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2 years ago

Help me on the last question please asap

Mathematics

1 answer:

bogdanovich [222]2 years ago

7 0

Is the question your asking the one at the bottom

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Nathan is jogging on a circular trail in a state park. The diameter of the trail is 650 meters. What approximate distance ( circ

erica [24]

If using the radius, you would multiply the radius by 2*pi, to calculate the circumference, which should be the same as the answer i got.

3 0

2 years ago

HELP ASAP!!<br>The equation (blank) has no solution

KengaRu [80]

Answer:

Just to recap, an equation has no solution when it results in an incorrect "equation".

For example:

Equation: x+3 = x+4

Subtract x: 3 = 4???

But clearly, 3 is not equal to 4, so this equation has NO SOLUTION.

Now onto our problem:

13y+2-2y = 10y+3-y

11y+2 = 9y+3

2y=1

y=1/2

9(3y+7)-2 = 3(-9y+9)

27y+61 = -27y+27

54y = -34

y = -34/54

32.1y+3.1+2.4y-8.2=34.5y-5.1

34.5-5.1=34.5y-5.1

5.1=5.1

infinite solutions

5(2.2y+3.4) = 5(y-2)+6y

11y+17 = 11y-10

17 = -10??

That's not true, so the option "5(2.2y+3.4) = 5(y-2)+6y" has no solution.

Let me know if this helps

6 0

2 years ago

Solve the following integral.<br><br> <img src="https://tex.z-dn.net/?f=%5Cint4x%5Ccos%282-3x%29dx" id="TexFormula1" title="\int

bagirrra123 [75]

Hi there!

$\boxed{-\frac{4x}{3}sin(2-3x) + \frac{4}{9}cos(2-3x) + C}$

To find the indefinite integral, we must integrate by parts.

Let "u" be the expression most easily differentiated, and "dv" the remaining expression. Take the derivative of "u" and the integral of "dv":

u = 4x

du = 4

dv = cos(2 - 3x)

v = 1/3sin(2 - 3x)

Write into the format:

∫udv = uv - ∫vdu

Thus, utilize the solved for expressions above:

4x · (-1/3sin(2 - 3x)) -∫ 4(1/3sin(2 - 3x))dx

Simplify:

-4x/3 sin(2 - 3x) - ∫ 4/3sin(2 - 3x)dx

Integrate the integral:

∫4/3(sin(2 - 3x)dx

u = 2 - 3x

du = -3dx ⇒ -1/3du = dx

-1/3∫ 4/3(sin(2 - 3x)dx ⇒ -4/9cos(2 - 3x) + C

Combine:

$-\frac{4x}{3}sin(2-3x) + \frac{4}{9}cos(2-3x) + C$

7 0

2 years ago

Read 2 more answers

Establish the identity.<br> (2 cos 0-6 sin 0)² + (6 cos 0+2 sin 0)2 = 40

kogti [31]

Rewriting the left-hand side as follows,

$(2\cos\theta-6\sin \theta)^2 +(6\cos \theta+2\sin \theta)^2\\\\=4\cos^2 \theta-24\cos \theta \sin \theta+36 \sin^2 \theta+36 \cos^2 \theta+24 \cos \theta \sin \theta+4 \sin^2 \theta\\\\=40\cos^2 \theta+40 \sin^2 \theta\\\\=40(\cos^2 \theta+\sin^2 \theta)\\\\=40$

8 0

1 year ago

In which quadrant is the ordered pair ( 7, -8 ) located? *

Paul [167]

Answer:

i think IV

Step-by-step explanation:

because of the picture

4 0

2 years ago

Read 2 more answers