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3 years ago

Find the slope of the line.

Mathematics

1 answer:

nadya68 [22]3 years ago

5 0

Answer:

Formula: Y=mx+b...

Step-by-step explanation:

Y=mx+b

M= Rise over run = 1/1
b=y-intercept=0

Answer: Y=1x

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Question 6 of 9 (3•8) • 1 = 3 • (8.1) O A. True B. False

olga nikolaevna [1]

Answer:

No, (false). The question I assume was: $(3*8)*1=3*(8.1)$

Step-by-step explanation:

1. Multiple: 3 * 8 = 24

2. Multiple: The result of step No. 1: * 1 = 24 * 1 = 24

3. Conversion a decimal number to a fraction: 8.1 = $\frac{81}{10}$ = $\frac{81}{10}$

a) Write down the decimal 8.1 divided by 1: 8.1 = $\frac{81}{1}$

b) Multiply both top and bottom by 10 for every number after the

decimal point. (For example, if there are two numbers after the decimal

point, then use 100, if there are three then use 1000, etc.)

$\frac{8.1}{1}$ = $\frac{81}{10}$

Note: $\frac{81}{10}$ is called a decimal fraction.

c) Simplify and reduce the fraction

$\frac{81}{10} = \frac{81*1}{10*1} = \frac{81}{10}$

4. Multiple: 3 * 8.1 = $\frac{3*81}{1*10}$ = $\frac{243}{10}$

Multiply both numerators and denominators. Result fraction keep to lowest

possible denominator GCD(243, 10) = 1. In the next intermediate step the

fraction result cannot be further simplified by canceling.

In words - three multiplied by eighty-one tenths = two hundred forty-three

tenths.

5. Compare: The result of step No. 2 vs The result of step No. 4: = 24 vs $\frac{243}{10}$ = $\frac{24}{1}$ vs $\frac{243}{10}$ = $\frac{24*10}{1*10}$ vs $\frac{243}{10}$ = $\frac{240}{10}$ vs $\frac{243}{10}$ = 240 vs 243 = No

5 0

3 years ago

Solve the following absolute value equation.

cestrela7 [59]

Answer:

x= 8

x= -8

Step-by-step explanation:

|x+1| /3 = 3

|x+1| = 9

|x| + |1| = 9

|x| = 8

x= 8

x= -8

4 0

2 years ago

Use mathematical induction to prove the statement is true for all positive integers n, or show why it is false:

kondaur [170]

$\text{Proof by induction:}$
$\text{Test that the statement holds or n = 1}$

$LHS = (3 - 2)^{2} = 1$
$RHS = \frac{6 - 4}{2} = \frac{2}{2} = 1 = LHS$
$\text{Thus, the statement holds for the base case.}$

$\text{Assume the statement holds for some arbitrary term, n= k}$
$1^{2} + 4^{2} + 7^{2} + ... + (3k - 2)^{2} = \frac{k(6k^{2} - 3k - 1)}{2}$

$\text{Prove it is true for n = k + 1}$
$RTP: 1^{2} + 4^{2} + 7^{2} + ... + [3(k + 1) - 2]^{2} = \frac{(k + 1)[6(k + 1)^{2} - 3(k + 1) - 1]}{2} = \frac{(k + 1)[6k^{2} + 9k + 2]}{2}$

$LHS = \underbrace{1^{2} + 4^{2} + 7^{2} + ... + (3k - 2)^{2}}_{\frac{k(6k^{2} - 3k - 1)}{2}} + [3(k + 1) - 2]^{2}$
$= \frac{k(6k^{2} - 3k - 1)}{2} + [3(k + 1) - 2]^{2}$
$= \frac{k(6k^{2} - 3k - 1) + 2[3(k + 1) - 2]^{2}}{2}$
$= \frac{k(6k^{2} - 3k - 1) + 2(3k + 1)^{2}}{2}$
$= \frac{k(6k^{2} - 3k - 1) + 18k^{2} + 12k + 2}{2}$
$= \frac{k(6k^{2} - 3k - 1 + 18k + 12) + 2}{2}$
$= \frac{k(6k^{2} + 15k + 11) + 2}{}$
$= \frac{(k + 1)[6k^{2} + 9k + 2]}{2}$
$= \frac{(k + 1)[6(k + 1)^{2} - 3(k + 1) - 1]}{2}$
$= RHS$

Since it is true for n = 1, n = k, and n = k + 1, by the principles of mathematical induction, it is true for all positive values of n.

3 0

3 years ago

I wear my snow boots If and only if it snows.

hoa [83]

Yes u only we're it if itsnows

6 0

3 years ago

Yo someone answer this because if I get a bad grade my xbox is getting taken away

blsea [12.9K]

You just have to substitute the x for the given number and complete the equation. Leave the x-box and laziness and study

8 0

3 years ago