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3 years ago

What is the area of a triangle with vertices at (0, −2) , (8, −2) , and (9, 1) ?

Mathematics

2 answers:

Oxana [17]3 years ago

7 0

Check the picture below.

recall that A = 1/2 bh.

Kay [80]3 years ago

3 0

Answer:

12 square units.

Step-by-step explanation:

When we plot the points representing the vertices of this triangle, we see that we do not have a segment perpendicular to any side; this means we do not have a height. What we do have is a horizontal line segment from (0, -2) to (8, -2), a segment running diagonally from (0, -2) to (9, 1) and a segment running diagonally from (9, 1) to (8, -2).

In order to find the height, we can drop a vertical line segment from the highest point, (9, 1), straight down adjacent to our horizontal line segment. This will be perpendicular. It runs from (9, 1) to (9, -2); its length would be 1--2 = 3 units. This is the height.

We will treat the horizontal segment as the base of the triangle; its length is 8-0 = 8.

Our formula for the area of a triangle is

A = 1/2bh

Using our information, we have

A = 1/2(8)(3) = 4(3) = 12

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Farmer Jones, and his wife, Dr. Jones, decide to build a fence in their field, to keep the sheep safe. Since Dr. Jones is a math

alex41 [277]

Answer: $\frac{8}{3}\times \sqrt{\frac{2}{5}}$

Step-by-step explanation:

Given two upward facing parabolas with equations

$y=6x^2 & y=x^2+2$

The two intersect at

$6x^2=x^2+2$

$5x^2=2$

$x^2$ = $\frac{2}{5}$

x= $\pm \sqrt{\frac{2}{5}}$

area enclosed by them is given by

A= $\int_{-\sqrt{\frac{2}{5}}}^{\sqrt{\frac{2}{5}}}\left [ \left ( x^2+2\right )-\left ( 6x^2\right ) \right ]dx$

A= $\int_{\sqrt{-\frac{2}{5}}}^{\sqrt{\frac{2}{5}}}\left ( 2-5x^2\right )dx$

A= $4\left [ \sqrt{\frac{2}{5}} \right ]-\frac{5}{3}\left [ \left ( \frac{2}{5}\right )^\frac{3}{2}-\left ( -\frac{2}{5}\right )^\frac{3}{2} \right ]$

A= $\frac{8}{3}\times \sqrt{\frac{2}{5}}$

7 0

3 years ago

1/2+1 1/2 i need the answer fast

iogann1982 [59]

Answer:

the answer is 2

Step-by-step explanation:

1/2 plus 1 is 1 1/2 then add another 1/2

3 0

3 years ago

Read 2 more answers

What is the inverse matrix that can be used to solve this systems of equations

Gala2k [10]

Step-by-step explanation:

The answer is OPTION C

Find the Inverse of a 3x3 Matrix.

First

Find the Determinant of A(The coefficients of e

Proceed towards finding the CO FACTOR of the 3x3 Matrix.

+. - +

A= [ 1 -1 -1 ]

[ -1 2 3 ]

[ 1 1 4 ]

The determinant of this is 1.

Find the co factor

| 2 3 | |-1 3 | |-1 2 |

| 1. 4. | |1 4 | |1. 1 |

|-1. -1 | |1 -1 | |1 -1

| 1. 4 | |1. 4| |1 1|

|-1. -1 | |1 -1 | |1. -1

|2. 3| |-1. 3| |-1 2|

After Evaluating The Determinant of each 2x 2 Matrix

You'll have

[ 5 7 -3]

[3 5 -2 ]

[-1 -2 1]

Reflect this along the diagonal( Keep 5,5 -2)

Then switching positions of other value

No need of Multiplying by the determinant because its value is 1 from calculation.

After this

Our Inverse Matrix Would be

[ 5 3 -1 ]

[7 5 -2 ]

[ -3 -2 1]

THIS IS OUR INVERSE.

OPTION C

5 0

3 years ago

Find the area of this shape <br><br> please

Elena-2011 [213]

<h3> $Area \:of\: rectangle = l \times b \\ = 3 \times 6 \\ = 18 {cm}^{2}$ </h3><h3> $Area\: of\: triangle = \frac{ab}{2} \\ = \frac{4 \times 6}{2} \\ = 12 {cm}^{2}$ </h3>

Total area = 12 + 18 = 30cm²

7 0

2 years ago

Read 2 more answers

-5=x/3-8 plz someone help me plz

Flura [38]

Isolate x by adding 8 to both sides.
3 = x/3
Then, multiply by 3 on both sides.
9 = x
To prove that this is right, you can imput 9 into the equation and it equals -5.

5 0

3 years ago

What is the area of a triangle with vertices at (0, −2) , ​ (8, −2) ​ , and ​ (9, 1) ​ ?

What is the area of a triangle with vertices at (0, −2) , (8, −2) , and (9, 1) ?