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3 years ago

LII Animals Seen in the Forest 20 Number of Animals Deer Birds Bears Squirrels

Mathematics

1 answer:

Darya [45]3 years ago

4 0

What’s the question?

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I need help with this <br>

horrorfan [7]

The display got this answer because you forgot to put the decimal point on 42.8 and typed in 428 instead this is why the display shows the wrong answer. You can estimate the answer by seeing the numbers are really low and the answer can’t be that big.

4 0

3 years ago

The interior angles formed by the sides of a quadrilateral have measures that sum to 360°.

Svetach [21]

Let

$angle 1=2x$ °

$angle 2=2x$ °

$angle 3=124$ °

$angle 4=124$ °

we know that

The interior angles formed by the sides of a quadrilateral have measures that sum to 360°

$angle 1+angle 2+angle 3+angle 4=360$ °

$2x+2x+124+124=360\\ 4x+248=360\\ 4x=112\\ x=28degrees$

therefore

the answer is

$x=28degrees$

4 0

3 years ago

Read 2 more answers

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

3 years ago

HELP!!!!

Harman [31]

The lateral area is the area of everything but the base. A shortcut formula for this area is Ph. (where P is the perimeter of the base and h is the height of the prism).

So the lateral area is (3+9+9.49)x37 = 795.13 m^2 or 795 m^2

The surface area is the lateral area plus the area of the 2 bases. The base is a triangle. (area = 1/2 bh)

Since there are two identical bases, the total area of the base is just 3x9 = 27

So the surface area is 795.13 + 27 = 822.13 m^2 or 822 m^2.

3 0

3 years ago

Simplify 24 ÷ (-2)(3) + 7Simplify 24 ÷ (-2)(3) + 7

Mnenie [13.5K]

Solving the expression $24\div (-2)(3) + 7$ we get -29