HELP!!!! Position and label the triangle on the coordinate plane

Mathematics

1 answer:

Darya [45]3 years ago

4 0

Answer:

Option (2).

Step-by-step explanation:

It is given in the question,

ΔLMN is a right triangle with base LM = 3a units

Hypotenuse MN = 5a

By applying Pythagoras theorem in ΔLMN,

MN² = LM² + NM²

(5a)² = (3a)² + MN²

25a² - 9a² = MN²

MN = √16a²

MN = 4a

Therefore, vertices of the triangle will be L(0, 0), M(3a, 0) and N(0, 4a).

Option (2) will be the answer.

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Which two points satisfy y=-x2(^)+2x+4 and x+y=4?

Wewaii [24]

Use subtitution method to solve the problem.

First, change y to the value of x in order to find the exact value of x.
x + y = 4
y = 4 - x

Second, subtitute y with 4 - x from the first equation
y = -x² + 2x + 4
4 - x = -x² + 2x + 4
move all terms to the left side
x² - 2x - x + 4 - 4 = 0
x² - 3x = 0
x(x - 3) = 0
x = 0 or x = 3

Third, now we have 2 values of x. Find the value of y for each of x
For x = 0
y = 4 - x
y = 4 - 0
y = 4

For x = 3
y = 4 - x
y = 4 - 3
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The solution is (0,4) and (3,1). The answer is option b

5 0

3 years ago

Please help, I didn't learn this in class!

NemiM [27]

Solve the following system using elimination:
{-2 x + 2 y + 3 z = 0 | (equation 1)
{-2 x - y + z = -3 | (equation 2)
{2 x + 3 y + 3 z = 5 | (equation 3)

Subtract equation 1 from equation 2:
{-(2 x) + 2 y + 3 z = 0 | (equation 1)
{0 x - 3 y - 2 z = -3 | (equation 2)
{2 x + 3 y + 3 z = 5 | (equation 3)

Multiply equation 2 by -1:
{-(2 x) + 2 y + 3 z = 0 | (equation 1)
{0 x+3 y + 2 z = 3 | (equation 2)
{2 x + 3 y + 3 z = 5 | (equation 3)

Add equation 1 to equation 3:
{-(2 x) + 2 y + 3 z = 0 | (equation 1)
{0 x+3 y + 2 z = 3 | (equation 2)
{0 x+5 y + 6 z = 5 | (equation 3)

Swap equation 2 with equation 3:
{-(2 x) + 2 y + 3 z = 0 | (equation 1)
{0 x+5 y + 6 z = 5 | (equation 2)
{0 x+3 y + 2 z = 3 | (equation 3)

Subtract 3/5 × (equation 2) from equation 3:
{-(2 x) + 2 y + 3 z = 0 | (equation 1)
{0 x+5 y + 6 z = 5 | (equation 2)
{0 x+0 y - (8 z)/5 = 0 | (equation 3)

Multiply equation 3 by 5/8:
{-(2 x) + 2 y + 3 z = 0 | (equation 1)
{0 x+5 y + 6 z = 5 | (equation 2)
{0 x+0 y - z = 0 | (equation 3)

Multiply equation 3 by -1:
{-(2 x) + 2 y + 3 z = 0 | (equation 1)
{0 x+5 y + 6 z = 5 | (equation 2)
{0 x+0 y+z = 0 | (equation 3)

Subtract 6 × (equation 3) from equation 2:
{-(2 x) + 2 y + 3 z = 0 | (equation 1)
{0 x+5 y+0 z = 5 | (equation 2)
{0 x+0 y+z = 0 | (equation 3)

Divide equation 2 by 5:
{-(2 x) + 2 y + 3 z = 0 | (equation 1)
{0 x+y+0 z = 1 | (equation 2)
{0 x+0 y+z = 0 | (equation 3)

Subtract 2 × (equation 2) from equation 1:
{-(2 x) + 0 y+3 z = -2 | (equation 1)
{0 x+y+0 z = 1 | (equation 2)
v0 x+0 y+z = 0 | (equation 3)

Subtract 3 × (equation 3) from equation 1:
{-(2 x)+0 y+0 z = -2 | (equation 1)
{0 x+y+0 z = 1 | (equation 2)
{0 x+0 y+z = 0 | (equation 3)

Divide equation 1 by -2:
{x+0 y+0 z = 1 | (equation 1)
{0 x+y+0 z = 1 | (equation 2)
{0 x+0 y+z = 0 | (equation 3)

Collect results:

Answer: {x = 1, y = 1, z = 0

7 0

3 years ago

Help me please, Find a, b, and c.

Mama L [17]

Answer:

a = $6*\sqrt{3}$

b = 12

c = $6\sqrt{2}$

Step-by-step explanation:

Since the triangles are right triangles with 60 and 45 degree angles, their side lengths follow special triangles.

A 45-45-90 right triangle has side lengths $1-1-\sqrt{2}$ .

A 30-60-90 right triangle has side lengths $1 - \sqrt{3} -2$ .

Starting with the top triangle which has a 60 degree angle, its side length 6 corresponds to a side length of 1 in the special triangle. It is 6 times bigger so its remaining sides will be 6 times bigger too.

Side a corresponds to side length $\sqrt{3}$ . Therefore, $a = 6*\sqrt{3}$ .

Side b corresponds to side length 2, b = 2*6 = 12.

The bottom triangle has a 45 degree angle, its side length b= 12 corresponds to $\sqrt{2}$ . This means $\sqrt{2}$ was multiplied by $12 = \sqrt{2} * \sqrt{72}$ . This means that side c is $\sqrt{72}=6\sqrt{2}$ .