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4 years ago

How do i solve this?

Mathematics

1 answer:

Aloiza [94]4 years ago

5 0

Okay on here it has to deal with a certain theorem or postulate. The theorem states that the two angles opposite of the corresponding angle to the extended angle equal the extended angle. In other words the two given angles inside the triangle equal the outside angle so (62)+(7x+14)=15x-4

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20 points if correct.

Andrei [34K]

Answer:

the answer is A the mean

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3 years ago

Read 2 more answers

Guys pretty pls help :3

dedylja [7]

Answer:

See description.

Step-by-step explanation:

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3 years ago

What is the sum of 3(x + 3) and -2(x - 5)?<br> A. X-2<br> B. x-1<br> C. X + 8<br> D. X + 19

Umnica [9.8K]

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7 0

3 years ago

You owe $225.45 on your credit card. You make a $55.00 payment on your credit card, and then purchase $87.98 worth of clothes at

zubka84 [21]

Answer:

The balance amount owed on the credit card altogether = $258.43

Step-by-step explanation:

Here, the current amount owed on the credit card = $225.45

Now, the amount of payment made to the credit card = $55.00

So, the remaining amount owed = Current Debt - Paid Amount

= $225.45 - 55 = $170. 45

Again, the purchase amount at Macy's = $87.98

⇒ New amount owed on credit card = Previous Debt + New Purchase

= $170. 45 + $87.98 = $258.43

Hence, the balance amount owed on the credit card = $258.43

5 0

4 years ago

Am i correct? calculus

Advocard [28]

Check the picture below.

now, the "x" is a constant, the rocket is going up, so "y" is changing and so is the angle, but "x" is always just 15 feet from the observer. That matters because the derivative of a constant is zero.

now, those are the values when the rocket is 30 feet up above.

$\bf tan(\theta )=\cfrac{y}{x}\implies tan(\theta )=\cfrac{y}{15}\implies tan(\theta )=\cfrac{1}{15}\cdot y
\\\\\\
\stackrel{chain~rule}{sec^2(\theta )\cfrac{d\theta }{dt}}=\cfrac{1}{15}\cdot \cfrac{dy}{dt}\implies \cfrac{1}{cos^2(\theta )}\cdot\cfrac{d\theta }{dt}=\cfrac{1}{15}\cdot \cfrac{dy}{dt}
\\\\\\
\boxed{\cfrac{d\theta }{dt}=\cfrac{cos^2(\theta )\frac{dy}{dt}}{15}}\\\\
-------------------------------\\\\
$

$\bf cos(\theta )=\cfrac{adjacent}{hypotenuse}\implies cos(\theta )=\cfrac{15}{15\sqrt{5}}\implies cos(\theta )=\cfrac{1}{\sqrt{5}}\\\\
-------------------------------\\\\
\cfrac{d\theta }{dt}=\cfrac{\left( \frac{1}{\sqrt{5}} \right)^2\cdot 11}{15}\implies \cfrac{d\theta }{dt}=\cfrac{\frac{1}{5}\cdot 11}{15}\implies \cfrac{d\theta }{dt}=\cfrac{\frac{11}{5}}{15}\implies \cfrac{d\theta }{dt}=\cfrac{11}{5}\cdot \cfrac{1}{15}
\\\\\\
\cfrac{d\theta }{dt}=\cfrac{11}{75}$

8 0

3 years ago