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2 years ago

A triangle has an area of 80 yards and a height of 10 yards how long is the base of the triangle

1 answer:

5 0

16 yards is your answer for this!
its 2 with area/hight.
if you need the steps of this just ask ill show you how to do it step by step

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1/2 is poured out and 1 1/2 is left

Step-by-step explanation:

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2 years ago

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iogann1982 [59]

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3 years ago

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Andrei [34K]

Answer:

Circle 1=6x+2y

Rectangle 4 = 3x+y

Circle 4 = 5x

Step-by-step explanation:

4x+3y and 2x-y can be added to find result for the circle.

Let C1 represent circle 1

4x+3y+2x-y=C1\\

Combining\:like\:terms\\

4x+2x+3y-y=C1

6x+2y=C1

So, The result is: C1=6x+2y

Now, We need to solve:

Let R1 represent rectangle 4

x+4y + R1 = 4x+5y

R1=4x+5y-(x+4y)

R1=4x+5y-x-4y

R1=4x-x+5y-4y

R1=3x+y

So, Solving x+4y + R1 = 4x+5y, We get R1 = 3x+y

Now, We need to solve the equation:

Let C4= Circle 4

2x-y+3x+y=C4

Combining the like terms

2x+3x-y+y=C4

5x=C4

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2 years ago

X + y = -1<br> X - y = 7

enyata [817]

1: Solve for either x or y in one of the equations. So x + y = -1 is y = -x -1
2: substitute the new equation in the opposite equation. So x - (-x - 1) = 7
3: distribute the negative. X + x + 1 = 7
4: combine like terms. 2x + 1 = 7
5: solve for x. Subtract 1 on both sides. 2x = 6
6: divide by 2 to get x by itself. X = 3
7: plug the new value of x into one of the ORIGINAL equations. 3 + y = -1
8: solve for y. Subtract 3 on both sides.
Y = -4
9: the solution is written as (x,y) so the solution would be (3, -4)

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3 years ago

2x+3y=1 and y=-2 - 9 solve using linear combination method. Please explain steps.

kaheart [24]

Answer:

$x=17$

Step-by-step explanation:

$2x+3y=1$

$y=-2-9$

In order to combine these two equations, an idea you need to keep in mind is finding a way of setting these equations as equal to each other. I saw that each equation shared a common value, $y$ . In this case, we need to isolate $y$ in the first equation so that both equations $=y$ .

$2x+3y=1$

$3y=-2x+1$

$\frac{3y}{3}=\frac{-2x+1}{3}$

$y=-\frac{2}{3}x+\frac{1}{3}$

With this, we now know that both $-2-9$ and $-\frac{2}{3}x+\frac{1}{3}$ are equal to $y$ , so we can set them equal to each other.

$y=-\frac{2}{3}x+\frac{1}{3}$

$y=-11$

$-\frac{2}{3}x+\frac{1}{3}=-11$

Reply to this if anything I'm saying or doing is confusing in any way, or if you find a mistake. :) Solve for $x$ .

$-\frac{2}{3}x+\frac{1}{3}=-11$

$-\frac{2}{3}x=-11-\frac{1}{3}$

$-\frac{2}{3}x =-\frac{33}{3}-\frac{1}{3}$

$-\frac{2}{3}x =-\frac{34}{3}$

$-\frac{2}{3}x(-\frac{3}{2})=-\frac{34}{3}(-\frac{3}{2})$

$x=\frac{102}{6}=17$

$x=17$

Hopefully this answer is correct AND makes sense in terms of how I achieved it. Again, reply to this with any questions or mistakes I made and I'll do my best to answer or fix them.

4 0

2 years ago