What is another way to write this number 300+70+5/10+8/100

P and q are integers that are multiple of 5. Which of this

zhuklara [117]

Answer:

we conclude that the only option (a) is true.

Step-by-step explanation:

As we know that the multiples of 5 are the numbers which we get when we multiply by 5.

i.e.

5×1=5

5×2=10

Here, 5 and 10 are multiples of 5.

Let p and q are integers that are multiples of 5.

Let us consider

p=5

q=10

so

p+q=5+10

= 15

A number is divisible by 5 if it ends in 5 or 0.

i.e. 15/5 = 3

so p+q is divisible by 3 as there is no remainder left.

Therefore, option (a) is true.

Checking the other options:

(b) P –q is divisible by 10

As

p=5

q=10

so

p-q=5-10

= -5

Numbers that are divisible by 10 need to be even or divisible by 2 and divisible by 5.

As -5 is not divisible by 2.

So, option b is NOT true.

(c) P +q is divisible by 20

As

p=5

q=10

so

p-q=5+10

= 15

Divisibility rule of 20 implies that the last two digits of the number are either 00 or divisible by 20 .

Therefore, P + q= 20 is not divisible by 20 as we don't get the whole number.

(d) P + q is divisible by 25

As

p=5

q=10

so

p-q=5+10

= 15

p+q=15 is not divisible by 25 as it does not end with 00, 25, 50, or 75.

so, option d is NOT correct.

Therefore, we conclude that the only option (a) is true.

5 0

2 years ago

What is the sum of the first seven terms of the geometric series? 3 12 48 192 ...

Ad libitum [116K]

Hence, the sum of the given GP is 16347

<h2>What is a series?</h2>

a series is the cumulative sum of a given sequence of terms. Typically, these terms are real or complex numbers, but much more generality is possible.

<h3>How to solve?</h3>

we can identify from the given geomertric series.

first term = a(say) = 3

geometric factor of progression = r(say) = 4

Sum of the first 7 terms = a( $\frac{r^n-1}{r-1}$ ) ,where n is 7

Sum = 3( $\frac{4^7-1}{3}$ ) = 16348-1

= 16347

Hence, the sum of the given GP is 16347.

to learn more about series: brainly.com/question/12578626

#SPJ4

8 0

1 year ago

Find the product of (x + 5)(x − 5).

AfilCa [17]

Answer:

x² - 30 + 5x

Step-by-step explanation:

(x + 5)(x − 5)

=x(x − 5)+5(x − 5)

=x² - 5 + 5x - 25

=x² - 30 + 5x

6 0

2 years ago

Evaluate the summation from n equals 1 to infinity of the quotient of 3 and 2 raised the quantity n minus 1 power

Shalnov [3]

Answer:

3*∑ (1/2)^n = 6

Step-by-step explanation:

We have the summation from n = 1 to n = ∞, for:

∑ 3/(2)^(n - 1)

We know that the summation between k = 0, and k = N - 1 for:

∑ r^k = (1 - r^N)/(1 - r)

if we have the summation between k = 0 and k = ∞ - 1 = ∞

(here we used that ∞ is really big, then ∞ - 1 = ∞)

In the numerator we will have the term r^∞, if 0 < r < 1, then r^∞ = 0.

Then if we assume that 0 < r < 1 we can write:

∑ r^k = 1/(1 - r)

In our case, we can rewrite our summation as:

3*∑ (1/2)^n

for n = 0 to n = ∞

You can see that i changed the limits for n, this does not really matter because we are summing between a number and infinity, the only thing you need to take care is that now the power is n, instead of (n - 1)

Then we have r = (1/2) which is clearly smaller than 1.

then (1/2)

Then this summation is equal to:

3*∑ (1/2)^n = 3*( 1/(1 - 1/2)) = 3*( 1/( 2/2 - 1/2)) = 3*( 1/(1/2)) = 3*2 = 6

3*∑ (1/2)^n = 6

5 0

3 years ago

1.

Alex73 [517]

Answer:

2x + 7= 4x + 17

7=2x+17

-10=2x

-10/2=2x/2

x=−5

5 0

2 years ago