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Mila [183]
3 years ago
9

I’m not good at graphs please help

Mathematics
2 answers:
topjm [15]3 years ago
7 0

Answer:

I cant see the picher its dark

Step-by-step explanation:

I would help u but cant see it

pantera1 [17]3 years ago
4 0

It’s pitch black I can’t help without seeing it I’m sorry :(
You might be interested in
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
I need help asap:) thanks
Len [333]

Answer:

More than one triangle can be formed by three angles measuring 75°, 45° and 60°. Solution : We know that, The sum of the measures of all angles of a triangle should be equal to 180

Step-by-step explanation:

7 0
3 years ago
Jorge wants to buy new vinyl flooring for his kitchen. The kitchen floor is 12 feet by 15 feet. How many square yards is the flo
adoni [48]

Answer:

180 square yards

Step-by-step explanation:

5 0
4 years ago
Solve for w.<br> - 2w – 14= 2(w+1)<br> Simplify your answer as much
snow_tiger [21]

Answer:

w = -4

Step-by-step explanation:

<u>Step 1:  Distribute</u>

-2w - 14 = 2(w + 1)

-2w - 14 = 2w + 2

<u>Step 2:  Add 2w to both sides</u>

-2w - 14 + 2w = 2w + 2 + 2w

-14 = 4w + 2

<u>Step 3:  Subtract 2 from both sides</u>

-14 - 2 = 4w + 2 - 2

-16 = 4w

<u>Step 4:  Divide both sides by 4</u>

-16 / 4 = 4w / 4

<em>-4 = w</em>

<em />

Answer:  w = -4

7 0
4 years ago
Read 2 more answers
What two factors that can be multiply by 19
Vsevolod [243]
19 is a prime number so 1, 19
hope this helps!
5 0
3 years ago
Read 2 more answers
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