(question is on picture linked)

URGENT HELP! URGENT I WILL GIVE ALL OF MY POINTS

anastassius [24]

Answer:

false, true, false

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

Convert 3,250 mL to L. 3.25 L 0.325 L 3,250,000 L 325,000 L

Sergio [31]

Answer:

3.25 L

Step-by-step explanation:

1L=1000mL

3250ml x1L/1000mL = 3.25L

6 0

3 years ago

At the zoo on monday, 65 percent of the people in attendance were children. if 520 children were in attendance, how many people

inna [77]

Children = .65 * people
people = 520 / .65
800 people were in attendance

7 0

3 years ago

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A punch glass is in the shape of a hemisphere with a radius of 5 cm. If the punch is being poured into the glass so that the cha

Galina-37 [17]

Answer:

28.27 cm/s

Step-by-step explanation:

Though Process:

The punch glass (call it bowl to have a shape in mind) is in the shape of a hemisphere
the radius $r=5cm$
Punch is being poured into the bowl
The height at which the punch is increasing in the bowl is $\frac{dh}{dt} = 1.5$
the exposed area is a circle, (since the bowl is a hemisphere)
the radius of this circle can be written as $'a'$

what is being asked is the rate of change of the exposed area when the height $h = 2 cm$
the rate of change of exposed area can be written as $\frac{dA}{dt}$ .
since the exposed area is changing with respect to the height of punch. We can use the chain rule: $\frac{dA}{dt} = \frac{dA}{dh} . \frac{dh}{dt}$

and since $A = \pi a^2$ the chain rule above can simplified to $\frac{da}{dt} = \frac{da}{dh} . \frac{dh}{dt}$ -- we can call this Eq(1)

Solution:

the area of the exposed circle is

$A =\pi a^2$

the rate of change of this area can be, (using chain rule)

$\frac{dA}{dt} = 2 \pi a \frac{da}{dt}$ we can call this Eq(2)

what we are really concerned about is how $a$ changes as the punch is being poured into the bowl i.e $\frac{da}{dh}$

So we need another formula: Using the property of hemispheres and pythagoras theorem, we can use:

$r = \frac{a^2 + h^2}{2h}$

and rearrage the formula so that a is the subject:

$a^2 = 2rh - h^2$

now we can derivate a with respect to h to get $\frac{da}{dh}$

$2a \frac{da}{dh} = 2r - 2h$

simplify

$\frac{da}{dh} = \frac{r-h}{a}$

we can put this in Eq(1) in place of $\frac{da}{dh}$

$\frac{da}{dt} = \frac{r-h}{a} . \frac{dh}{dt}$

and since we know $\frac{dh}{dt} = 1.5$

$\frac{da}{dt} = \frac{(r-h)(1.5)}{a}$

and now we use substitute this $\frac{da}{dt}$ . in Eq(2)

$\frac{dA}{dt} = 2 \pi a \frac{(r-h)(1.5)}{a}$

simplify,

$\frac{dA}{dt} = 3 \pi (r-h)$

This is the rate of change of area, this is being asked in the quesiton!

Finally, we can put our known values:

$r = 5cm$

$h = 2cm$ from the question

$\frac{dA}{dt} = 3 \pi (5-2)$

$\frac{dA}{dt} = 9 \pi cm/s// or//\frac{dA}{dt} = 28.27 cm/s$

5 0

2 years ago

Which expression is equivalent to StartFraction 2 a + 1 Over 10 a minus 5 Endfraction divided by StartFraction 10 a Over 4 a squ

const2013 [10]

Answer:

$\frac{(2a + 1)^2}{50a}$

Step-by-step explanation:

Given

$\frac{2a + 1}{10a - 5} / \frac{10a}{4a^2 - 1}$

Required

Find the equivalent

We start by changing the / to *

$\frac{2a + 1}{10a - 5} / \frac{10a}{4a^2 - 1}$

$\frac{2a + 1}{10a - 5} * \frac{4a^2 - 1}{10a}$

Factorize 10a - 5

$\frac{2a + 1}{5(2a - 1)} * \frac{4a^2 - 1}{10a}$

Expand 4a² - 1

$\frac{2a + 1}{5(2a - 1)} * \frac{(2a)^2 - 1}{10a}$

$\frac{2a + 1}{5(2a - 1)} * \frac{(2a)^2 - 1^2}{10a}$

Express (2a)² - 1² as a difference of two squares

Difference of two squares is such that: $a^2- b^2= (a+b)(a-b)$

The expression becomes

$\frac{2a + 1}{5(2a - 1)} * \frac{(2a - 1)(2a + 1)}{10a}$

Combine both fractions to form a single fraction

$\frac{(2a + 1)(2a - 1)(2a + 1)}{5(2a - 1)10a}$

Divide the numerator and denominator by 2a - 1

$\frac{(2a + 1)((2a + 1)}{5*10a}$

Simplify the numerator

$\frac{(2a + 1)^2}{5*10a}$

$\frac{(2a + 1)^2}{50a}$

Hence,

$\frac{2a + 1}{10a - 5} / \frac{10a}{4a^2 - 1}$ = $\frac{(2a + 1)^2}{50a}$

4 0

3 years ago