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2 years ago

Name a segment parallel to bc

Mathematics

2 answers:

Archy [21]2 years ago

6 0

Do you have a picture?

Len [333]2 years ago

4 0

Try AD put a picture next time

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If it takes a leopard to run 500 yards in 1/2 a min how how many miles could it run in 2 hours

qaws [65]

Answer:120000

Step-by-step explanation:

bc in a minute their is 60 seconds and if the leoperd runs 500 yards every 30 seconds u would multiply 60 times 2 bc their is 60 min in a hour and u would get 120.then u would multiply that by 2 getting u 240 which is two hours then u would do 500 times 240

6 0

2 years ago

PLEASE HURRY ASAP I NEED IT ASAPP PLEASE SHOW YOUR WORKKK I WILL MAKE YOU BRAINLIEST

Arturiano [62]

Answer:

BD = √97 cm ≈ 9.849 cm

Step-by-step explanation:

Diagonal BD of rectangle ABCD is the hypotenuse of right triangle ABD. Opposite sides of the rectangle are the same length, so we have ...

AB = 4 cm

AD = 9 cm

The sides are related to the diagonals by the Pythagorean theorem.

<h3>Pythagorean theorem</h3>

The Pythagorean theorem tells you the relation between sides and hypotenuse of a right triangle:

AB² +AD² = BD²

4² +9² = BD² = 16 +81 . . . . . evaluating the squares

BD = √97 . . . . . take the square root

The length of BD is √97 cm, about 9.849 cm.

6 0

1 year ago

Find value of x for 7(3x+4)+2=156

mr_godi [17]

21x + 28 = 156 - 2
21x = 154 - 28
21x/21 = 126/21
X = 6

4 0

2 years ago

Read 2 more answers

A family donates 15% of their income to the church if they're anywhere income is 45800 how much is given to the church

schepotkina [342]

Answer:

2,748?

Step-by-step explanation:

6 0

2 years ago

The radius of a cone is increasing at a constant rate of 7 meters per minute, and the volume is decreasing at a rate of 236 cubi

storchak [24]

Answer:

The rate of change of the height is 0.021 meters per minute

Step-by-step explanation:

From the formula

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

Differentiate the equation with respect to time t, such that

$\frac{d}{dt} (V) = \frac{d}{dt} (\frac{1}{3}\pi r^{2}h)$

$\frac{dV}{dt} = \frac{1}{3}\pi \frac{d}{dt} (r^{2}h)$

To differentiate the product,

Let r² = u, so that

$\frac{dV}{dt} = \frac{1}{3}\pi \frac{d}{dt} (uh)$

Then, using product rule

$\frac{dV}{dt} = \frac{1}{3}\pi [u\frac{dh}{dt} + h\frac{du}{dt}]$

Since $u = r^{2}$

Then, $\frac{du}{dr} = 2r$

Using the Chain's rule

$\frac{du}{dt} = \frac{du}{dr} \times \frac{dr}{dt}$

∴ $\frac{dV}{dt} = \frac{1}{3}\pi [u\frac{dh}{dt} + h(\frac{du}{dr} \times \frac{dr}{dt})]$

Then,

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

Now,

From the question

$\frac{dr}{dt} = 7 m/min$

$\frac{dV}{dt} = 236 m^{3}/min$

At the instant when $r = 99 m$

and $V = 180 m^{3}$

We will determine the value of h, using

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

$180 = \frac{1}{3}\pi (99)^{2}h$

$180 \times 3 = 9801\pi h$

$h =\frac{540}{9801\pi }$

$h =\frac{20}{363\pi }$

Now, Putting the parameters into the equation

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

$236 = \frac{1}{3}\pi [(99)^{2} \frac{dh}{dt} + (\frac{20}{363\pi }) (2(99)) (7)]$

$236 \times 3 = \pi [9801 \frac{dh}{dt} + (\frac{20}{363\pi }) 1386]$

$708 = 9801\pi \frac{dh}{dt} + \frac{27720}{363}$

$708 = 30790.75 \frac{dh}{dt} + 76.36$

$708 - 76.36 = 30790.75\frac{dh}{dt}$

$631.64 = 30790.75\frac{dh}{dt}$

$\frac{dh}{dt}= \frac{631.64}{30790.75}$

$\frac{dh}{dt} = 0.021 m/min$

Hence, the rate of change of the height is 0.021 meters per minute.

3 0

2 years ago