All categories

3 years ago

A right triangle has a hypotenuse of 8 feet. One leg has a length of 5 feet. How long is the other leg?

Mathematics

2 answers:

posledela3 years ago

7 0

THEOREM:

<u>Pythagorean theorem</u>:– In a right angled triangle, the sum of squares of two legs is equal to the square of hypotenuse.

ANSWER:

Let the other leg be p, by pythagorean theorem

p² = h² - b²

p² = (8)² - (5)²

p² = 64 - 25

p² = 39

p = √39 ft.

So, <u>Correct choice</u> - [B] √39 ft.

liraira [26]3 years ago

3 0

Answer:

Step-by-step explanation:

The formula a^2 + b^2 = c^2 can be used to find any of the legs of the triangle. a and b are legs of the triangle and c is the hypotenuse. We can substitute known values into the equation:

a^2 + b^2 = c^2

One leg is 5 so we can put that in as a. And c, the hypotenuse, is 8.

5^2 + b^2 = 8^2

Now solve for b.

First, square 5 and 8

25 + b^2 = 64

Subtract 25 from both sides

b^2 = 39

Take the square root of both sides

b = √39

You might be interested in

Plz explain step by step, I'm taking Adv. Math (Algebra) It's confusing If can't explain, plz at least give right answer ✌

Mashcka [7]

Question 4:

OOOOOHHH a breakthrough! I was confused when I read this question but now I get it!

Okay when dividing this equation I would do this

5^4

Since 25 is 5 x 5 I would mark it as 5^2, and then I can subtract!

5^4

5^2 =

5^2

If you do it like this then it would be the first choice.

7 0

3 years ago

What is the equation for the plane illustrated below?

TiliK225 [7]

Answer:

Hence, none of the options presented are valid. The plane is represented by $3 \cdot x + 3\cdot y + 2\cdot z = 6$ .

Step-by-step explanation:

The general equation in rectangular form for a 3-dimension plane is represented by:

$a\cdot x + b\cdot y + c\cdot z = d$

Where:

$x$ , $y$ , $z$ - Orthogonal inputs.

$a$ , $b$ , $c$ , $d$ - Plane constants.

The plane presented in the figure contains the following three points: (2, 0, 0), (0, 2, 0), (0, 0, 3)

For the determination of the resultant equation, three equations of line in three distinct planes orthogonal to each other. That is, expressions for the xy, yz and xz-planes with the resource of the general equation of the line:

xy-plane (2, 0, 0) and (0, 2, 0)

$y = m\cdot x + b$

$m = \frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

Where:

$m$ - Slope, dimensionless.

$x_{1}$ , $x_{2}$ - Initial and final values for the independent variable, dimensionless.

$y_{1}$ , $y_{2}$ - Initial and final values for the dependent variable, dimensionless.

$b$ - x-Intercept, dimensionless.

If $x_{1} = 2$ , $y_{1} = 0$ , $x_{2} = 0$ and $y_{2} = 2$ , then:

Slope

$m = \frac{2-0}{0-2}$

$m = -1$

x-Intercept

$b = y_{1} - m\cdot x_{1}$

$b = 0 -(-1)\cdot (2)$

$b = 2$

The equation of the line in the xy-plane is $y = -x+2$ or $x + y = 2$ , which is equivalent to $3\cdot x + 3\cdot y = 6$ .

yz-plane (0, 2, 0) and (0, 0, 3)

$z = m\cdot y + b$

$m = \frac{z_{2}-z_{1}}{y_{2}-y_{1}}$

Where:

$m$ - Slope, dimensionless.

$y_{1}$ , $y_{2}$ - Initial and final values for the independent variable, dimensionless.

$z_{1}$ , $z_{2}$ - Initial and final values for the dependent variable, dimensionless.

$b$ - y-Intercept, dimensionless.

If $y_{1} = 2$ , $z_{1} = 0$ , $y_{2} = 0$ and $z_{2} = 3$ , then:

Slope

$m = \frac{3-0}{0-2}$

$m = -\frac{3}{2}$

y-Intercept

$b = z_{1} - m\cdot y_{1}$

$b = 0 -\left(-\frac{3}{2} \right)\cdot (2)$

$b = 3$

The equation of the line in the yz-plane is $z = -\frac{3}{2}\cdot y+3$ or $3\cdot y + 2\cdot z = 6$ .

xz-plane (2, 0, 0) and (0, 0, 3)

$z = m\cdot x + b$

$m = \frac{z_{2}-z_{1}}{x_{2}-x_{1}}$

Where:

$m$ - Slope, dimensionless.

$x_{1}$ , $x_{2}$ - Initial and final values for the independent variable, dimensionless.

$z_{1}$ , $z_{2}$ - Initial and final values for the dependent variable, dimensionless.

$b$ - z-Intercept, dimensionless.

If $x_{1} = 2$ , $z_{1} = 0$ , $x_{2} = 0$ and $z_{2} = 3$ , then:

Slope

$m = \frac{3-0}{0-2}$

$m = -\frac{3}{2}$

x-Intercept

$b = z_{1} - m\cdot x_{1}$

$b = 0 -\left(-\frac{3}{2} \right)\cdot (2)$

$b = 3$

The equation of the line in the xz-plane is $z = -\frac{3}{2}\cdot x+3$ or $3\cdot x + 2\cdot z = 6$

After comparing each equation of the line to the definition of the equation of the plane, the following coefficients are obtained:

$a = 3$ , $b = 3$ , $c = 2$ , $d = 6$

Hence, none of the options presented are valid. The plane is represented by $3 \cdot x + 3\cdot y + 2\cdot z = 6$ .

8 0

3 years ago

Draw and label a tape diagram to represent the following (choose two to do):

Aloiza [94]

Answer:

ask ur teacher

Step-by-step explanation:

6 0

3 years ago

BRO PLEASE HELP ASAP

KiRa [710]

Answer:

bro i think its 10

Step-by-step explanation:

3 0

3 years ago

Pleaseeeeeeee friends

IgorC [24]

B(x)=x2+3
I believe so

3 0

3 years ago