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3 years ago

If h(x)=6-x what is the value of (hxh)(10)?

Mathematics

2 answers:

Elanso [62]3 years ago

7 0

For this case we have by definition, that given two functions, then:

$(f * g) (x) = f (x) + g (x)$

Here, we want to find ( $(h * h) (x):$

$(h * h) (x) = (6-x) (6-x)\\(h * h) (x) = 36-6x-6x + x ^ 2\\(h * h) (x) = 36-12x + x ^ 2$

We want to find the value of the function when $x = 10$ .

$(h * h) (10) = 36-12 (10) + (10) ^ 2\\(h * h) (10) = 36-120 + 100\\(h * h) (x) = 16$

Answer:

$(h * h) (x) = 16$

Ghella [55]3 years ago

5 0

Answer:

Step-by-step explanation:

h(x) × h(x) = (6 - x)²

(h × h)(10) = (6 - 10)² = (- 4)² = 16

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Please help me with Part C of this question!!!

anzhelika [568]

Answer:

a. 3/4 inches per minute

b. -1 1/8 inches per minute

c. B is fastest; 1 1/8 is more than 3/4

Step-by-step explanation:

A <em>change</em> is a <em>difference</em>. A <em>rate of change</em> is <em>one difference divided by another</em>, usually the change in y-value divided by the change in x-value.

The change in elevation is the difference between the elevation at the end of the period (6 inches) and the elevation at the beginning of the period (3 inches). The change in time period is the difference between the end time (8 min) and the beginning time (4 min).

change in elevation per minute = (6 -3 inches)/(8 -4 min)

= (3 inches)/(4 min) = 3/4 inches/minute

Similarly, ...

change in elevation per minute = (3 -7 1/2 inches)/(18 -14 min)

= (-4 1/2 inches)/(4 min) = -1 1/8 inches/minute

We know that 3/4 is more than -1 1/8, but when we talk about the "fastest rate of change", we're generally interested in the magnitude--the value without the sign. That means we understand a rate of change of -1 1/8 inches per minute to be "faster" than a rate of change of 3/4 inches per minute.

The rate of change from Part B is fastest. 1 1/8 inches per minute is more than 3/4 inches per minute.

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1 year ago

If a1 = 4 and an -4an-1 then find the value of an.

IRISSAK [1]

1 = 4
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-4an -(4)
an 1

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2 years ago

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Step-by-step explanation:

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3 years ago

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Lelechka [254]

Answer: 0.0007

Step-by-step explanation:

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3 years ago

For what values of θ on the polar curve r=θ, with 0≤θ≤2π , are the tangent lines horizontal? Vertical?

Bond [772]

Given that $r=\theta$ , then $r'=1$

The slope of a tangent line in the polar coordinate is given by:

$m= \frac{r'\sin\theta+r\cos\theta}{r'\cos\theta-r\sin\theta}$

Thus, we have:

$m= \frac{\sin\theta+\theta\cos\theta}{\cos\theta-\theta\sin\theta}$

Part A:

For horizontal tangent lines, m = 0.

Thus, we have:

$\sin\theta+\theta\cos\theta=0 \\ \\ \theta\cos\theta=-\sin\theta \\ \\ \theta=- \frac{\sin\theta}{\cos\theta} =-\tan\theta$

Therefore, the <span>values of θ on the polar curve r = θ, with 0 ≤ θ ≤ 2π, such that the tangent lines are horizontal are:

</span><span>θ = 0

</span>θ = <span>2.02875783811043
</span>
θ = <span>4.91318043943488

Part B:

For vertical tangent lines, $\frac{1}{m} =0$

Thus, we have:

$\cos\theta-\theta\sin\theta=0 \\ \\ \Rightarrow\theta\sin\theta=\cos\theta \\ \\ \Rightarrow\theta= \frac{\cos\theta}{\sin\theta} =\sec\theta$

</span>Therefore, the <span>values of θ on the polar curve r = θ, with 0 ≤ θ ≤ 2π, such that the tangent lines are vertical are:

</span>θ = <span>4.91718592528713</span>

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3 years ago