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notka56 [123]
3 years ago
12

FOR 90 POINTS!!! PLEASE HELP!!! I'M STUCK ON THIS!

Mathematics
2 answers:
spin [16.1K]3 years ago
7 0
(a) y = -7/5x + 17.8
(b) it decreased by 7/5 per day.

Hope this helps!!!   :)
valina [46]3 years ago
5 0
A = y = -7/5x + 17.8

B = Decreased by 7/5 in a day.

Hope this helps!
STSN
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Ten minus One-third of a number is 4
zloy xaker [14]
Ok, so you automatically know that you are subtracting 6 from 10 to get 4. So, in order to figure out how it is 1/3 of that number, it is going to be multiplied by 3, getting 18. So, the number would be 18.
6 0
4 years ago
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CALCULUS - Find the values of in the interval (0,2pi) where the tangent line to the graph of y = sinxcosx is
Rufina [12.5K]

Answer:

\{\frac{\pi}{4}, \frac{3\pi}{4},\frac{5\pi}{4},\frac{7\pi}{4}\}

Step-by-step explanation:

We want to find the values between the interval (0, 2π) where the tangent line to the graph of y=sin(x)cos(x) is horizontal.

Since the tangent line is horizontal, this means that our derivative at those points are 0.

So, first, let's find the derivative of our function.

y=\sin(x)\cos(x)

Take the derivative of both sides with respect to x:

\frac{d}{dx}[y]=\frac{d}{dx}[\sin(x)\cos(x)]

We need to use the product rule:

(uv)'=u'v+uv'

So, differentiate:

y'=\frac{d}{dx}[\sin(x)]\cos(x)+\sin(x)\frac{d}{dx}[\cos(x)]

Evaluate:

y'=(\cos(x))(\cos(x))+\sin(x)(-\sin(x))

Simplify:

y'=\cos^2(x)-\sin^2(x)

Since our tangent line is horizontal, the slope is 0. So, substitute 0 for y':

0=\cos^2(x)-\sin^2(x)

Now, let's solve for x. First, we can use the difference of two squares to obtain:

0=(\cos(x)-\sin(x))(\cos(x)+\sin(x))

Zero Product Property:

0=\cos(x)-\sin(x)\text{ or } 0=\cos(x)+\sin(x)

Solve for each case.

Case 1:

0=\cos(x)-\sin(x)

Add sin(x) to both sides:

\cos(x)=\sin(x)

To solve this, we can use the unit circle.

Recall at what points cosine equals sine.

This only happens twice: at π/4 (45°) and at 5π/4 (225°).

At both of these points, both cosine and sine equals √2/2 and -√2/2.

And between the intervals 0 and 2π, these are the only two times that happens.

Case II:

We have:

0=\cos(x)+\sin(x)

Subtract sine from both sides:

\cos(x)=-\sin(x)

Again, we can use the unit circle. Recall when cosine is the opposite of sine.

Like the previous one, this also happens at the 45°. However, this times, it happens at 3π/4 and 7π/4.

At 3π/4, cosine is -√2/2, and sine is √2/2. If we divide by a negative, we will see that cos(x)=-sin(x).

At 7π/4, cosine is √2/2, and sine is -√2/2, thus making our equation true.

Therefore, our solution set is:

\{\frac{\pi}{4}, \frac{3\pi}{4},\frac{5\pi}{4},\frac{7\pi}{4}\}

And we're done!

Edit: Small Mistake :)

5 0
3 years ago
Luis does chores at home to earn extra money. Each week he earns $20. He already has $320 in his savings account, and is trying
Setler [38]

Answer:

30.5 Weeks

Step-by-step explanation:

The phone costs $930, and he already has $320 saved in his bank account so subtract 320 from 930. You will get 610. Then, you take 610 and divide it by 20 because he earns $20 a week. I hope this helped.

8 0
3 years ago
Two fair coins are flipped. Given that at least one coin lands on a head, calculate the probability of one head and one tail. Wr
Greeley [361]

The probability of one head and one tail is 2/3.

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

  • The possibilities for flipping two fair coins are {T,T}, {H,H}, {H,T}, {T,H}
  • Given the case that at least one coin lands on a head, So the total possibilities are {H,H}, {H,T}, {T,H} = 3 possibilities
  • Required event is 1 head and 1 tail= {H,T}, {T,H} = 2 possibilities

To calculate the probability of one head and one tail,

Probability = required events / Total events

Probability = 2/3

8 0
3 years ago
Required
snow_lady [41]

Answer:

19

Step-by-step explanation:

70 - 51 = 19

To check:

19 + 51 = 70

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