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3 years ago

Find the lateral area for the given cylinder. use 3.14 for pi and round to the nearest whole number

2 answers:

OverLord2011 [107]3 years ago

5 0

Formula for lateral area of cylinder:
A $A_{L}=2 \pi rh$
Here r=6 units and h=26 units
Put values in formula.
$A _{L} =2(3.14)(6)(26) \\ A_{L}=980\ square \ units$ \\
(rounded to nearest whole number)

Answer: Lateral area of cylinder= 980 square units

iris [78.8K]3 years ago

3 0

Answer:

D. 904 yd2

Step-by-step explanation:

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What’s the answer helpp

padilas [110]

Answer is the second option, the one that says X+1/1

8 0

3 years ago

Use decimals and fractions in the same equation showing the Commutative Property. Repeat for the Associative Property.

Anna71 [15]

For Commutative Property of Addition

Let us take a decimal number $2.14$ and a fraction $\frac{5}{20}$

Now, according to the Commutative property of Addition:

For any two numbers $a$ and $b$ :

$a+b = b+a$

So, for $2.14$ and $\frac{5}{20}$

Let us add

$2.14+\frac{5}{20} = \frac{214}{100} +\frac{5}{20}$

$=\frac{214}{100} + \frac{5 \times 5}{20 \times 5} \\ \\=\frac{214}{100} + \frac{25}{100} \\ \\= 2.14 + 0.25 \\ \\ = 2.39$

Also,

$\frac{5}{20} + 2.14 = \frac{5}{20} + \frac{214}{100}$

$= \frac{5 \times 5}{20 \times 5} +\frac{214}{100} \\ \\= \frac{25}{100} + \frac{214}{100} \\ \\= 0.25 + 2.14 \\ \\ = 2.39$

Therefore, $2.14+\frac{5}{20} = \frac{5}{20} + 2.14$

Hence, Commutative Property of Addition is satisfied.

For Associated Property of Addition

Let us take two same decimal numbers $2.14$ , $7.25$ and a fraction $\frac{5}{20}$

Now, according to the Associated property of Addition:

For any three numbers $a$ , $b$ and $c$

$a + (b+c) =(a+b) + c$

So, for $2.14$ , $7.25$ and $\frac{5}{20}$

The Left hand side:

$a + (b+c)$

$2.14 + (7.25 + \frac{5}{20}) = 2.14 + (\frac{725}{100} + \frac{5 \times 5}{20\times 5})$

$= 2.14 + (\frac{725}{100} + \frac{25 }{100})$

$= 2.14 + (\frac{750}{100} )$

$= \frac{214}{100} + \frac{750}{100}$

$= \frac{964}{100}$

$=9.64$

The Right hand side:

$(a + b )+c$

$(2.14 + 7.25 )+ \frac{5}{20}= ( 9.39 ) + \frac{5 }{20}$

$= 9.39 + \frac{5 \times 5 }{20 \times 5}$

$= 9.39 + \frac{25}{100}$

$= 9.39 + 0.25$

$= 9.64$

Thus,

$(2.14 + (7.25 + \frac{5}{20} )= (2.14 + 7.25 )+ \frac{5}{20}$

Therefore, $2.14+\frac{5}{20} = \frac{5}{20} + 2.14$

Hence, Associative Property of Addition is satisfied.

8 0

2 years ago

A credit union client deposits $3,300 in an account earning 6.5% interest, compounded annually. What will the balance of the acc

astra-53 [7]

Answer:

The balance of account at the end of 36 years is $31,849.29

Step-by-step explanation:

We are given the following in the question:

P = $3,300

r = 6.5% = 0.065

t = 36 years

The compound interest is given by:

$A = p\bigg(1+\dfrac{r}{n}\bigg)^{nt}$

where A is the amount, p is the principal, r is the interest rate, t is the time in years and n is the nature of compound interest.

Since interest is compounded annualy we use n = 1

$A = 3300\bigg(1+\dfrac{0.065}{1}\bigg)^{36}\\\\A = \$31,849.29$

Thus, balance of account at the end of 36 years is $31,849.29

6 0

2 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

1 year ago

Read 2 more answers

CAN SOMEONE PLEASE HELP ME WITH THIS FUNCTION PROBLEM

ycow [4]

Answer:

Choose f(x) = 11x + 1

Step-by-step explanation:

Note that we will simply plug the value of 2 into the equation for February:

f(2) = 11(2) + 1 = 23

And plug the value of 6 into the equation for June:

f(6) = 11(6) + 1 = 67

Note how the points on the graph seem to match up with these values. If we evaluate following the same style for each:

f(3) = 11(3) + 1 = 34

f(4) = 11(4) + 1 = 45

f(5) = 11(5) + 1 = 56

Note, these values seems to be very close in approximation to the graph points for each month.

The other three functions return values that are just to far away from what are represented in the graph.

Cheers.

8 0

3 years ago