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Ronch [10]
3 years ago
5

Which of the following is the independent variable of F(G(X))?

Mathematics
1 answer:
VashaNatasha [74]3 years ago
3 0

Answer:

x

Step-by-step explanation:

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M2 -5m - 24 pls answer my questions<br>​
Gwar [14]

Answer:

(m+3) (m-8)

Step-by-step explanation:

...,...............

7 0
3 years ago
The principal has a budget of $225 and expects at least 16 people to attend.
ale4655 [162]

Answer:

$48

Step-by-step explanation:

16 x 3 = 48

7 0
3 years ago
Read 2 more answers
Braydon is at the 10-mile Marker at the park to run. He can run at a pace of 3 miles per hour. Lauren is at the 12-mile marker a
madreJ [45]

it will take 1.33 hours for Braydon and Lauren to  get to the same mile marker on the path in the  park .

<u>Step-by-step explanation:</u>

Here we have , Braydon can run at 3 miles per hour , he's initially at 10 mile marker . Lauren is at the 12-mile marker at the park, She is walking at a pace of 1.5 miles per hour. We need to find How long will it take for Braydon and Lauren to  get to the same mile marker on the path in the  park .Let's find out:

Let after time t they meet each other so ,  Braydon can run at 3 miles per hour , he's initially at 10 mile marker . Distance traveled is given by :

⇒ 3t+10

Now , Lauren is at the 12-mile marker at the park, She is walking at a pace of 1.5 miles per hour , Distance traveled is given by :

⇒ 1.5t + 12

Equating both we get :

⇒ 3t+10=1.5t+12

⇒ 1.5t=2

⇒ t=\frac{2}{1.5}

⇒ t=1.33

Therefore , it will take 1.33 hours for Braydon and Lauren to  get to the same mile marker on the path in the  park .

7 0
3 years ago
2/3 of 16 <br><br><br><br><br><br> ...........................
maxonik [38]

Answer:

2/3 * 16

32/3 = 10.66

hope it helps

5 0
3 years ago
Read 2 more answers
Cos4theta+cos2theta/ cos4theta-cos2theta= _____
vovangra [49]

\bf \textit{Sum to Product Identities} \\\\ cos(\alpha)+cos(\beta)=2cos\left(\cfrac{\alpha+\beta}{2}\right)cos\left(\cfrac{\alpha-\beta}{2}\right) \\\\\\ cos(\alpha)-cos(\beta)=-2sin\left(\cfrac{\alpha+\beta}{2}\right)sin\left(\cfrac{\alpha-\beta}{2}\right) \\\\[-0.35em] \rule{34em}{0.25pt}

\bf \cfrac{cos(4\theta )+cos(2\theta )}{cos(4\theta )-cos(2\theta )}\implies \cfrac{2cos\left( \frac{4\theta +2\theta }{2} \right)cos\left( \frac{4\theta -2\theta }{2} \right)}{-2sin\left( \frac{4\theta +2\theta }{2} \right)sin\left( \frac{4\theta -2\theta }{2} \right)} \implies \cfrac{cos\left( \frac{6\theta }{2} \right)cos\left( \frac{2\theta }{2} \right)}{-sin\left( \frac{6\theta }{2} \right)sin\left( \frac{2\theta }{2} \right)}

\bf \cfrac{cos(3\theta )cos(\theta )}{-sin(3\theta )sin(\theta )}\implies -\cfrac{cos(3\theta )}{sin(3\theta )}\cdot \cfrac{cos(\theta )}{sin(\theta )}\implies -cot(3\theta )cot(\theta )

8 0
3 years ago
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