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3 years ago

The multiplicative inverse of - 5 and 10v - 5

Mathematics

1 answer:

Digiron [165]3 years ago

3 0

The answer is 1/5 hope this helps

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• a helicopter hovering at 10 m above the ground is shining a spotlight at a person 30 m away1 . if the person is 2 m tall how l

pshichka [43]

As shown in the figure

Height of helicopter above the ground=30 m

Distance from the base to the point where the person is standing=30 m

height of the person=2 m

Now the person cast the shadow of length x m.

Triangle ABC and triangle EDC are similar.

∵ ∠B= ∠D=90°

∠C is common.

So by AA similarity ΔABC and ΔEDC are similar.

As we know when triangles are similar their sides are proportional.

AB/AD =BC/DC

Let DC=x meter

⇒ $\frac{10}{x}=\frac{30+x}{x}$

⇒5=\frac{30+x}{x}[/tex]

⇒5x=30+x

⇒4x = 30

⇒ x=30/4

⇒ x=7.5 meter

So length of shadow=7.5 meter

2. volume of a sphere of radius r is v (r) = $\frac{4}{3}π r^{3}$

$\frac{\mathrm{d} }{\mathrm{d} r}V$ =4/3π×3 $r^2$

=4π $r^2$

surface area of a sphere of radius r is 4π $r^2$

b) Ratio of derivative of volume of sphere to surface area of sphere= $\frac{4\pi r^2}{4\pi r^2}$

=1 [ incomplete question but you have written few words ]

4 0

3 years ago

PLEASE HELP WITH GEOMETRY WILL GIVE BRAINLIEST

sammy [17]

Answer:

Option D

Step-by-step explanation:

In the first three options we can evaluate the values of x with the help of Sine Rule for all the triangles

The Rule says

$\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}$

Where a , b and c are the sides opposite to the angles A, B and C of any ΔABC.

Hence , in order to determine unknown values of sides or angles , we need any 3 values from all 3 sides and 3 angles in a triangle. First Three options give us three values but the last option gives only 2.

Example

in first option we can apply Sine Rule as

$\frac{\sin 84}{x}=\frac{\sin 63}{3}$

$\frac{0.994}{x}=\frac{0.891}{3}$

$x=\frac{0.991 \times 3}{0.891}$

$x=\frac{2.973}{0.891}$

$x=3.336$

8 0

4 years ago

Determine whether the pair of equations represent parallel lines, perpendicular lines, or neither.

Ivahew [28]

Answer:

Parallel

Step-by-step explanation:

Parallel lines have the same slope but different y-intercepts. If you multiply the top equation by 2, you get:

2(12x + 4y = 16)

24x + 8y = 32

This shows that both lines have the same slope, but then you find the y-intercepts, they are different:

1st equation y-int = 4

2nd equation y-int = 9/2 or 4.5

6 0

3 years ago

How do you find the limit?

coldgirl [10]

Answer:

2/5

Step-by-step explanation:

Hi! Whenever you find a limit, you first directly substitute x = 5 in.

$\displaystyle \large{ \lim_{x \to 5} \frac{x^2-6x+5}{x^2-25}}\\

\displaystyle \large{ \lim_{x \to 5} \frac{5^2-6(5)+5}{5^2-25}}\\

\displaystyle \large{ \lim_{x \to 5} \frac{25-30+5}{25-25}}\\

\displaystyle \large{ \lim_{x \to 5} \frac{0}{0}}$

Hm, looks like we got 0/0 after directly substitution. 0/0 is one of indeterminate form so we have to use another method to evaluate the limit since direct substitution does not work.

For a polynomial or fractional function, to evaluate a limit with another method if direct substitution does not work, you can do by using factorization method. Simply factor the expression of both denominator and numerator then cancel the same expression.

From x²-6x+5, you can factor as (x-5)(x-1) because -5-1 = -6 which is middle term and (-5)(-1) = 5 which is the last term.

From x²-25, you can factor as (x+5)(x-5) via differences of two squares.

After factoring the expressions, we get a new Limit.

$\displaystyle \large{ \lim_{x\to 5}\frac{(x-5)(x-1)}{(x-5)(x+5)}}$

We can cancel x-5.

$\displaystyle \large{ \lim_{x\to 5}\frac{x-1}{x+5}}$

Then directly substitute x = 5 in.

$\displaystyle \large{ \lim_{x\to 5}\frac{5-1}{5+5}}\\

\displaystyle \large{ \lim_{x\to 5}\frac{4}{10}}\\

\displaystyle \large{ \lim_{x\to 5}\frac{2}{5}=\frac{2}{5}}$

Therefore, the limit value is 2/5.

L’Hopital Method

I wouldn’t recommend using this method since it’s too easy but only if you know the differentiation. You can use this method with a limit that’s evaluated to indeterminate form. Most people use this method when the limit method is too long or hard such as Trigonometric limits or Transcendental function limits.

The method is basically to differentiate both denominator and numerator, do not confuse this with quotient rules.

So from the given function:

$\displaystyle \large{ \lim_{x \to 5} \frac{x^2-6x+5}{x^2-25}}$

Differentiate numerator and denominator, apply power rules.

Differential (Power Rules)

$\displaystyle \large{y = ax^n \longrightarrow y\prime= nax^{n-1}$

Differentiation (Property of Addition/Subtraction)

$\displaystyle \large{y = f(x)+g(x) \longrightarrow y\prime = f\prime (x) + g\prime (x)}$

Hence from the expressions,

$\displaystyle \large{ \lim_{x \to 5} \frac{\frac{d}{dx}(x^2-6x+5)}{\frac{d}{dx}(x^2-25)}}\\

\displaystyle \large{ \lim_{x \to 5} \frac{\frac{d}{dx}(x^2)-\frac{d}{dx}(6x)+\frac{d}{dx}(5)}{\frac{d}{dx}(x^2)-\frac{d}{dx}(25)}}$

Differential (Constant)

$\displaystyle \large{y = c \longrightarrow y\prime = 0 \ \ \ \ \sf{(c\ \ is \ \ a \ \ constant.)}}$

Therefore,

$\displaystyle \large{ \lim_{x \to 5} \frac{2x-6}{2x}}\\

\displaystyle \large{ \lim_{x \to 5} \frac{2(x-3)}{2x}}\\

\displaystyle \large{ \lim_{x \to 5} \frac{x-3}{x}}$