c-a)^2]$

We have to prove that $a^3+b^3+c^3-3abc=\frac{a+b+c}{2}[(a-b)^2+(b-c)^2+(c-a)^2]$

That is to prove that LHS=RHS

Now taking RHS

$\frac{a+b+c}{2}[(a-b)^2+(b-c)^2+(c-a)^2]$

$=\frac{a+b+c}{2}[a^2-2ab+b^2+b^2-2bc+c^2+c^2-2ac+a^2]$ (using $(a-b)^2=a^2-2ab+b^2$ )

$=\frac{a+b+c}{2}[2a^2-2ab+2b^2-2bc+2c^2-2ac]$ (adding the like terms)

$=\frac{a+b+c}{2}[2a^2+2b^2+2c^2-2ab-2bc-2ac]$

$=\frac{a+b+c}{2}\times 2[a^2+b^2+c^2-ab-bc-ac]$

$=a+b+c[a^2+b^2+c^2-ab-bc-ac]$

Now multiply the each term to another each term in the factor

$=a^3+ab^2+ac^2-a^2b-abc-a^2c+ba62+b^3+bc^2-ab^2-b^2c-abc+ca^2+cb^2+c^3-abc-bc^2-ac^2]$

$=a^3+b^3+c^3-3abc$ (adding the like terms and other terms getting cancelled)

$=a^3+b^3+c^3-3abc$ =LHS

Therefore LHS=RHS

Therefore $a^3+b^3+c^3-3abc=\frac{a+b+c}{2}[(a-b)^2+(b-c)^2+(c-a)^2]$

Hence proved.

8 0

4 years ago