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givi [52]
3 years ago
13

Need Help please stuck on this one.

Mathematics
1 answer:
ArbitrLikvidat [17]3 years ago
3 0

Answer:

option D

Step-by-step explanation:

According to Pascal's triangle

1

1 1

1 2 1

1 3 3 1

it is clear the second term will have an integer coefficient of 3

Therefore 3(x)²(-1)¹

3x²(-1)=-3x²

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If she doesn't include herself the no but if she includes herself then yes
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3 years ago
Enter T or F depending on whether the statement is true or not. (You must enter T or F -- True and False will not work.) If it i
slava [35]

Answer:

See below

Step-by-step explanation:

If a is divisible by 3 then a is divisible by 9

<u>FALSE </u>

Counter-example

6 is divisible by 3 but not by 9

The subtraction of 2 rational numbers is rational.

<u>TRUE </u>

<em>Proof </em>

If a, b are two rational numbers

\large a=\frac{p}{q}\;b=\frac{r}{s}

for some integers p, q, r, s.

Then

\large a-b=\frac{p}{q}-\frac{r}{s}=\frac{ps-qr}{qs}

since ps-qr and qs are integers, a-b is rational

A sufficient condition for an integer to be divisible by 8 is that it is divisible by 2

<u>FALSE </u>

<em>Counter-example </em>

4 is divisible by 2 but not by 8

A sufficient condition for an integer to be divisible by 6 is that it is divisible by 2

<u>FALSE </u>

<em>Counter-example </em>

4 is divisible by 2 but not by 6

If a is divisible by 9 then a is divisible by 3.

<u>TRUE </u>

<em>Proof</em>

If a is divisible by 9, then a = 9k for some integer k, but 9=3*3, so a = 3*(3k).

Since 3k is integer, a is divisible also by 3.

The product of 2 consecutive integers is even.

<u>TRUE </u>

<em>Proof </em>

Let p, q be two consecutive integers, then either p is even or odd.

Suppose first p is even. Then

p = 2n and q = 2n+1, so p*q=2n(2n+1)=2n*2n+2n=2(n*2n+n)

since (n*2n+n) is integer p*q is even.

Suppose now p is odd

p = 2n+1 q = 2n+2, then

p*q=(2n+1)(2n+2)=2n*2n+2*2n+2n+2=2(n*2n+2n+n+1)

since (n*2n+2n+n+1) is integer p*q is also even.

Answer the following questions:

30 division 3 =

10

30 mod 3 =

0 (the remainder when dividing 30 by 3)

-26 division 5 =

-5 (plus remainder -1)

-26 mod 5 =

-1 (the remainder when dividing -26 by 5)

28 division 4 =

7

28 mod 4 =

0

-29 division 10 =

-2 (plus remainder -9)

-29 mod 10 =

-9

24 division 9 =

2 (plus remainder 6)

24 mod 9 =

6

-28 division 6=

-4 (plus remainder -4)

-28 mod 6 =

-4

965255471 mod 101 =

87 (the remainder when dividing 965255471 by 101)

630153353 mod 101 =

11 (the remainder when dividing 630153353 by 101)

8 0
4 years ago
9. Combine the radical expression, if possible.
Anastaziya [24]

Answer:

we conclude that:

3\sqrt{125}+4\sqrt{20}=23\sqrt{5}    

Hence, the last option i.e. 23\sqrt{5} is correct.                      

Step-by-step explanation:

Given the expression

3\sqrt{125}+4\sqrt{20}

Combining the radical expressions

3\sqrt{125}+4\sqrt{20}

let us first solve

3\sqrt{125}

=3\sqrt{5^3}

=3\sqrt{5^2\cdot \:5}

Apply radical rule:  \sqrt{ab}=\sqrt{a}\sqrt{b},\:\quad \:a\ge 0,\:b\ge 0

=3\sqrt{5^2}\sqrt{5}

Apply radical rule:   \sqrt{a^2}=a,\:\quad \:a\ge 0

=3\cdot \:5\sqrt{5}

=15\sqrt{5}

Thus,

3\sqrt{125}=15\sqrt{5}

similarly solving

4\sqrt{20}

=4\sqrt{2^2\cdot \:5}

Apply radical rule:  \sqrt{ab}=\sqrt{a}\sqrt{b},\:\quad \:a\ge 0,\:b\ge 0

=4\sqrt{2^2}\sqrt{5}

Apply radical rule:   \sqrt{a^2}=a,\:\quad \:a\ge 0

=4\cdot \:2\sqrt{5}

=8\sqrt{5}

Thus,

4\sqrt{20}=8\sqrt{5}

so we get

3\sqrt{125}=15\sqrt{5}

4\sqrt{20}=8\sqrt{5}

so the expression becomes

3\sqrt{125}+4\sqrt{20}=15\sqrt{5}+8\sqrt{5}        ∵ 3\sqrt{125}=15\sqrt{5} , 4\sqrt{20}=8\sqrt{5}

                        =23\sqrt{5}                   ∵ 15\sqrt{5}+8\sqrt{5}=23\sqrt{5}

Therefore, we conclude that:

3\sqrt{125}+4\sqrt{20}=23\sqrt{5}    

Hence, the last option i.e. 23\sqrt{5} is correct.                    

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3 years ago
What is 75 plus something equals 250
stepan [7]
I'll let you do the work.

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