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3 years ago

30 pts please answer

Mathematics

2 answers:

ddd [48]3 years ago

4 0

I think it's d but can't be sure

qaws [65]3 years ago

3 0

Answer:

The answer is D

Step-by-step explanation:

I hope this helps

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Over the past ten years, the town's population doubled in size. The population is currently 12,000. What was the population ten

Alekssandra [29.7K]

6,000 right? I think

7 0

3 years ago

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it rained 2.79 inches in july. What was the average daily rainfall in July? (Hint: July has 31 days.)

natali 33 [55]

Rainfall in July = 2.79 inches.

July has 31 days.

2.79 : 31 = x : 1 Set up the proportion

2.79 / 31 = x/1 Cross multiply and switch.

2.79*1 / 31 = x Do the division

0.09 inches = x Answer

3 0

3 years ago

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What are the values a, b, and c in the following quadratic equation?<br><br> −6x = −8x2 − 13

lesya692 [45]

Answer:

Quadratic equation follows form: ax^2 + bx + c = 0

a = 8

b = - 6

c = 13

4 0

3 years ago

Identify az of this sequence: 0.25, 0.5, 0.75, 1, 1.25. 1.5

4vir4ik [10]

<h3><u>Question:</u></h3>

Identify a3 of this sequence: 0.25, 0.5, 0.75, 1, 1.25, 1.5,...a3=

<h3><u>Answer:</u></h3>

The third term of sequence is 0.75

$a_3 = 0.75$

<h3><u>Solution:</u></h3>

Given that, sequence is:

0.25, 0.5, 0.75, 1, 1.25. 1.5

Let us find the difference between terms

$0.5 - 0.25 = 0.25\\\\0.75-0.5 = 0.25\\\\1.25-1=0.25\\\\1.5-1.25=0.25$

This is a arithmetic sequence

Because the difference between any term and its immediately preceding term is always 0.25

In a arithmetic sequence,

$a_1 = \text{ first term of sequnece }\\\\a_2 = \text{second term of sequnce }\\\\a_3 = \text{third term of sequnece }\\\\a_n = \text{ nth term of sequnce }$

Thus, in the sequence 0.25, 0.5, 0.75, 1, 1.25. 1.5

$a_3 = \text{ third term } = 0.75$

Thus the third term of sequence is 0.75

5 0

3 years ago

*Find the formula of the series 1+2²+3²+4²+....+n²*<br>

Sophie [7]

The formula is $\frac{n(n+1)(2n+1)}{6}$

What are series?

In mathematics, we can describe a series as adding infinitely many numbers or quantities to a given starting number or amount.

We will find the formula as shown as below:

Let $S=1+2^2+3^2+4^2+................+n^2$

We know $(n+1)^3=n^3+3n^2+3n+1$

$(1+1)^3=1^3+3(1)^2+3(1)+1$

$(2+1)^3=2^3+3(2)^2+3(2)+1$

$(3+1)^3=3^3+3(3)^2+3(3)+1$

$(n+1)^3=n^3+3(n)^2+3(n)+1$

On adding

$2^3+3^3+4^3......(n+1)^3=(1^3+2^3+3^3+.....+n^3)+3(1^2+2^2+.....+n^2)+3(1+2+3....n)+(1+1+1+....+1)$

$2^3+3^3+4^3......(n+1)^3-(1^3+2^3+3^3+.....+n^3)=3S+\frac{3n(n+1)}{2}+(1+1+1+....+1)$

$(n+1)^3-1^3=3S+\frac{3n(n+1)}{2} +n$

$n^3+3(n)^2+3(n)+1-1=3S+\frac{3n(n+1)}{2} +n$

$2n^3+6n^2+6n=6S+3n^2+3n+2n$

$6S=2n^3+3n^2+n$

$6S=2n^2(n+1)+n(n+1)$

$6S=(n+1)(2n^2+n)$

$6S=n(n+1)(2n+1)$

$S=\frac{n(n+1)(2n+1)}{6}$

Hence, the formula is $\frac{n(n+1)(2n+1)}{6}$

Learn more about Series here:

brainly.com/question/24643676

#SPJ1

3 0

2 years ago