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3 years ago

Please answer ASAP!!!.........................................................

Mathematics

2 answers:

Olin [163]3 years ago

6 0

L=11.5m
I really hope this will help u

Mnenie [13.5K]3 years ago

3 0

11.5 is your answer

hope this helps

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The midpoint of AB is M (2,0). If the coordinates of A are (7, -2), what

Lubov Fominskaja [6]

Answer:

The coordinates of point B are (-3,2)

Step-by-step explanation:

we know that

The formula to calculate the midpoint between two points is equal to

$M(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$

we have

$M(2,0)\\(x_1,y_1)=(7,-2)$

substitute

$(2,0)=(\frac{7+x_2}{2},\frac{-2+y_2}{2})$

Find the value of x_2

we have that

$2=\frac{7+x2}{2}$

$4=7+x_2\\x_2=-3$

Find the value of y_2

we have that

$0=\frac{-2+y_2}{2}$

$0=-2+y_2\\y_2=2$

therefore

The coordinates of point B are (-3,2)

5 0

3 years ago

What is the equation of the following line? Be sure to scroll down first to see

valina [46]

Answer:

40-2?

Step-by-step explanation:

well it sort of makes sense because there are 2, 0's and then a 4 and after -2

i hope i helped

4 0

4 years ago

Work out the bearings

mamaluj [8]

Answer:

Step-by-step explanation:

Use Pythagorean Theorem: a^2+b^2=c^2 (a squared plus b squared equals c squared), where a and b are legs from the 90 degree angle and c is the opposite side of the triangle.

So for part a: a=AB=30 b=AC=30 and c=BC

a^2 +b^2 = c^2 Plug in the knowns

30^2 +30^2=c^2 Use Pemdas, Exponents first.

30^2=900 Simplify the equation

900+900=c^2 Simplify again

1800=c^2 Now to get rid of the ^2 we need to know the square root of 1800.

It is 42.426406871.........(There are more numbers in the decimal. Choose your stopping point and round) I'm going to round to the hundredths.

So, the answer to part a is 42.43 kilometers.

Now for part b. a=AD=45 b=AB=30 c=DB

Same equation

a^2 + b^2 =c^2 Plug in the knowns

45^2 + 30^2 =c^2 Simplify

2025 + 900 =c^2 Simplify

2925 =c^2 Find the square root

54.3083269 Round to the decimal place needed. I'm going with Hundredths.

Part b answer is 54.31 kilometers.

5 0

4 years ago

The difference between number 20 and its opposite is what percentage of 200? (if you do not remember: opposite of 15 is −15; opp

Lesechka [4]

The opposite of 20 is -20 and the difference is 40. If we put 40/200 we get 0.2, or 20%.

4 0

3 years ago

Read 3 more answers

\lim _{x\to 0}\left(\frac{2x\ln \left(1+3x\right)+\sin \left(x\right)\tan \left(3x\right)-2x^3}{1-\cos \left(3x\right)}\right)

Vinvika [58]

$\displaystyle \lim_{x\to 0}\left(\frac{2x\ln \left(1+3x\right)+\sin \left(x\right)\tan \left(3x\right)-2x^3}{1-\cos \left(3x\right)}\right)$

Both the numerator and denominator approach 0, so this is a candidate for applying L'Hopital's rule. Doing so gives

$\displaystyle \lim_{x\to 0}\left(2\ln(1+3x)+\dfrac{6x}{1+3x}+\cos(x)\tan(3x)+3\sin(x)\sec^2(x)-6x^2}{3\sin(3x)}\right)$

This again gives an indeterminate form 0/0, but no need to use L'Hopital's rule again just yet. Split up the limit as

$\displaystyle \lim_{x\to0}\frac{2\ln(1+3x)}{3\sin(3x)} + \lim_{x\to0}\frac{6x}{3(1+3x)\sin(3x)} \\\\ + \lim_{x\to0}\frac{\cos(x)\tan(3x)}{3\sin(3x)} + \lim_{x\to0}\frac{3\sin(x)\sec^2(x)}{3\sin(3x)} \\\\ - \lim_{x\to0}\frac{6x^2}{3\sin(3x)}$

Now recall two well-known limits:

$\displaystyle \lim_{x\to0}\frac{\sin(ax)}{ax}=1\text{ if }a\neq0 \\\\ \lim_{x\to0}\frac{\ln(1+ax)}{ax}=1\text{ if }a\neq0$

Compute each remaining limit:

$\displaystyle \lim_{x\to0}\frac{2\ln(1+3x)}{3\sin(3x)} = \frac23 \times \lim_{x\to0}\frac{\ln(1+3x)}{3x} \times \lim_{x\to0}\frac{3x}{\sin(3x)} = \frac23$

$\displaystyle \lim_{x\to0}\frac{6x}{3(1+3x)\sin(3x)} = \frac23 \times \lim_{x\to0}\frac{3x}{\sin(3x)} \times \lim_{x\to0}\frac{1}{1+3x} = \frac23$

$\displaystyle \lim_{x\to0}\frac{\cos(x)\tan(3x)}{3\sin(3x)} = \frac13 \times \lim_{x\to0}\frac{\cos(x)}{\cos(3x)} = \frac13$

$\displaystyle \lim_{x\to0}\frac{3\sin(x)\sec^2(x)}{3\sin(3x)} = \frac13 \times \lim_{x\to0}\frac{\sin(x)}x \times \lim_{x\to0}\frac{3x}{\sin(3x)} \times \lim_{x\to0}\sec^2(x) = \frac13$

$\displaystyle \lim_{x\to0}\frac{6x^2}{3\sin(3x)} = \frac23 \times \lim_{x\to0}x \times \lim_{x\to0}\frac{3x}{\sin(3x)} \times \lim_{x\to0}x = 0$

So, the original limit has a value of

2/3 + 2/3 + 1/3 + 1/3 - 0 = 2

6 0

3 years ago