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Leno4ka [110]
2 years ago
14

The graph of a line goes up and to the right when:

Mathematics
2 answers:
olasank [31]2 years ago
8 0

Answer:b

Step-by-step explanation:

Afina-wow [57]2 years ago
4 0
<h3><u>Answer:</u></h3>

Option: C is the correct answer.

c) the coefficient of x is positive

<h3><u>Step-by-step explanation:</u></h3>

We know that graph of a line goes up and to the right when the coefficient of x is positive.

Since we know that when a line goes up and to the right this means that the line is increasing and hence we will get a positive slope of the line and we know that the slope intercept of a line is given as:

y=mx+c

as m is positive this means that:

the coefficient of x is positive.

Hence, the correct answer is:

c) the coefficient of x is positive

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You play basketball at your school's
vaieri [72.5K]

Answer:

20 + 2x = 4x

Step-by-step explanation:

So you are setting the two expressions equal to each other.

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20 + 2x = 4x

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8 0
3 years ago
Can somebody prove this mathmatical induction?
Flauer [41]

Answer:

See explanation

Step-by-step explanation:

1 step:

n=1, then

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So, for j=1 this statement is true

2 step:

Assume that for n=k the following statement is true

\sum \limits_{j=1}^k2^j=2(2^k-1)

3 step:

Check for n=k+1 whether the statement

\sum \limits_{j=1}^{k+1}2^j=2(2^{k+1}-1)

is true.

Start with the left side:

\sum \limits _{j=1}^{k+1}2^j=\sum \limits _{j=1}^k2^j+2^{k+1}\ \ (\ast)

According to the 2nd step,

\sum \limits_{j=1}^k2^j=2(2^k-1)

Substitute it into the \ast

\sum \limits _{j=1}^{k+1}2^j=\sum \limits _{j=1}^k2^j+2^{k+1}=2(2^k-1)+2^{k+1}=2^{k+1}-2+2^{k+1}=2\cdot 2^{k+1}-2=2^{k+2}-2=2(2^{k+1}-1)

So, you have proved the initial statement

4 0
3 years ago
How do you solve this equation√8k=k
kondor19780726 [428]

the answer is                        

<span>[√8k ]² =k², so  8k=k² which implies k² - 8k= 0, and (k – 8)k= 0, so k = 8 or k =0</span>

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2 years ago
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