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2 years ago

PLSSS HELP IF YOU TURLY KNOW THISS

Mathematics

2 answers:

maria [59]2 years ago

7 0

Answer:

Step-by-step explanation:

hope this helps

RUDIKE [14]2 years ago

6 0

Answer:

Step-by-step explanation:

because 3 x 3 = 9

1 x 5 = 5

simplest form will be 1 4/5

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A city planner makes a scale drawing of a proposed playground. The length of the actual playground is 78 feet, and the width is

Allushta [10]

Answer:

B) 6.5 inches by 3.5 inches

Step-by-step explanation:

Multiplying the actual dimensions by the scale gives the scale dimensions:

(0.5 in)/(6 ft) × {78 ft, 42 ft}

= {39/6 in, 21/6 in}

= {6.5 in, 3.5 in}

The size on the scale drawing is 6.5 inches by 3.5 inches.

5 0

3 years ago

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PLS HELP Find the volume.

bazaltina [42]

Answer:

660

Step-by-step explanation:

V= l×w×h/3

V= 24×11×7.5/3

V=1 980÷3

V= 660

8 0

3 years ago

Which set of parametric equations represents the following function y=square root of x+3

Irina18 [472]

The given equation is:

$y= \sqrt{x+3}$

We have to find, which of the given set of parametric equations given in the options, result in the above equation:

The correct answer would be option A.

The equations in option A are:

$x(t) = 5t \\ 
y(t)= \sqrt{5t+3}$

From first equation we can see that 5t is equal to x. Using the value of 5th in second equation, we get the equation as:

$y= \sqrt{x+3}$

Therefore, the correct answer is option A

3 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}}$

Now we can substitute x = 5 in.

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

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

Thus, the limit value is 2/5 same as the first method.

Notes:

If you still get an indeterminate form 0/0 as example after using l’hopital rules, you have to differentiate until you don’t get indeterminate form.

8 0

3 years ago

I need help plz 99 points waiting for the person that answers the question

amid [387]

Answer:

Step-by-step explanation:

This is from a website so you might have to rephrase it but Direct variation describes a simple relationship between two variables . We say y varies directly with x (or as x , in some textbooks) if:

y=kx

for some constant k , called the constant of variation or constant of proportionality . (Some textbooks describe direct variation by saying " y varies directly as x ", " y varies proportionally as x ", or " y is directly proportional to x .")

This means that as x increases, y increases and as x decreases, y decreases—and that the ratio between them always stays the same.

The graph of the direct variation equation is a straight line through the origin.

4 0

3 years ago

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