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andrezito [222]
3 years ago
15

Question: Find The Sum Of The Integers From -6 To 58 Of Ss. O 1,500 O 1,734 O 1,690 O 1,621​

Mathematics
1 answer:
REY [17]3 years ago
5 0

Answer:

The Sum Of The Integers From -6 To 58 is <u>1690.</u>

Step-by-step explanation:

Given,

a=-6

T_n=58

We have to find out the sum of integers from -6 To 58.

Firstly we will find out the total number of terms that is 'n'.

Here a_1=-6\ and\ a_2=-5

\therefore d=a_2-a_1=-5-(-6)=-5+6=1

Now we use the formula of A.P.

T_n=a+(n-1)d

On substituting the values, we get;

58=-6+(n-1)1\\\\n-1=58+6\\\\n-1=64\\\\n=64+1=65

So there are 65 terms in between  -6 To 58.

That means we have to find the sum of 65 terms in between  -6 To 58.

Now we use the formula of Sum of n_terms.

S_n=\frac{n}{2}(2a-(n-1)d)

On substituting the values, we get;

S_{65}=\frac{65}{2}(2\times-6+(65-1)1)\\\\S_{65}=\frac{65}{2}(-12+64)\\\\S_{65}=\frac{65}{2}\times52\\\\S_{65}=65\times26=1690

Hence The Sum Of The Integers From -6 To 58 is <u>1690.</u>

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Which point on the number line is the best approximation for √59?
Elina [12.6K]

Answer:

6^2 = 36

Step-by-step explanation:

The number that best approximates the square of 6 is 36, its actual value. The point so labelled is the appropriate point on the number line.

7 0
2 years ago
The sum of three consecutive even numbers is 114. what is the smallest of the three numbers?​
saul85 [17]

Answer:

36

Step-by-step explanation:

Let's let n be any number.

If we multiply n by 2, we get 2n. Since anything multiplied by 2 is even, 2n <em>must</em> be even. Therefore, let 2n be our first number.

So, the consecutive even number will be (2n+2) followed by (2n+4).

We know that their sum must be 114. Therefore, we can write the following equation:

(2n)+(2n+2)+(2n+4)=114

Solve for n. Combine like terms on the left:

6n+6=114

Subtract 6 from both sides:

6n=108

Finally, divide both sides by 6. Therefore, the value of n is:

n=108/6=18

Therefore, our first even number is 2n or 2(18) which equates to 36.

And the consecutive even numbers will be 38 and 40.

So, the smallest of them, the first term, is 36.

4 0
3 years ago
Read 2 more answers
A spherical hot air balloon has a diameter of 55 feet when the balloon is inflated the radius increases at a rate of 1.5 feet pe
Nuetrik [128]

Answer: 46.90mins

Step-by-step explanation:

The given data:

The diameter of the balloon = 55 feet

The rate of increase of the radius of the balloon when inflated = 1.5 feet/min.

Solution:

dr/dt = 1.5 feet per minute = 1.5 ft/min

V = 4/3·π·r³

The maximum volume of the balloon

= 4/3 × 3.14 × 55³

= 696556.67 ft³

When the volume 2/3 the maximum volume

= 2/3 × 696556.67 ft³

= 464371.11 ft³

The radius, r₂ at the point is

= 4/3·π·r₂³

= 464371.11 ft³

r₂³ = 464371.11 ft³ × 3/4

= 348278.33 ft³

348278.333333

r₂ = ∛(348278.33 ft³) ≈ 70.36 ft

The time for the radius to increase to the above length = Length/(Rate of increase of length of the radius)

The time for the radius to increase to the

above length

Time taken for the radius to increase the length.

= is 70.369 ft/(1.5 ft/min)

= 46.90 minutes

46.90mins is the time taken to inflate the balloon.

5 0
3 years ago
It would be enchanting if you could help me out on this question, pease and crackers. thank you!(don't mind my weird language lo
Alina [70]

Answer:

615.44cm

Step by Step Answer:

A=πr^{2}=π·14^{2}≈615.75216

5 0
3 years ago
Zachary purchased a computer for 1800 on a payment plan. Three months after he purchased the computer, his balance was 1350. Fiv
zubka84 [21]

<u>Solution-</u>

Zachary purchased a computer for 1800 on a payment plan.  (Initial Money)

3 months after he bought the computer, his balance was 1350. (Money after 3 months)

Total money paid in 3 months = 1800-1350 = 450

Money paid per month = 450/3 = 150

5 months after he bought the computer, his balance was 1050.

Total spent = 1800-1050 = 750 = (5× 150)

So the equation that models the balance b after m months,

b = 1800 - m(150)

∴ Here, the slope signifies the constant monthly deduction of $150.


3 0
3 years ago
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