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ch4aika [34]
4 years ago
12

9. Consider a pattern that begins with 28 and each consecutive number is six more than the previous term. What are the first thr

ee terms of this pattern?
Mathematics
1 answer:
Art [367]4 years ago
6 0

Answer:

The first three numbers in the pattern would be 6, 12, and 18 since the rule pattern goes by six. I'm not sure if this is what you were looking for but I hope I helped!

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your parents are renting an apartment for you when you go away to college. An annual contract is $502.00/month with a 2-month pe
11111nata11111 [884]
114 per month, That is what you meant right?
8 0
4 years ago
Find the equation of the line with x-intercept 4 and y-intercept -1 in slope-intercept<br> form.
Nastasia [14]

Answer:

Step-by-step explanation:

The slope-intercept form is y=mx+b, where m is the slope and b is the y-coordinate of the y-intercept, so we can already conclude that b=-1.

This results in y = mx - 1

We also have the x-intercept, and we know that the y-coordinate of an x-intercept is always 0. This means that we have the point (4,0). Substitute this into the formula to generate m:

0 = 4m - 1

4m = 1

m = 1/4

thus y = \frac{1}{4}x - 1

7 0
3 years ago
9:45 on Tuesday Questions
LenaWriter [7]

Answer:  -2x + 5y = 8

Step-by-step explanation:

Since the lines must be parallel, the slopes must be the same, so keep the coefficients of x and y

Substitute the values of x and y from the given coordinate  (6,4) to find the new constant, b, the y-intercept.

-2x + 5y = b

-2(6) + 5(4) = b

-12 + 20 = b

8 = b

Rewrite the original equation substituting the new b for the original -15

-2x + 5y = 8

5 0
4 years ago
Calculate the limit values:
Nataliya [291]
A) This particular limit is of the indeterminate form,
\frac{ \infty }{ \infty }
if we plug in infinity directly, though it is not a number just to check.

If a limit is in this form, we apply L'Hopital's Rule.

's
Lim_{x \rightarrow \infty } \frac{ ln(x ^{2} + 1 ) }{x} = Lim_ {x \rightarrow \infty } \frac{( ln(x ^{2} + 1 ) ) '}{x ' }
So we take the derivatives and obtain,

Lim_ {x \rightarrow \infty } \frac{ ln(x ^{2} + 1 ) }{x} = Lim_{x \rightarrow \infty } \frac{ \frac{2x}{x^{2} + 1} }{1}

Still it is of the same indeterminate form, so we apply the rule again,

Lim_{x \rightarrow \infty } \frac{ ln(x ^{2} + 1 ) }{x} = Lim_{x \rightarrow \infty } \frac{ 2 }{2x}

This simplifies to,

Lim_{x \rightarrow \infty } \frac{ ln(x ^{2} + 1 ) }{x} = Lim_{x \rightarrow \infty } \frac{ 1 }{x} = 0

b) This limit is also of the indeterminate form,

\frac{0}{0}
we still apply the L'Hopital's Rule,

Lim_ {x \rightarrow0 }\frac{ tanx}{x} = Lim_ {x \rightarrow0 } \frac{ (tanx)'}{x ' }

Lim_ {x \rightarrow0 }\frac{ tanx}{x} = Lim_ {x \rightarrow0 } \frac{ \sec ^{2} (x) }{1 }

When we plug in zero now we obtain,

Lim_ {x \rightarrow0 }\frac{ tanx}{x} = Lim_ {x \rightarrow0 } \frac{ \sec ^{2} (0) }{1 } = \frac{1}{1} = 1
c) This also in the same indeterminate form

Lim_ {x \rightarrow0 }\frac{ {e}^{2x} - 1 - 2x}{ {x}^{2} } = Lim_ {x \rightarrow0 } \frac{ ({e}^{2x} - 1 - 2x)'}{( {x}^{2} ) ' }

Lim_ {x \rightarrow0 }\frac{ {e}^{2x} - 1 - 2x}{ {x}^{2} } = Lim_ {x \rightarrow0 } \frac{ (2{e}^{2x} - 2)}{ 2x }

It is still of that indeterminate form so we apply the rule again, to obtain;

Lim_ {x \rightarrow0 }\frac{ {e}^{2x} - 1 - 2x}{ {x}^{2} } = Lim_ {x \rightarrow0 } \frac{ (4{e}^{2x} )}{ 2 }

Now we have remove the discontinuity, we can evaluate the limit now, plugging in zero to obtain;

Lim_ {x \rightarrow0 }\frac{ {e}^{2x} - 1 - 2x}{ {x}^{2} } = \frac{ (4{e}^{2(0)} )}{ 2 }

This gives us;

Lim_ {x \rightarrow0 }\frac{ {e}^{2x} - 1 - 2x}{ {x}^{2} } =\frac{ (4(1) )}{ 2 }=2

d) Lim_ {x \rightarrow +\infty }\sqrt{x^2+2x}-x

For this kind of question we need to rationalize the radical function, to obtain;

Lim_ {x \rightarrow +\infty }\frac{2x}{\sqrt{x^2+2x}+x}

We now divide both the numerator and denominator by x, to obtain,

Lim_ {x \rightarrow +\infty }\frac{2}{\sqrt{1+\frac{2}{x}}+1}

This simplifies to,

=\frac{2}{\sqrt{1+0}+1}=1
5 0
4 years ago
Mode and median mean and range of <br><br> 14,19,16,13,16,14
valentinak56 [21]
Mean = 15 1/3
Median = 15
Mode = 14,16
Range = 6
5 0
3 years ago
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