the base and height of a triangle are in the ratio 4:5 . if the area is 640 sq.cm find the the base and height

A sequence is defined by f(0) = -20, f(n) = f(n-1) - 5 forn > 1.

bulgar [2K]

Answer:

1. Proved down

2. proved down

3. f(10) = -20 - 5 - 5 - 5 - 5 - 5 - 5 - 5 - 5 - 5 - 5

Step-by-step explanation:

Let us explain how to solve the question

∵ f(0) = -20, f(n) = f(n - 1) - 5 for n > 1

→ That means we have an arithmetic sequence with constant

difference -5 and first term -20

1. → f(1) means we need to find the second term, which equal the

term - 5

∵ f(1) means n = 1

∴ f(1) = f(1 - 1) - 5

∴ f(1) = f(0) - 5

∵ f(0) = -20

∴ f(1) = -20 - 5 → Proved

2. → f(3) means we need to find the third term, which equal the

second term - 5

∵ f(3) means n = 3

∴ f(3) = f(3 - 1) - 5

∴ f(3) = f(2) - 5

→ f(2) = f(1) - 5

∵ f(1) = -20 - 5

∴ f(2) = [-20 - 5] - 5 = -20 - 5 - 5

∴ f(3) = [-20 - 5 - 5] - 5

∴ f(3) = -20 - 5 - 5 - 5 → Proved

3. → From 1 and 2 we notice that the number of -5 is equal to n,

at n = 1 there is one (-5), when n= 3 there are three (-5)

∵ n = 10

∴ There are ten (-5)

∴ f(10) = -20 - 5(10)

∴ f(10) = -20 - 5 - 5 - 5 - 5 - 5 - 5 - 5 - 5 - 5 - 5 → Proved

6 0

3 years ago

A sphere is inscribed in a cube with a volume of 64 cubic inches. What is the volume of the sphere? Round your answer to the nea

erma4kov [3.2K]

Answer:

the volume of the sphere is

$33.51 in^{3}$

Step-by-step explanation:

This problem bothers on the mensuration of solid shapes, sphere and cube.

Given data

Volume of cube v = 64 cubic inches

since we are dealing with a cube the height and the radius of the sphere is same as the sides of the cube,

we know that volume of cube is expressed as

$v= l*b*h$

$v=l^{3}$

$64= l^{3}$

$l= \sqrt[3]{64}$

$l= 4 in$

also diameter d=length l

Diameter d= $4in$

Radius r = $\frac{d}{2}$ = $\frac{4}{2}$ = ${2 in}$

Height h= $4in$

we know that the volume of a sphere is given by

$v= \frac{4}{3} \pi r^{3}$

substituting into the formula we have

$v= \frac{4}{3} \ *3.142*2^{3} \\v=\frac{4*3.142*8}3} \\v= \frac{100.54}{3} \\v= 33.52in^{3}$

8 0

2 years ago

Solve by any method from this unit. y2 - 5y = 3 Please help

ExtremeBDS [4]

Answer:

-1

Step-by-step explanation:

First we combine like terms, so y2 - 5y = - 3y

Now we get - 3y = 3

Then we divide both sides by - 3

Now we get y =-1

7 0

2 years ago

How to fill in the prime factor tree?????? Answer???

dem82 [27]

Step-by-step explanation:

On the 8 branch, to the left is 2, and on the right is 4. On the left branch, and on the left, it is 2, and there is a 2 for the 6 branch.

5 0

2 years ago

If 1st and 4tg terms of G.p are 500 and 32 respectively it's second term is ?

bixtya [17]

Answer:

$T_{2} = 200$

Step-by-step explanation:

Given

Geometry Progression

$T_1 = 500$

$T_4 = 32$

Required

Calculate the second term

First, we need to write out the formula to calculate the nth term of a GP

$T_n = ar^{n-1}$

For first term: Tn = 500 and n = 1

$500 = ar^{1-1}$

$500 = ar^{0}$

$500 = a$

$a = 500$

For fought term: Tn = 32 and n = 4

$32 = ar^{4-1}$

$32 = ar^3$

Substitute 500 for a

$32 = 500 * r^3$

Make r^3 the subject

$r^3 = \frac{32}{500}$

$r^3 = 0.064$

Take cube roots

$\sqrt[3]{r^3} = \sqrt[3]{0.064}$

$r = \sqrt[3]{0.064}$

$r = 0.4$

Using: $T_n = ar^{n-1}$

$n = 2$ $r = 0.4$ and $a = 500$

$T_{2} = 500 * 0.4^{2-1}$

$T_{2} = 500 * 0.4^1$

$T_{2} = 500 * 0.4$

$T_{2} = 200$

<em>Hence, the second term is 200</em>

5 0

3 years ago