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3 years ago

AFGH is a right triangle. A. True B. False

Mathematics

2 answers:

Kay [80]3 years ago

7 0

Answer:false

Step-by-step explanation:a p e x

Scrat [10]3 years ago

6 0

Option B:

False

Solution:

Given GF = 10, FH = 6, GH = $\sqrt{63}$

To verify that ΔFGH is right triangle or not:

Pythagoras theorem:

If the square of the hypotenuse is equal to the sum of the squares of the other two sides, then the triangle is a right triangle.

Using Pythagoras theorem,

$\text{Hypotenuse}^2={GF}^2$

= 10²

= 100

$\text {(sum of the sides)}^2$ = $GH^2+FH^2$

= $6^2+(\sqrt{63} )^2$

= 36 + 63

= 99

100 ≠ 99

$GF^2\neq GH^2+FH^2$

Hence ΔFGH is not a right triangle.

The given statement is false.

Option B is the correct answer.

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Vikki [24]

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3 years ago

3 3/5−1 1/2 subtract

otez555 [7]

Answer:

2 1/10

Steps:

You need both fractions to have the same denominator, you get 6/10 and 5/10

Subtract your fraction:

6/10 - 5/10 = 1/10

Then your whole numbers:

3 - 1 = 2

Then put your answers together:

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3 years ago

A manufacturing process makes rods that vary slightly in length but follow a normal distribution with mean length 25 cm and stan

Galina-37 [17]

Answer:

The probability of randomly selecting a rod that is shorter than 22 cm

P(X<22) = 0.1251

Step-by-step explanation:

Step(i):-

Given mean of the Population = 25cm

Given standard deviation of the Population = 2.60

Let 'x' be the random variable in normal distribution

Given x=22

$Z = \frac{x-mean }{S.D} = \frac{22-25}{2.60} = -1.15$

Step(ii):-

The probability of randomly selecting a rod that is shorter than 22 cm

P(X<22) = P( Z<-1.15)

= 1-P(Z>1.15)

= 1-( 0.5+A(1.15)

= 0.5 - A(1.15)

= 0.5 - 0.3749

= 0.1251

The probability of randomly selecting a rod that is shorter than 22 cm

P(X<22) = 0.1251

7 0

3 years ago

What is the simplified form of (4ab)^2

Alex

$Use\ (a\cdot b)^n=a^nb^n\\-----------------\\\\(4ab)^2=4^2a^2b^2=\huge\boxed{16a^2b^2}$

7 0

4 years ago

The radius of a cone is decreasing at a constant rate of 7 inches per second, and the volume is decreasing at a rate of 948 cubi

inessss [21]

Answer:

The height of cone is decreasing at a rate of 0.085131 inch per second.

Step-by-step explanation:

We are given the following information in the question:

The radius of a cone is decreasing at a constant rate.

$\displaystyle\frac{dr}{dt} = -7\text{ inch per second}$

The volume is decreasing at a constant rate.

$\displaystyle\frac{dV}{dt} = -948\text{ cubic inch per second}$

Instant radius = 99 inch

Instant Volume = 525 cubic inches

We have to find the rate of change of height with respect to time.

Volume of cone =

$V = \displaystyle\frac{1}{3}\pi r^2 h$

Instant volume =

$525 = \displaystyle\frac{1}{3}\pi r^2h = \frac{1}{3}\pi (99)^2h\\\\\text{Instant heigth} = h = \frac{525\times 3}{\pi(99)^2}$

Differentiating with respect to t,

$\displaystyle\frac{dV}{dt} = \frac{1}{3}\pi \bigg(2r\frac{dr}{dt}h + r^2\frac{dh}{dt}\bigg)$

Putting all the values, we get,

$\displaystyle\frac{dV}{dt} = \frac{1}{3}\pi \bigg(2r\frac{dr}{dt}h + r^2\frac{dh}{dt}\bigg)\\\\-948 = \frac{1}{3}\pi\bigg(2(99)(-7)(\frac{525\times 3}{\pi(99)^2}) + (99)(99)\frac{dh}{dt}\bigg)\\\\\frac{-948\times 3}{\pi} + \frac{2\times 7\times 525\times 3}{99\times \pi} = (99)^2\frac{dh}{dt}\\\\\frac{1}{(99)^2}\bigg(\frac{-948\times 3}{\pi} + \frac{2\times 7\times 525\times 3}{99\times \pi}\bigg) = \frac{dh}{dt}\\\\\frac{dh}{dt} = -0.085131$

Thus, the height of cone is decreasing at a rate of 0.085131 inch per second.

3 0

3 years ago