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zalisa [80]
3 years ago
8

10. A 266-hp output is required for an engine. If there is a 5% loss of power, what amount of input

Mathematics
1 answer:
marta [7]3 years ago
7 0

Answer:

Amount of input horsepower needed is 279.3 hp.

Step-by-step explanation:

if there is a loss of power of 5% the base horsepower will need to be increased by same percentage so that same amount of power can be delivered.

Now to calculate 5% of 266 hp, we can write

266 ×\frac{5}{100} = 13.3 hp

Now to deliver same amount of power we will have to add 5% of 266 hp to its actual power.

So real base power which is needed to deliver 266hp output even at a loss of 5% will be = (266+13.3) hp = 279.3 hp

To know more about horsepower and percentage problem:

brainly.com/question/24585630

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Answer:

SteIf -11 + N = -11 then N is the additive inverse because if we do process of elimination (POE) we can cross out C and D because this is a addition problem not a multiplication, and we know its not B because that's like 99 + 0 = 99 set, but instead its A because its just like  -8 + 8 = 0. (If this was not a multiple choice question then tell me)



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3 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 :)

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Answer:

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Step-by-step explanation:

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Answer:

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