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serious [3.7K]
2 years ago
10

What is the area of a triangle with vertices of (2,5),(4,-2) and (-2,0). Area= square units

Mathematics
1 answer:
Reptile [31]2 years ago
3 0

Answer:

A=bh/2

Step-by-step explanation:

use the distance formula to find the length of the base and the height. Then, multiply those numbers and divide by 2

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A plot of land is in the shape of a triangle. If one side is X meters, a second side Is (5x-2) meters and a third side is (7x-4)
lukranit [14]
Perimeter = x + 5x - 2 + 7x - 4

=  13x - 6  meters answer
6 0
3 years ago
Read 2 more answers
In the diagram, GH bisects FGI. A. Solve for x and find m FGH. B. Find m HGI. C.find m FGI. A. X=_(simplify your answer)
Sophie [7]

Answer:

Part A) x=9 and m∠FGH=29°

Part B) m∠HGI=29°

Part C) m∠FGI=58°

Step-by-step explanation:

Part A) Solve for x and find m∠FGH

we know that

m∠FGI=m∠FGH+m∠HGI

If GH bisects FGI

then

m∠FGH=m∠HGI

substitute the values

4x-7=5x-16

solve for x

5x-4x=16-7

x=9

m∠FGH=4x-7

m∠FGH=4(9)-7=29°

Part B) Find the measure of angle HGI

m∠HGI=5x-16

Remember that

m∠FGH=m∠HGI

Verify

m∠HGI=5(9)-16=29° -----> is correct

Part C) Find the measure of angle FGI

m∠FGI=m∠FGH+m∠HGI

substitute the values

m∠FGI=29°+29°=58°

8 0
2 years ago
Pic below help ASAP (not exam)
mel-nik [20]
Divide 9.3 by 3.1:

9.3 / 3.1 = 3

subtract the exponents: 34 -17 = 17

then turn those into scientific notation:

3 x 10^17
6 0
3 years ago
The equation 7^2=a^2 shows the relationship between a planet’s orbital period, T, and the planet’s mean distance from the sun, A
jeyben [28]

Answer:

The period of Y increases by a factor of k^ {3/2} with respect to the period of X

Step-by-step explanation:

The equation T ^ 2 = a ^ 3 shows the relationship between the orbital period of a planet, T, and the average distance from the planet to the sun, A, in astronomical units, AU. If planet Y is k times the average distance from the sun as planet X, at what factor does the orbital period increase?  

For the planet Y:  

T_y ^ 2 = a_y ^ 3

For planet X:  

T_x ^ 2 = a_x ^ 3

To know the factor of aumeto we compared T_x with T_y

We know that the distance "a" from planet Y is k times larger than the distance from planet X to the sun. So:  

a_y ^ 3 = (a_xk) ^ 3

So

\frac{T_y ^ 2}{T_x ^ 2}=\frac{a_y ^ 3}{a_x^ 3}\\\\\frac{T_y ^ 2}{T_x ^ 2}=\frac{(a_xk)^3}{a_x ^ 3}\\\\\frac{T_y^ 2}{T_x^ 2}=\frac{k ^ {3}a_{x}^ 3}{a_{x}^ 3}\\\\\frac{T_{y}^ 2}{T_{x}^ 2}=k ^ 3\\\\T_{y}^ 2 = T_{x}^{2}k^{3}\\\\T_{y} =k^{\frac{3}{2}}T_x

Then, the period of Y increases by a factor of k^ {3/2} with respect to the period of X



4 0
3 years ago
PLEASE HELP ME PLEASE
Anna [14]

Answer:

We conclude that:

  • f(2) = -5

Step-by-step explanation:

Given the function

y = f(x)

In order to determine the value f(2), we need to check the value of y at x = 2.

In other words, we need to check:

at x = 2, what is the value of y?

From the graph, it is clear that at x = 2, the value of y = -5

In other words,

f(2) = -5

Therefore, we conclude that:

  • f(2) = -5

4 0
2 years ago
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