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3 years ago

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a photographer wants to enlarge a 5x7 photo for a wall mural without distorting its width-to- length ratio. What are the largest

dimensions to which he can enlarge the photo without having its perimeter exceed 840 inches?

1 answer:

krok68 [10]3 years ago

6 0

The largest dimension of the enlarged photo will be: Width= 175 inches and Length= 245 inches.

<u><em>Explanation</em></u>

Dimension of the original photo is 5×7 . That means the ratio of the width to the length is 5 : 7

So, lets assume the width is $5x$ and length is $7x$ for the enlarged photo.

Formula for the perimeter of a rectangle $=2(length+width)$

As the perimeter of the enlarged photo should not exceed 840 inches, so...

$2(7x+5x)\leq 840\\ \\ 2(12x)\leq 840\\ \\ 24x\leq 840\\ \\ x\leq \frac{840}{24}\\ \\ x\leq 35$

That means the maximum value of $x$ must be 35.

So, the largest possible width $=5x=5*35=175$ inches

and the largest possible length $= 7x= 7*35=245$ inches

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The diagram shows how cos θ, sin θ, and tan θ relate to the unit circle. Copy the diagram and show how sec θ, csc θ, and cot θ r

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<span>Copy the diagram and show how sec θ, csc θ, and cot θ relate to the unit circle.

The representation of the diagram is shown if Figure 1. There's a relationship between </span>sec θ, csc θ, and cot θ related the unit circle. Lines green, blue and pink show the relationship.

a.1 First, find in the diagram a segment whose length is sec θ.

The segment whose length is sec θ is shown in Figure 2, this length is the segment $\overline{OF}$ , that is, the line in green.

a.2 <span>Explain why its length is sec θ.

We know these relationships:

(1) $sin \theta=\frac{\overline{BD}}{\overline{OB}}=\frac{\overline{BD}}{r}=\frac{\overline{BD}}{1}=\overline{BD}$

(2) </span> $cos \theta=\frac{\overline{OD}}{\overline{OB}}=\frac{\overline{OD}}{r}=\frac{\overline{OD}}{1}=\overline{OD}$
<span>
(3) </span> $tan \theta=\frac{\overline{FD}}{\overline{OC}}=\frac{\overline{FC}}{r}=\frac{\overline{FC}}{1}=\overline{FC}$
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Triangles </span>ΔOFC and ΔOBD are similar, so it is true that:

$\frac{\overline{FC}}{\overline{OF}}= \frac{\overline{BD}}{\overline{OB}}$ <span>

</span>∴ $\overline{OF}= \frac{\overline{FC}}{\overline{BD}}= \frac{tan \theta}{sin \theta}= \frac{1}{cos \theta} \rightarrow \boxed{sec \theta= \frac{1}{cos \theta}}$ <span>

b.1 </span>Next, find cot θ

The segment whose length is cot θ is shown in Figure 3, this length is the segment $\overline{AR}$ , that is, the line in pink.

b.2 <span>Use the representation of tangent as a clue for what to show for cotangent.
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It's true that:

$\frac{\overline{OS}}{\overline{OC}}= \frac{\overline{SR}}{\overline{FC}}$

But:

$\overline{SR}=\overline{OA}$
$\overline{OS}=\overline{AR}$

Then:

$\overline{AR}= \frac{1}{\overline{FC}}= \frac{1}{tan\theta} \rightarrow \boxed{cot \theta= \frac{1}{tan \theta}}$

b.3 Justify your claim for cot θ.

As shown in Figure 3, θ is an internal angle and ∠A = 90°, therefore ΔOAR is a right angle, so it is true that:

$cot \theta= \frac{\overline{AR}}{\overline{OA}}=\frac{\overline{AR}}{r}=\frac{\overline{AR}}{1} \rightarrow \boxed{cot \theta=\overline{AR}}$

c. find csc θ in your diagram.

The segment whose length is csc θ is shown in Figure 4, this length is the segment $\overline{OR}$ , that is, the line in green.

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4 years ago

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4 years ago

A ball is thrown into the air from a height of 4 feet at time t = 0. The function that models this situation is h(t) = -16t2 + 6

galina1969 [7]

<h2>Hello!</h2>

The answers are:

a) The height of the ball after 3 seconds is 49 feet.

b) The maximum height of the ball is 66 feet.

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<h2>Why?</h2>

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$y=ax^{2} +bx+c$

- We can calculate the coordinates of the vertex of the parabola using the following formula:

$x_{vertex}=\frac{-b}{2a}$

- Evaluating a function means replacing the variable with the given value to evaluate.

The given function is:

$h(t)=-16x^{2}+63t+4$

Where,

$a=-16\\b=63\\c=4$

Now, calculating we have:

a) What is the height of the ball after 3 seconds?

We need to evaluate the time of 3 seconds into the function, so:

$h(3)=-16(3)^{2} +63(3)+4=49feet$

So, the height of the ball after 3 seconds is 49 feet.

b) What is the maximum height of the ball?

Since the function is describing the motion of a ball thrown into the air, we can find the maximum height by finding the y-coordinate of the vertex. If the parabola opens downward or upward, the vertex will be always the highest or the lowest point of the parabola.

So, calculating the vertex we have:

$x_{vertex}=\frac{-b}{2a}=\frac{-63}{2*-16}\\\\x_{vertex}=\frac{-63}{2*-16}=\frac{-63}{-32}=1.97$

Then, replacing "x" into the equation of the parabola, we find the y-coordinate of the vertex:

$y=-16(1.97)^{2}+63(1.97)+4=-16*3.88+63*1.97+4\\y=-16*3.88+63*1.97+4=-62.08+124.11+4=66.03$

So, if the y-coordinate is 66.03, the maximum height of the ball is 66.03 feet, or 66 feet (rounded to the nearest foot).

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We can find the time when the ball hit the ground by making equal to 0 the function and finding the roots (zeroes)

Since it's a quadratic function, we can find the zeroes using the quadratic equation:

$\frac{-b+-\sqrt{b^{2}-4ac } }{2a}$

Substituting a, b and c, we have:

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Now, since negative time does not exists, we can conclude that the ball hit the ground after 4 seconds.

d) what domain makes sense for the function?

Since the function represents the motion of the thrown ball at "t" time, the domain would be only the positive real numbers, from 0 to 4, since we found that the ball hit the ground at t equal to 4 seconds. However, if we were talking about a quadratic function with no time involved, the domain would be all the real numbers,

Note: I've attached the graph of the function.

Have a nice day!

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Delicious77 [7]

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