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2 years ago

Will literally make BRAINLIEST and marry you if you do this for me

Mathematics

1 answer:

Tanya [424]2 years ago

6 0

Answer:

0.44

Step-by-step explanation:

See the picture below.

Remember that the sine is the ratio of the length of the opposite leg to the length of the hypotenuse.

Perhaps you know SOHCAHTOA, where SOH means sine, opposite, hypotenuse.

For angle J, leg LK is the opposite leg.

The hypotenuse of the triangle is the side opposite the right angle, teh angle of 90°, which is angle L, so the hypotenuse is side KJ.

$\sin J = \dfrac{opposite~leg}{hypotenuse}$

$\sin J = \dfrac{39}{89}$

$\sin J = 0.43820...$

Answer: 0.44

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Find the value of Y<br> .....

vekshin1

Answer:

$y = 6 \sqrt{3} \: units$

Step-by-step explanation:

In order to find the value of y, first we need to find the length of the perpendicular dropped from one of the vertices of the triangle to its opposite side.

By geometric mean theorem:

Length of the perpendicular

$=\sqrt{9\times 3}$

$=\sqrt{27}\: units$

Next, by Pythagoras theorem:

${y}^{2} = {9}^{2} + {( \sqrt{27} )}^{2} \\ \\ {y}^{2} = 81 + 27 \\ \\ {y}^{2} = 108 \\ \\ y = \sqrt{108} \\ \\ y = \sqrt{36 \times 3} \\ \\ y = \sqrt{ {6}^{2} \times 3} \\ \\ y = 6 \sqrt{3} \: units$

4 0

2 years ago

Oliver traveled 124.25 miles in 3.5 at a constant speed. What is the constant of proportionality in miles per hour?

Darina [25.2K]

Oliver traveled 124.25 miles in 3.5 hours at a constant speed

Rate = 124.25 / 3.5 = 35.5 miles per hour (MPH)

Answer:

The constant of proportionality is 35.5 miles per hour (MPH)

5 0

3 years ago

1) Use Newton's method with the specified initial approximation x1 to find x3, the third approximation to the root of the given

neonofarm [45]

Answer:

Check below, please

Step-by-step explanation:

Hello!

1) In the Newton Method, we'll stop our approximations till the value gets repeated. Like this

$x_{1}=2\\x_{2}=2-\frac{f(2)}{f'(2)}=2.5\\x_{3}=2.5-\frac{f(2.5)}{f'(2.5)}\approx 2.4166\\x_{4}=2.4166-\frac{f(2.4166)}{f'(2.4166)}\approx 2.41421\\x_{5}=2.41421-\frac{f(2.41421)}{f'(2.41421)}\approx \mathbf{2.41421}$

2) Looking at the graph, let's pick -1.2 and 3.2 as our approximations since it is a quadratic function. Passing through theses points -1.2 and 3.2 there are tangent lines that can be traced, which are the starting point to get to the roots.

We can rewrite it as: $x^2-2x-4=0$

$x_{1}=-1.1\\x_{2}=-1.1-\frac{f(-1.1)}{f'(-1.1)}=-1.24047\\x_{3}=-1.24047-\frac{f(1.24047)}{f'(1.24047)}\approx -1.23607\\x_{4}=-1.23607-\frac{f(-1.23607)}{f'(-1.23607)}\approx -1.23606\\x_{5}=-1.23606-\frac{f(-1.23606)}{f'(-1.23606)}\approx \mathbf{-1.23606}$

As for

$x_{1}=3.2\\x_{2}=3.2-\frac{f(3.2)}{f'(3.2)}=3.23636\\x_{3}=3.23636-\frac{f(3.23636)}{f'(3.23636)}\approx 3.23606\\x_{4}=3.23606-\frac{f(3.23606)}{f'(3.23606)}\approx \mathbf{3.23606}\\$

3) Rewriting and calculating its derivative. Remember to do it, in radians.

$5\cos(x)-x-1=0 \:and f'(x)=-5\sin(x)-1$

$x_{1}=1\\x_{2}=1-\frac{f(1)}{f'(1)}=1.13471\\x_{3}=1.13471-\frac{f(1.13471)}{f'(1.13471)}\approx 1.13060\\x_{4}=1.13060-\frac{f(1.13060)}{f'(1.13060)}\approx 1.13059\\x_{5}= 1.13059-\frac{f( 1.13059)}{f'( 1.13059)}\approx \mathbf{ 1.13059}$

For the second root, let's try -1.5

$x_{1}=-1.5\\x_{2}=-1.5-\frac{f(-1.5)}{f'(-1.5)}=-1.71409\\x_{3}=-1.71409-\frac{f(-1.71409)}{f'(-1.71409)}\approx -1.71410\\x_{4}=-1.71410-\frac{f(-1.71410)}{f'(-1.71410)}\approx \mathbf{-1.71410}\\$

For x=-3.9, last root.

$x_{1}=-3.9\\x_{2}=-3.9-\frac{f(-3.9)}{f'(-3.9)}=-4.06438\\x_{3}=-4.06438-\frac{f(-4.06438)}{f'(-4.06438)}\approx -4.05507\\x_{4}=-4.05507-\frac{f(-4.05507)}{f'(-4.05507)}\approx \mathbf{-4.05507}\\$

5) In this case, let's make a little adjustment on the Newton formula to find critical numbers. Remember their relation with 1st and 2nd derivatives.

$x_{n+1}=x_{n}-\frac{f'(n)}{f''(n)}$

$f(x)=x^6-x^4+3x^3-2x$

$\mathbf{f'(x)=6x^5-4x^3+9x^2-2}$

$\mathbf{f''(x)=30x^4-12x^2+18x}$

For -1.2

$x_{1}=-1.2\\x_{2}=-1.2-\frac{f'(-1.2)}{f''(-1.2)}=-1.32611\\x_{3}=-1.32611-\frac{f'(-1.32611)}{f''(-1.32611)}\approx -1.29575\\x_{4}=-1.29575-\frac{f'(-1.29575)}{f''(-4.05507)}\approx -1.29325\\x_{5}= -1.29325-\frac{f'( -1.29325)}{f''( -1.29325)}\approx -1.29322\\x_{6}= -1.29322-\frac{f'( -1.29322)}{f''( -1.29322)}\approx \mathbf{-1.29322}\\$

For x=0.4

$x_{1}=0.4\\x_{2}=0.4\frac{f'(0.4)}{f''(0.4)}=0.52476\\x_{3}=0.52476-\frac{f'(0.52476)}{f''(0.52476)}\approx 0.50823\\x_{4}=0.50823-\frac{f'(0.50823)}{f''(0.50823)}\approx 0.50785\\x_{5}= 0.50785-\frac{f'(0.50785)}{f''(0.50785)}\approx \mathbf{0.50785}\\$

and for x=-0.4

$x_{1}=-0.4\\x_{2}=-0.4\frac{f'(-0.4)}{f''(-0.4)}=-0.44375\\x_{3}=-0.44375-\frac{f'(-0.44375)}{f''(-0.44375)}\approx -0.44173\\x_{4}=-0.44173-\frac{f'(-0.44173)}{f''(-0.44173)}\approx \mathbf{-0.44173}\\$

These roots (in bold) are the critical numbers

3 0

2 years ago

A class of 20 boys and 15 girls is divided into n groups so that each group has x boys and y girls. Find x, y and n. what values

melamori03 [73]

20 = 2 x 2 x 5
15 = 3 x 5

common divisor (CD) of 20 and 15 is 5 so n should be 5 and each group has 4 boys and 3 girls.

If there are more than one CD, then there are more than one answer.

If they ask for the largest number of groups, greatest common divisor (GCD) will be the answer.

4 0

3 years ago

Help I don't know how to do this<br>

Vilka [71]

Answer:

P(t) = 282.2(1.009)^t

Step-by-step explanation:

Look at the attached image.

Hope you can read my handwriting. the image cut off the right side, b = 1.009213324... but question asks to round to nearest thousandth so it's 1.009

For the second part just use the equation to find P when t = -1 and see if P is less than (underpredicts) the number from the question or greater than (overpredicts) the number. I haven't calculated it but I think it will be smaller and thus underpredicts just from looking at the numbers when t = -1

6 0

3 years ago