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2 years ago

HELP ME OUT WITH THIS PLEASE I GIVE BRAINLIST and 100 points ❤️

Mathematics

2 answers:

Galina-37 [17]2 years ago

6 0

Answer:

incorrect length is 12.5

actual length is 11

12.5-11

1.5

1.5/12.5 ×100

20%

Step-by-step explanation:

hope it really helps you ^o^

Harman [31]2 years ago

5 0

Answer:

incorrect length is 12.5

actual length is 11

12.5-11

1.5

1.5/12.5 ×100

20%

Step-by-step explanation:

hope it helps

You might be interested in

If sinA =3/5 and A+B =90 find cosB

-Dominant- [34]

$A+B=90^o\to B=90^o-A\\\\\cos B=\cos(90^o-A)=\cos90^o\cos A+\sin90^o\sin A\\\\=\cos A+1\cdot\dfrac{3}{5}=0+\dfrac{3}{5}=\dfrac{3}{5}\\\\\boxed{\cos B=\dfrac{3}{5}}$

$Used:\\\\\cos(x+y)=\cos x\cos y+\sin x\sin y\\\\\cos90^o=0,\ sin90^o=1$

Other method:

Look at the picture.

$\sin\theta=\dfrac{opposite}{hypotenuse}\\\\\sin A=\dfrac{3}{5}\to opposite=3\ and\ hypotenuse=5$

$\cos\theta=\dfrac{adjacent}{hypotenuse}\\\\adjacent=3\ and\ hypotenuse=5\to\cos B=\dfrac{3}{5}$

8 0

2 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

2 years ago

A bounce house can hold 15 children. Seven children go in the bounce house. Write and solve an inequality that represents the ad

Ksivusya [100]

15 per child 7 go in so 15-7=8 the answer is 8 more children can jump is the bounce house

Hope this helped

8 0

2 years ago

Read 2 more answers

in a school ,the number of boys who play cricket is 3 times as much as a number of boys who play tennis. if 12 boys, who play cr

Ierofanga [76]

Answer:

The number of boys who play tennis is 36.

Step-by-step explanation:

Let the number of boys who play tennis is x.

number of boys who play cricket = y

x = 3 y..... (1)

12 boys plays both cricket and tennis.

According to the question,

x - 12 = y + 1 2

3 y - 12 = y + 12

2 y = 24

y = 12

So, the number of boys who plays tennis is x = 3 x 12 = 36.

8 0

2 years ago

Evaluate the integral ∫2−1|x−1|dx

defon

I think you might be referring to the definite integral,

$\displaystyle \int_{-1}^2|x-1|\,\mathrm dx$

Recall the definition of absolute value:

$|x| = \begin{cases}x&\text{if }x\ge0\\-x&\text{if }x$

Then $|x-1|=x-1$ if $x\ge1$ , and $|x-1|=1-x$ is $x$ . So spliting up the integral at x = 1, we have

$\displaystyle \int_{-1}^2|x-1|\,\mathrm dx = \int_{-1}^1(1-x)\,\mathrm dx + \int_1^2(x-1)\,\mathrm dx$

The rest is simple:

$\displaystyle \int_{-1}^2|x-1|\,\mathrm dx = \left(x-\dfrac{x^2}2\right)\bigg|_{-1}^1 + \left(\dfrac{x^2}2-x\right)\bigg|_1^2 \\\\ = \left(\left(1-\frac12\right)-\left(-1-\frac12\right)\right) + \left(\left(2-2\right)-\left(\frac12-1\right)\right) \\\\ = \boxed{\frac52}$

5 0

2 years ago