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Ierofanga [76]
3 years ago
7

The graph below represents the function f(x) and the translated function g(x) .

Mathematics
1 answer:
Grace [21]3 years ago
6 0

Answer:  D.  g(x) = (x-b)^2

Explanation:

If we want to shift the curve some distance b to the right, then we'll need to replace x with x-b. What this effectively does is shift the xy axis itself b units to the left while the curve is fixed in place. This gives the illusion of the curve moving b units to the right.

Put another way: the old input x is updated to the new input x-b which is smaller by b units. This helps explain why the xy axis is shifted to the left this amount.

So that's how we go from f(x) = x^2 to g(x) = (x-b)^2. You can think of it as g(x) = f(x-b) = (x-b)^2

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SIZIF [17.4K]
The correct answer is B: y ≥ 1/3x - 1

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0 ≥ 0 - 1
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Please help me on 17-18!<br> I have no clue on what to do on the problems at hand.
nika2105 [10]

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7 0
3 years ago
Simplify each expression to a single trig function or number cosx(secx-cosx)
AlladinOne [14]
I did this test  b4, yours is answer #number 12

Convert things to their basic forms. 
<span>Remember a few identities </span>
<span>sin^2 + cos^2 = 1 so </span>
<span>sin^2 = 1 - cos^2 and </span>
<span>cos^2 = 1 - sin^2 </span>

<span>I'm going to skip typing the theta symbol, just to make things faster. Just assume it is there and fill it in as you work the problems. </span>

<span>Follow along to see how each problem was worked out. You'll catch on to the general technique. </span>

<span>====== </span>
<span>1. sec θ sin θ </span>
<span>1/cos * sin = sin/cos = tan </span>

<span>2. cos θ tan θ </span>
<span>cos * sin/cos = sin </span>

<span>3. tan^2 θ- sec^2 θ </span>
<span>sin^2 / cos^2 - 1/cos^2 </span>
<span>(sin^2 - 1)/cos^2 </span>
<span>-(1-sin^2)/cos^2 </span>
<span>-cos^2/cos^2 </span>
<span>-1 </span>

<span>4. 1- cos^2θ </span>
<span>sin^2 </span>

<span>5. (1-cosθ)(1+cosθ) </span>
<span>Remember (a+b)(a--b) = a^2 - b^2 </span>
<span>1-cos^2 = sin^2 </span>

<span>6. (secx-1) (secx+1) </span>
<span>sec^2 -1 </span>
<span>1/cos^2 - 1 </span>
<span>1/cos^2 - cos^2/cos^2 </span>
<span>(1-cos^2)/cos^2 </span>
<span>sin^2/cos62 </span>
<span>tan^2 </span>

<span>7. (1/sin^2A)-(1/tan^2A) </span>
<span>1/sin^2 - 1/(sin^2/cos^2) </span>
<span>1/sin^2 - cos^2/sin^2 </span>
<span>(1-cos^2)/sin^2 </span>
<span>sin^2/sin^2 </span>
<span>1 </span>

<span>8. 1- (sin^2θ/tan^2θ) </span>
<span>1-sin^2/(sin^2/cos^2) </span>
<span>1 - sin^2*cos^2/sin^2 </span>
<span>1-cos^2 </span>
<span>sin^2 </span>


<span>9. (1/cos^2θ)-(1/cot^2θ) </span>
<span>1/cos^2 - 1/(cos^2/sin^2) </span>
<span>1/cos^2 - sin^2/cos^2 </span>
<span>(1-sin^2)/cos^2 </span>
<span>cos^2/cos^2 </span>
<span>1 </span>

<span>10. cosθ (secθ-cosθ) </span>
<span>cos *(1/cos - cos) </span>
<span>1-cos^2 </span>
<span>sin^2 </span>

<span>11. cos^2A (sec^2A-1) </span>
<span>cos^2 * (1/cos^2 - 1) </span>
<span>1 - cos^2 </span>
<span>sin^2 </span>


<span>12. (1-cosx)(1+secx)(cosx) </span>
<span>(1-cos)(1+1/cos)cos </span>
<span>(1-cos)(cos + 1) </span>
<span>-(cos-1)(cos+1) </span>
<span>-(cos^2 - 1) </span>
<span>-(-sin^2) </span>
<span>sin^2 </span>

<span>13. (sinxcosx)/(1-cos^2x) </span>
<span>sin*cos/sin^2 </span>
<span>cos/sin </span>
<span>cot </span>

<span>14. (tan^2θ/secθ+1) +1 </span>
<span>(sin^2/cos^2)/(1/cos) + 2 </span>
<span>sin^2/cos + 2 </span>
<span>sin*tan + 2 </span>
5 0
3 years ago
I'm so lost on this work
kozerog [31]

Answer:

Typist A can type faster

Step-by-step explanation:

Replace the t in w = 50t with any numbers and graph the points you get

You can see that typist A can type 50 words per minute when typist B can only type 40

6 0
3 years ago
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