All categories

2 years ago

Do the following lengths form a right triangle? (Also how to tell that they do form a right triangle)

Mathematics

1 answer:

ololo11 [35]2 years ago

4 0

Answer: Yes this is a right triangle

We can determine this by using the converse of the pythagorean theorem.

a = 5, b = 12, c = 13

$a^2 + b^2 = c^2\\\\5^2 + 12^2 = 13^2\\\\25 + 144 = 169\\\\169 = 169 \ \ \checkmark\\\\$

Since those a,b,c values work in the pythagorean theorem, this proves we have a right triangle.

You might be interested in

Evaluate the triple integral ∭EzdV where E is the solid bounded by the cylinder y2+z2=81 and the planes x=0,y=9x and z=0 in the

dem82 [27]

Answer:

I = 91.125

Step-by-step explanation:

Given that:

$I = \int \int_E \int zdV$ where E is bounded by the cylinder $y^2 + z^2 = 81$ and the planes x = 0 , y = 9x and z = 0 in the first octant.

The initial activity to carry out is to determine the limits of the region

since curve z = 0 and $y^2 + z^2 = 81$

∴ $z^2 = 81 - y^2$

$z = \sqrt{81 - y^2}$

Thus, z lies between 0 to $\sqrt{81 - y^2}$

GIven curve x = 0 and y = 9x

$x =\dfrac{y}{9}$

As such,x lies between 0 to $\dfrac{y}{9}$

Given curve x = 0 , $x =\dfrac{y}{9}$ and z = 0, $y^2 + z^2 = 81$

y = 0 and

$y^2 = 81 \\ \\ y = \sqrt{81} \\ \\ y = 9$

∴ y lies between 0 and 9

Then $I = \int^9_{y=0} \int^{\dfrac{y}{9}}_{x=0} \int^{\sqrt{81-y^2}}_{z=0} \ zdzdxdy$

$I = \int^9_{y=0} \int^{\dfrac{y}{9}}_{x=0} \begin {bmatrix} \dfrac{z^2}{2} \end {bmatrix} ^ {\sqrt {{81-y^2}}}_{0} \ dxdy$

$I = \int^9_{y=0} \int^{\dfrac{y}{9}}_{x=0} \begin {bmatrix} \dfrac{(\sqrt{81 -y^2})^2 }{2}-0 \end {bmatrix} \ dxdy$

$I = \int^9_{y=0} \int^{\dfrac{y}{9}}_{x=0} \begin {bmatrix} \dfrac{{81 -y^2} }{2} \end {bmatrix} \ dxdy$

$I = \int^9_{y=0} \begin {bmatrix} \dfrac{{81x -xy^2} }{2} \end {bmatrix} ^{\dfrac{y}{9}}_{0} \ dy$

$I = \int^9_{y=0} \begin {bmatrix} \dfrac{{81(\dfrac{y}{9}) -(\dfrac{y}{9})y^2} }{2}-0 \end {bmatrix} \ dy$

$I = \int^9_{y=0} \begin {bmatrix} \dfrac{{81 \ y -y^3} }{18} \end {bmatrix} \ dy$

$I = \dfrac{1}{18} \int^9_{y=0} \begin {bmatrix} {81 \ y -y^3} \end {bmatrix} \ dy$

$I = \dfrac{1}{18} \begin {bmatrix} {81 \ \dfrac{y^2}{2} - \dfrac{y^4}{4}} \end {bmatrix}^9_0$

$I = \dfrac{1}{18} \begin {bmatrix} {40.5 \ (9^2) - \dfrac{9^4}{4}} \end {bmatrix}$

$I = \dfrac{1}{18} \begin {bmatrix} 3280.5 - 1640.25 \end {bmatrix}$

$I = \dfrac{1}{18} \begin {bmatrix} 1640.25 \end {bmatrix}$

I = 91.125

4 0

3 years ago

What is the slope of the line that passes through the points (9, 1) and (17, 2)

Wewaii [24]

Answer:

Step-by-step explanation:

Slope of a line that goes trough the points (x1=9, y1=1) and (x2=17, y2=2)

m= (y2-y1)/(x2-x1) = (2-1) / (17-9) = 1/8

3 0

2 years ago

6.7z = 5.2z + 12.3 z=

vladimir1956 [14]

Answer:

z = 8.2

Step-by-step explanation:

6.7z = 5.2z + 12.3

combine like terms by subtracting 5.2z from both sides

1.5z = 12.3

divide by 1.5

z = 8.2

7 0

3 years ago

Read 2 more answers

Kevin calculated the product of 3.2 x 104 and 3.6 x 102 as 11.52 x 106. Which is the next step that Kevin should app

garri49 [273]

Answer:

the correct answer is C

Step-by-step explanation:

i was just on the question and the correct answer was C. or

Write in scientific notation, moving the decimal in the coefficient one place to the left, which adds 1 to the exponent value.

7 0

3 years ago

Read 2 more answers

Write an equation of the line in slope intercept form that passes through (3, 1/2) and has a slope of 4.

mash [69]

Answer:

Step-by-step explanation:

Write an equation of the line in slope intercept form that passes through (3, 1/2) and has a slope of 4.

7 0

2 years ago