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3 years ago

9

dakota runs a half marathon every 6 weeks. madison runs a half a marathon every 15 weeks. how many weeks untill they will run to

gether?

1 answer:

Rzqust [24]3 years ago

7 0

30 weeks they would both run30 miles

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What is 20 5/81 in simplest form

Marizza181 [45]

20 5/81 in its simplest form is 20 5/81.

20 5/81 originally looks like this ⇒ 1625/81

((20 * 81)+5)/81 = (1620+5)/81 = 1625/81 * this is an improper fraction because the numerator is greater than the denominator. Improper fractions must be converted to proper fractions. It will then become a mixed fraction.

<u> 0020 5/81</u>
81 | 1625
<u>162 </u> * 81 x 2
05

5 0

3 years ago

Please help me for the love of God if i fail I have to repeat the class

Elena-2011 [213]

$\theta$ is in quadrant I, so $\cos\theta>0$ .

$x$ is in quadrant II, so $\sin x>0$ .

Recall that for any angle $\alpha$ ,

$\sin^2\alpha+\cos^2\alpha=1$

Then with the conditions determined above, we get

$\cos\theta=\sqrt{1-\left(\dfrac45\right)^2}=\dfrac35$

and

$\sin x=\sqrt{1-\left(-\dfrac5{13}\right)^2}=\dfrac{12}{13}$

Now recall the compound angle formulas:

$\sin(\alpha\pm\beta)=\sin\alpha\cos\beta\pm\cos\alpha\sin\beta$

$\cos(\alpha\pm\beta)=\cos\alpha\cos\beta\mp\sin\alpha\sin\beta$

$\sin2\alpha=2\sin\alpha\cos\alpha$

$\cos2\alpha=\cos^2\alpha-\sin^2\alpha$

as well as the definition of tangent:

$\tan\alpha=\dfrac{\sin\alpha}{\cos\alpha}$

Then

1. $\sin(\theta+x)=\sin\theta\cos x+\cos\theta\sin x=\dfrac{16}{65}$

2. $\cos(\theta-x)=\cos\theta\cos x+\sin\theta\sin x=\dfrac{33}{65}$

3. $\tan(\theta+x)=\dfrac{\sin(\theta+x)}{\cos(\theta+x)}=-\dfrac{16}{63}$

4. $\sin2\theta=2\sin\theta\cos\theta=\dfrac{24}{25}$

5. $\cos2x=\cos^2x-\sin^2x=-\dfrac{119}{169}$

6. $\tan2\theta=\dfrac{\sin2\theta}{\cos2\theta}=-\dfrac{24}7$

7. A bit more work required here. Recall the half-angle identities:

$\cos^2\dfrac\alpha2=\dfrac{1+\cos\alpha}2$

$\sin^2\dfrac\alpha2=\dfrac{1-\cos\alpha}2$

$\implies\tan^2\dfrac\alpha2=\dfrac{1-\cos\alpha}{1+\cos\alpha}$

Because $x$ is in quadrant II, we know that $\dfrac x2$ is in quadrant I. Specifically, we know $\dfrac\pi2$ , so $\dfrac\pi4$ . In this quadrant, we have $\tan\dfrac x2>0$ , so

$\tan\dfrac x2=\sqrt{\dfrac{1-\cos x}{1+\cos x}}=\dfrac32$

8. $\sin3\theta=\sin(\theta+2\theta)=\dfrac{44}{125}$

6 0

3 years ago

Math help please please

Vika [28.1K]

Answer:

3(3x+1) (2x-1)

Step-by-step explanation:

18x^2 -3x-3

Factor out a 3

3(6x^2 -x-1)

3(3x+1) (2x-1)

6 0

3 years ago

Create an equivalent system of these equations and test your solution <br> x + y = 1<br> x 3y =9

ad-work [718]

An equvilent equation
remember you can do anything to an equation as long asyou do it to both sides

assuming yo have
x+y=1 and
x-3y=9
mulitply both by 2
2x+2y=2
2x-6y=18
those are equvilent

ok, solve initial

x+y=1
x-3y=9
multiply first equation by -1 and add to 2nd equation

-x-y=-1
<u>x-3y=9 +</u>
0x-4y=8

-4y=8
divide both sides by -4
y=-2

sub back
x+y=1
x-2=1
add 2
x=3

x=3
y=-2
(3,-2)

if we test it in other one

2x+2y=2
2(3)+2(-2)=2
6-4=2
2=2
yep

2x-6y=18
2(3)-6(-2)=18
6+12=18
18=18
yep

solution is (3,-2)

4 0

3 years ago

Read 2 more answers

The graph of a linear equation passes through (0, 5) and (−10, 0). What is the equation?

Gnesinka [82]

Answer:

-5/-10

Step-by-step explanation:

6 0

2 years ago