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3 years ago

5

What is the pi of S=(Πr^2)+(2Πr)h

1 answer:

Lemur [1.5K]3 years ago

5 0

Answer:

Step-by-step explanation:

That is the surface area of a cylinder and the pi is 22/7

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Mark the congruent parts of the triangles and complete the proof

eimsori [14]

1) given

2) definition of an angle bisector

3) reflexive

4) $\triangle ABD \cong \triangle CBD$ (ASA)

5) $\angle A \cong \angle C$ (CPCTC)

3 0

2 years ago

Emma will roll two number cubes, labeled 1 through 6. She will record the sum of the two numbers after each roll. She will roll

coldgirl [10]

Answer:

Emma should expect the sum to be equal to five "60 times"

Step-by-step explanation:

Please kindly check the attached file for explanation

5 0

3 years ago

How many computers must the AB Computer Company sell to break even? Let x be the number of computers.

Eduardwww [97]

Answer:

The Company AB must sell 116 computers to break even

Step-by-step explanation:

we have

$R(x)=1.5x$ -----> equation A

$C(x)=145+\frac{1}{4}x$ ----> equation B

where

C is the cost function

R is the revenue function

x is the number of computers sold

we know that

Break even is when the cost is equal to the revenue

so

equate equation A and equation B

$1.5x=145+\frac{1}{4}x$

Solve for x

subtract 1/4x both sides

$1.5x-\frac{1}{4}x=145$

$1.5x-0.25x=145$

$1.25x=145$

Divide by 1.25 both sides

$x=116$

therefore

The Company AB must sell 116 computers to break even

7 0

3 years ago

Find the two straight lines represented by x² + 5xy - 6y² = 0

Andreyy89

By factorizing, we have

$x^2 + 5xy - 6y^2 = (x - y) (x + 6y) = 0$

so that

$x - y = 0 \implies \boxed{y = x}$

or

$x + 6y = 0 \implies \boxed{y = -\dfrac x6}$

3 0

2 years ago

The population of Henderson City was 3,381,000 in 1994, and is growing at an annual rate 1.8%

liq [111]

<h2>In the year 2000, population will be 3,762,979 approximately. Population will double by the year 2033.</h2>

Step-by-step explanation:

Given that the population grows every year at the same rate( 1.8% ), we can model the population similar to a compound Interest problem.

From 1994, every subsequent year the new population is obtained by multiplying the previous years' population by $\frac{100+1.8}{100}$ = $\frac{101.8}{100}$ .

So, the population in the year t can be given by $P(t)=3,381,000\textrm{x}(\frac{101.8}{100})^{(t-1994)}$

Population in the year 2000 = $3,381,000\textrm{x}(\frac{101.8}{100})^{6}$ = $3,762,979.38$

Population in year 2000 = 3,762,979

Let us assume population doubles by year $y$ .

$2\textrm{x}(3,381,000)=(3,381,000)\textrm{x}(\frac{101.8}{100})^{(y-1994)}$

$log_{10}2=(y-1994)log_{10}(\frac{101.8}{100})$

$y-1994=\frac{log_{10}2}{log_{10}1.018}=38.8537$

$y$ ≈ $2033$

∴ By 2033, the population doubles.

4 0

3 years ago