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2 years ago

How do I do this, i am doing this and i need full points

Mathematics

1 answer:

sesenic [268]2 years ago

6 0

Answer:

D is correct ...

nind d d find her r d d d f FBI f

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BRIA is a rectangle. AN=5. NR=x-3. please solve for x, NR, and BI

Lena [83]

Answer:

x=8

NR=5 units

BI=10 units

Step-by-step explanation:

In a rectangle BRIA

AN=5 units

NR=x-3

We have to solve for x , NR and BI.

We know that

Diagonals of rectangle bisect to each other.

BI and AR are the diagonals of rectangle BRIA and intersect at point N.

AN=NR

$5=x-3$

$x=5+3=8$

$x=8$

Substitute the value of x

NR=8-3=5

By property of rectangle

BI=AR=AN+NR=5+5=10 unit

BI=10 units

7 0

3 years ago

Answer as a fraction 1 / 2 5 6 =

frez [133]

Answer:

Step-by-step explanation:

4 0

3 years ago

A developer has 7 1/2 acres of land to build houses on. If each house is to be built on -1/4 acre plot, how many houses can he b

mel-nik [20]

It’s either A.3 or C.30

6 0

3 years ago

Justin surveyed 280 of the students in his school about their favorite color. 98

Mamont248 [21]

Answer:

$\frac{98}{280} \times \frac{100}{1}$

$0.35 \times 100 \\ = 35\%$

8 0

3 years ago

Hello people ~

Luden [163]

Cone details:

height: h cm

radius: r cm

Sphere details:

radius: 10 cm

================

From the endpoints (EO, UO) of the circle to the center of the circle (O), the radius is will be always the same.

Using Pythagoras Theorem

(a)

TO² + TU² = OU²

(h-10)² + r² = 10² [insert values]

r² = 10² - (h-10)² [change sides]

r² = 100 - (h² -20h + 100) [expand]

r² = 100 - h² + 20h -100 [simplify]

r² = 20h - h² [shown]

r = √20h - h² ["r" in terms of "h"]

(b)

volume of cone = 1/3 * π * r² * h

===========================

$\longrightarrow \sf V = \dfrac{1}{3} * \pi * (\sqrt{20h - h^2})^2 \ ( h)$

$\longrightarrow \sf V = \dfrac{1}{3} * \pi * (20h - h^2) (h)$

$\longrightarrow \sf V = \dfrac{1}{3} * \pi * (20 - h) (h) ( h)$

$\longrightarrow \sf V = \dfrac{1}{3} \pi h^2(20-h)$

To find maximum/minimum, we have to find first derivative.

(c)

First derivative

$\Longrightarrow \sf V' =\dfrac{d}{dx} ( \dfrac{1}{3} \pi h^2(20-h) )$

apply chain rule

$\sf \Longrightarrow V'=\dfrac{\pi \left(40h-3h^2\right)}{3}$

Equate the first derivative to zero, that is V'(x) = 0

$\Longrightarrow \sf \dfrac{\pi \left(40h-3h^2\right)}{3}=0$

$\Longrightarrow \sf 40h-3h^2=0$

$\Longrightarrow \sf h(40-3h)=0$

$\Longrightarrow \sf h=0, \ 40-3h=0$

$\Longrightarrow \sf h=0,\:h=\dfrac{40}{3}$

maximum volume: when h = 40/3

$\sf \Longrightarrow max= \dfrac{1}{3} \pi (\dfrac{40}{3} )^2(20-\dfrac{40}{3} )$

$\sf \Longrightarrow maximum= 1241.123 \ cm^3$

minimum volume: when h = 0

$\sf \Longrightarrow min= \dfrac{1}{3} \pi (0)^2(20-0)$

$\sf \Longrightarrow minimum=0 \ cm^3$

6 0

2 years ago

Read 2 more answers