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3 years ago

Find the sum of all odd integers between 70 and 120

Mathematics

1 answer:

Evgen [1.6K]3 years ago

3 0

Answer:

It forms an AP

AP : 71 , 73 , 75 , .................. 119

a = 71

d = 2

$a_n$ = a + (n-1)d

119 = 71 + (n-1)2

n-1 = (119 - 71)/2

n = 24

$S_n$ = n/2 × (a + $a_n$ )

= 24/2 × (71 + 119)

= 12 × 190 = 2280

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A sequence is defined recursively using the formula f(n + 1) = -0.5f(n). If the first term of the sequence is 120, what is f(5)?

PolarNik [594]

Answer: 7.5

Step-by-step explanation:

The given formula tells us that the next term f(n+1) of the sequence is -0.5 times the previous term f(n)

First term of the sequence is f(1) = 120

Second term of the sequence is f(2) = f(1+1) = -0.5 f(1) = -0.5 (120) = -60

Third term of the sequence is f(3) = f(2+1) = -0.5 f(2) = -0.5 (-60) = 30

Fourth term of the sequence is f(4) = f(3+1) = -0.5 f(3) = -0.5 (30) = -15

Fifth term of the sequence is f(5) = f(4+1) = -0.5 f(4) = -0.5 (-15) = <em>7.5</em>

3 0

4 years ago

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Svetllana [295]

Answer:

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Step-by-step explanation:

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3 years ago

Monica [59]

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4 0

2 years ago

the diffrence of two numbers,a and b,is 21,the diffrence of five times a and two times b is 18 what are the values of a and b

Triss [41]

Answer:

a=-8

b=-29

Step-by-step explanation:

Let's assume

first number is a

second number is b

the difference of two numbers,a and b,is 21

so, we get

$a-b=21$

we can solve for 'a'

$a=b+21$

the difference of five times a and two times b is 18

so, we get

$5a-2b=18$

now, we can plug 'a'

$5(b+21)-2b=18$

now, we can solve for b

$5b+105-2b=18$

$3b+105=18$

$3b=-87$

$b=-29$

now, we can find 'a'

$a=-29+21$

$a=-8$

8 0

3 years ago

Read 2 more answers

Given: AB ∥ DC

RSB [31]

Answer:

$A=1,720.16\ units^2$

Step-by-step explanation:

we know that

The area of the trapezoid is equal to

$A=\frac{1}{2}(DC+AB)DE$

step 1

Find the measure of angle DAE

m∠ADC+m∠DAE=180° -----> by consecutive interior angles

we have

m∠ADC = 134°

substitute

134°+m∠DAE=180°

m∠DAE=180°-134°=46°

step 2

In the right triangle ADE

Find the length side AE

cos(∠DAE)=AE/AD

$AE=cos(46\°)(40)\\AE=27.79\ units$

step 3

In the right triangle ADE

Find the length side DE

sin(∠DAE)=DE/AD

$DE=sin(46\°)(40)\\DE=28.77\ units$

step 4

Find the area of ABCD

$A=\frac{1}{2}(DC+AB)DE$

we have

$DC=32\ units\\AB=DC+2(AE)=32+2(27.79)=87.58\ units\\DE=28.77\ units$

substitute

$A=\frac{1}{2}(32+87.58)28.77$

$A=1,720.16\ units^2$

5 0

4 years ago

Find the sum of all odd integers between 70 and 120​

Find the sum of all odd integers between 70 and 120