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3 years ago

What is the slope line No slope 1 0 -2

Mathematics

2 answers:

vlada-n [284]3 years ago

8 0

Answer:

Step-by-step explanation:

Horizontal lines always have 0 slope.

One brainliest away from virtuoso, help is appreaciated :)

REY [17]3 years ago

3 0

The correct answer is 0

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Simplify completely the quantity 4 times x to the third power plus 5 times x to the 2nd power plus 3 times x all over x. (2 poin

julia-pushkina [17]

We start with

$\dfrac{4x^3+5x^2+3x}{x}$

We can factor x at the numerator:

$\dfrac{x(4x^2+5x+3)}{x}$

So, assuming $x\neq 0$ (otherwise the expression would make no sense) we can simplify it:

$\dfrac{x(4x^2+5x+3)}{x} = 4x^2+5x+3$

This equation has no roots, so we can't simplify it any further.

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3 years ago

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3 years ago

Find the difference 5 2/3 - 3 1/5 A. 5 2/3 B. 8 13/15 C. 2 7/15 D. 3 1/5

RSB [31]

To find the difference of given expression:

$5\frac{2}{3} -3\frac{1}{5}$

First step is to subtract the whole number5 and 3.

So, 5 - 3 = 2.

Now we can find the difference of fractions.

To subtract the fractions , we need to find the common denominator.

Common denominator of 3 and 5 is 15.

So, we need to make both the denominator equal to 15.

$\frac{2}{3} =\frac{2*5}{3*5} =\frac{10}{15}$

Similarly, $\frac{1}{5} =\frac{1*3}{3*5} = \frac{3}{15}$

Therefore, $\frac{2}{3} -\frac{1}{5} =\frac{10}{15} -\frac{3}{15}$

= $\frac{10-3}{15}$

= $\frac{7}{15}$

So, the difference of given expression is $2\frac{7}{15}$ .

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3 years ago

Solve the following by using mathematical Induction. For >/ 1

cluponka [151]

Answer:

See explanation

Step-by-step explanation:

Prove that

$1^2+2^2+3^3+...+n^2=\dfrac{1}{6}n(n+1)(2n+1)$

1. When $n=1,$ we have

in left part $1^2=1;$

in right part $\dfrac{1}{6}\cdot 1\cdot (1+1)\cdot (2\cdot 1+1)=\dfrac{1}{6}\cdot 1\cdot 2\cdot 3=1.$

2. Assume that for all $k$ following equality is true

$1^2+2^2+3^3+...+k^2=\dfrac{1}{6}k(k+1)(2k+1)$

3. Prove that for $k+1$ the following equality is true too.

$1^2+2^2+3^3+...+(k+1)^2=\dfrac{1}{6}(k+1)((k+1)+1)(2(k+1)+1)$

Consider left part:

$1^2+2^2+3^2+...+(k+1)^2=\\ \\=(1^2+2^2+3^3+...+k^2)+(k+1)^2=\\ \\=\dfrac{1}{6}k(k+1)(2k+1)+(k+1)^2=\\ \\=(k+1)\left(\dfrac{1}{6}k(2k+1)+k+1\right)=\\ \\=(k+1)\dfrac{2k^2+k+6k+6}{6}=\\ \\=(k+1)\dfrac{2k^2+7k+6}{6}=\\ \\=(k+1)\dfrac{2k^2+4k+3k+6}{6}=\\ \\=(k+1)\dfrac{2k(k+2)+3(k+2)}{6}=\\ \\=(k+1)\dfrac{(k+2)(2k+3)}{6}$