If a triangle has 3 side one is 12 one is 72 what is the last sude

Answers

Answer 1
Answer:

this triangle has 3 sides one is 12, lets call this side a²

another side is 72, lets call this side c²

this triangle must follow Pythagoras therom if it is right angled, that is a²+b²=c²

where a and b are the two shorter sides and c is the longest side (called the hypotenuse) opposite the right angle

to find the last side you do 72² - 12² = 5040

but it is the square root of 5040 which is 70.99 (two decimal places) if you want it in decimal form, if you want it in surd form it is √5040

Answer 2
Answer: There is no way to answer that with only the information given. 
The 3rd side can be ANY length between 60 and 84.  You can't
solve a triangle if you only know 2 sides of it.

Related Questions

A fertilizer covers 5/8 square foot in 1/4 hour
What is the answer to this?
Find the perimeter od a rectangular frame with sides of 4x -1 and 6x+5
Factor completely: 2x^3 – 32xA. 2(x + 4)(x – 4)B. 2x(x + 4)(x – 4)C. x(x – 4)(x – 4)D. 2x(x – 16)My progress so far:2x^3 – 32xx^3 - 16
Solve for x.3(x)+10= 2x3+2x6

For what values of k do the lines 2x + ky = 3 and kx + 8y = 7 intersect?

Answers

The lines 2x + ky = 3 and kx + 8y = 7 intersect for all values of k except 4 and -4 the answer is R - {4, -4}.

What is a linear equation?

It is defined as the relation between two variables, if we plot the graph of the linear equation we will get a straight line.

If in the linear equation, one variable is present, then the equation is known as the linear equation in one variable.

It is given that:

Two linear equations in two variables are:

2x + ky = 3 and

kx + 8y = 7

As we know the condition for the lines to intersect or have unique solutions:

2/k ≠ k/8

After cross multiplication:

k² ≠ 16

k ≠ ±4

The values of k can be anything except 4 and - 4

Thus, the lines 2x + ky = 3 and kx + 8y = 7 intersect for all values of k except 4 and -4 the answer is R - {4, -4}.

Learn more about the linear equation here:

brainly.com/question/11897796

#SPJ2

Answer:

ky=3-2x

y=(3-2x)/k

y=(7-kx)/8

(3-2x)/k=(7-kx)/8

8(3-2x)=(7-kx)k

24-16x=7k-k*kx

24=7k-kkx+16x

24+7k=kkx+16x

24+7k=x(k^2+16)

What’s 9.3 x 10 to the power of 5 times 1.26 x 10 to the power of -2 in scientific notation?

Answers

9.3*10^5=930,000 meaning you moved you decimal space back 5 spaces to make it 9.3

1.26*10^-2=0.0126 meaning you moved your decimal space forward 2 spaces to make it 1.26

If marching bands vary from 21 to 49 players,which number of players can be arranged in the greatest number of rectangles?

Answers

It would be 48 players because if you can arrange players in 
2 by 24
3 by 16
4 by 12
6 by 8

Hope this helps!!

Answer: 48 or 36 players

Step-by-step explanation:

This is because the number that has the largest of pairs of factors is 48 or 36.

48

1x48

2x24

3x16

4x12

6x8

36

1x36

2x18

3x12

4x9

6x6

2x=(360/4) solve for x please

Answers

If 2x=(360/4)  then the The value of x is 45.

To solve for x, we can simplify the equation step by step:

Step 1: Simplify the right side of the equation:

360/4 = 90

Step 2: Substitute the value of (360/4) into the equation:

2x = 90

Step 3: Isolate x by dividing both sides of the equation by 2:

x = 90/2

x = 45

So, the value of x is 45.

To know more about equation   here

brainly.com/question/29174899

#SPJ2

Hi there! First, we have to simplify both sides of the equation, 2x=360/4=2x=90. Now, we divide both sides by 2, 2x/2=90/2, 90/2=45. Therefore, the answer is x=45.

The joint of meat needs to be cooked for 30 minutes per 500 g plus an extra 20 minutes. How long will it take to cook a joint of meat weighting 2.5 kilo grand?

Answers

The equation made from this is:
Let x = each 500 grams

Cooking time = 30x + 20

2.5 kg = 2500 grams
2500 / 5 = 5
X = 5

We put this in to the equation,
30*5 + 20 = 170 minutes

It will take 170 minutes to cook a joint of meat weighing 2.5 kilograms

Solve each quadratic equation by factoring. 6n^2-18n-18=6

Answers

Divide by 6 to get:

n^2-3n-3=1

Subtracting 1, we see that:

n^2-3n-4=0

Factoring, we have:

(n-4)(n+1)=0

Therefore, 4 and -1 are the solutions, since if either n-4 or n+1 is zero, the whole expression is zero.
6n^(2)-18n-18=6 \n  \n 6n^(2)-18n-24=0 \n  \n 6(n^(2)-3n-4)=0 \n  \n n^(2)-3n-4=0 \n  \n n^(2)-4n+n-4=0 \n  \n n(n-4)+(n-4)=0 \n  \n (n+1)(n-4)=0 \n  \n n+1=0 \ \vee \ n-4=0 \n  \n n=-1 \ \vee \ n=4 \n  \n n\in \lbrace -1,4 \rbrace

==================================

"Associate with people who are likely to improve you."