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3 years ago

What is the v intercept on the equatior y = 5%B

Mathematics

1 answer:

Alex73 [517]3 years ago

7 0

Answer:

Is that even a thing lol

Step-by-step explanation:

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Micah places a mirror on the ground 24 feet from the base of a tree. He walks

irina [24]

The height of tree is 8 feet

<h3><u>Solution:</u></h3>

Given, Micah places a mirror on the ground 24 feet from the base of a tree

At that point, Micah's eyes are 6 feet above the ground

And he is 9 feet from the image in the mirror.

To find : height of tree = ?

Let "n" be the height of tree

From the question, we can see there is a directly proportional relationship between heights and distances.

Proportional relationships are relationships between two variables where their ratios are equivalent.

$\frac{\text { height of micah eyes above ground }}{\text { distance of micah from mirror }}=\frac{\text { height of tree above the ground }}{\text { distance of tree from mirror }}$

$\begin{array}{l}{\frac{6 \text { feet }}{9 \text { feet }}=\frac{n \text { feet }}{24 \text { feet }}} \\\\ {\frac{6}{9}=\frac{n}{24}} \\\\ {\text { n }=\frac{1}{3} \times 24} \\\\ {\text { n }=8}\end{array}$

Hence, the height of the tree is 8 feet

6 0

3 years ago

Is the following function even, odd, or neither? f(x) = -2x + 6<br>NEED ANSWER ASAP !!!!!!!!!!!! :(

Vladimir [108]

Answer:

its an odd

Step-by-step explanation:

The leading coefficient is negative

hope this helps

5 0

2 years ago

Express the trig ratios as fractions in simplest terms.

Neporo4naja [7]

Answer:

See below

Step-by-step explanation:

$\cos M=\frac{adjacent}{hypotenuse}=\frac{56}{70}=\frac{4}{5}\\\\\sin L=\frac{opposite}{hypotenuse}=\frac{56}{70}=\frac{4}{5}$

Hence, both ratios are equal to each other

3 0

2 years ago

Distance between two ships At noon, ship A was 12 nautical miles due north of ship B. Ship A was sailing south at 12 knots (naut

frozen [14]

Answer:

a) $\sqrt{144-288t+208t^2}$ b.) -12knots, 8 knots c) No e) $4\sqrt{13}$

Step-by-step explanation:

We know that the initial distance between ships A and B was 12 nautical miles. Ship A moves at 12 knots(nautical miles per hour) south. Ship B moves at 8 knots east.

We know that at time t , the ship A has moved $12\dot t (n.m)$ and ship B has moved $8\dot t (n.m)$ . We also know that the ship A moves closer to the line of the movement of B and that ship B moves further on its line.

Using Pythagorean theorem, we can write the distance s as:

$\sqrt{(12-12\dot t)^2 + (8\dot t)^2}\\s=\sqrt{144-288t+144t^2+64t^2}\\s=\sqrt{144-288t+208t^2}$

We want to find $\frac{ds}{dt}$ for t=0 and t=1

$\sqrt{144-288t+208t^2}|\frac{d}{dt}\\\\\frac{ds}{dt}=\frac{1}{2\sqrt{144-288t+208t^2}}\dot (-288+416t)\\\\\frac{ds}{dt}=\frac{208t-144}{\sqrt{144-288t+208t^2}}\\\\\frac{ds}{dt}(0)=\frac{208\dot 0-144}{\sqrt{144-288\dot 0 + 209\dot 0^2}}=-12knots\\\\\frac{ds}{dt}(1)=\frac{208\dot 1-144}{\sqrt{144-288\dot 1 + 209\dot 1^2}}=8knots$

We know that the visibility was 5n.m. We want to see whether the distance s was under 5 miles at any point.

Ships have seen each other = $s\leq 5\\\\\sqrt{144-288t+208t^2}\leq 5\\\\144-288t+208t^2\leq 25\\\\199-288t+208t^2\leq 0$

Since function $f(x)=199-288x+208x^2$ is quadratic, concave up and has no real roots, we know that $199-288x+208x^2>0$ for every t. So, the ships haven't seen each other.

Attachedis the graph of s(red) and ds/dt(blue). We can see that our results from parts b and c were correct.

Function ds/dt has a horizontal asympote in the first quadrant if

$\lim_{t \to \infty} \frac{ds}{dt}$

So, lets check this limit:

$\lim_{t \to \infty} \frac{ds}{dt}=\lim_{t \to \infty} \frac{208t-144}{\sqrt{144-288t+208t^2}}\\\\=\lim_{t \to \infty} \frac{208-\frac{144}{t}}{\sqrt{\frac{144}{t^2}-\frac{288}{t}+208}}\\\\=\frac{208-0}{\sqrt{0-0+208}}\\\\=\frac{208}{\sqrt{208}}\\\\=4\sqrt{13}$

Notice that:

$4\sqrt{13}=\sqrt{12^2+5^2}$ =√(speed of ship A² + speed of ship B²)

5 0

3 years ago

Coach Jamison wants to celebrate the final win of the school's baseball season with a trip to the local fast food place. The tea

PIT_PIT [208]

So we have 2 variables here: tacos and orders of nachos.

When we translate the paragraphs into equation: $28t + 27n = 114.80 \\ \\ 12t + 3n = 37.20$

Now, in this situation we can make use the elimination method by converting 3n to -27n.

$- 108t - 27n = - 334.8$

Add both equations: $28t + 27n - 108t - 27n = 114.8 - 334.8 \\ \\ - 80t = - 220 \\ \\ t = 2.75$

So we find that one taco costs $2.75.

We can plug this into any of the first two equations to find n: $12 \times 2.75 + 3n = 37.2 \\ \\ 33 + 3n = 37.2 \\ \\ 3n = 4.2 \\ \\ n = 1.4$

So one order of nachos cost $1.40.

4 0

3 years ago