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3 years ago

11

Formula to find the lateral area of a right cone

1 answer:

BabaBlast [244]3 years ago

6 0

Answer:

This is the formula to find a lateral area of a right cone

A=1/2bh

Or A=1/2lw

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Which number line models the expression -3 + 5 ?

damaskus [11]

I think the answer would be B

3 0

3 years ago

Find the lateral surface area of this

Allushta [10]

Answer:226.2

Step-by-step explanation

Lateral= 2πrh

H=12

R=3

2*π*3*12=226.19

Round to nearest 10th is 226.2

6 0

2 years ago

At what rate per annum does the compound interest of Rs 576 becomes to Rs 153 in 2 years

ArbitrLikvidat [17]

Answer:

The value of rate per annum is R = 12.5 %

Step-by-step explanation:

Principal amount = 576

Interest = 153

Amount after 2 years = 576 + 153 = 729

We know that

$A = P [1 + \frac{R}{100} ] ^{T}$

Put all the values in above equation

$729 = 576 [1 + \frac{R}{100} ]^{2}$

$\frac{27}{24} = 1 + \frac{R}{100}$

R = 12.5 %

This is the value of rate per annum.

4 0

3 years ago

The sum of the first two terms of an infinite GP is 6 and each term is 5 times the sum of the succeeding termsthen the second te

kozerog [31]

Answer:

$\dfrac{6}{7}$ .

Step-by-step explanation:

It is given that the sum of first two terms of an infinite GP is 6.

nth term of a GP is

$a_n=ar^{n-1}$

$a+ar=6$ ...(1)

Each term is 5 times the sum of the succeeding terms.

$a_n=5(a_{n+1}+a_{n+2}+...+\infty)$

$ar^{n-1}=5(ar^n+ar^{n+1}+...+\infty)$

$ar^{n-1}=5ar^n(1+r+r^2+...+\infty)$

Divide both sides by a.

$r^{n-1}=5r^n(\dfrac{1}{1-r})$ $[\text{Sum of infinite GP}=\dfrac{a}{1-r}]$

$\dfrac{r^n}{r}=\dfrac{5r^n}{1-r}$

$\dfrac{1}{r}=\dfrac{5}{1-r}$

$1-r=5r$

$1=6r$

$\dfrac{1}{6}=r$

The common ratio is 1/6.

Put r=1/6 in (1).

$a+a(\dfrac{1}{6})=6$

$6a+a=36$

$7a=36$

$a=\dfrac{36}{7}$

Second term $ar=\dfrac{36}{7}\times \dfrac{1}{6}=\dfrac{6}{7}$

Therefore, the second term is $\dfrac{6}{7}$ .

4 0

2 years ago

Suppose the population of a certain city is 3031 thousand. It is expected to decrease to 2246 thousand in 50 years. Find the per

Kipish [7]

Answer:

The rate at which the population decrease is 59.8 %

Step-by-step explanation:

Given as :

The population of city = 30,31,000

Now The population of city decrease to 22,46,000

The time period in which population decrease is 50 years

Let the percentage rate of decrease = R%

So,

Final population = initial population × $(1-\frac{Rate}{100})^{Time}$

Or, 22,46,000 = 30,31,000 × $(1-\frac{Rate}{100})^{50}$

Or, $\frac{2246000}{3031000}$ = $(1-\frac{Rate}{100})^{50}$

Or, $\frac{2246}{3031}$ = $(1-\frac{Rate}{100})^{50}$

$(\frac{2246}{3031})^{\frac{1}{50}}$ = $( 1-\frac{Rate}{100})$

So , 0.99402 = $( 1-\frac{Rate}{100})$

Or, $\frac{Rate}{100}$ = 1 - 0.99402

So, Rate = $5.98\times 10^{-3}$ × 100

Or, Rate = 0.598 = 59.8 %

Hence The rate at which the population decrease is 59.8 % Answer

4 0

3 years ago