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UkoKoshka [18]
2 years ago
6

Analyze each step to identify if Andy made an error. Yes, he made an error in Step 1. He switched the x and y values. Yes, he ma

de an error in Step 2. He did not distribute StartFraction 3 Over 4 EndFraction properly. Yes, he made an error in Step 3. He should have subtracted 2 from both sides. No, his work is correct.
Mathematics
2 answers:
quester [9]2 years ago
6 0

Answer:

Step 3, because the values for x for which y^1 < y^2 also include values between 3 and 5.

Step-by-step explanation:

Just took the test.

mylen [45]2 years ago
3 0

Answer:

c

Step-by-step explanation:

it just is

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The distance that a object covered in time was measured and recorded in the table. What equation would describe this motion?
Cerrena [4.2K]
Sorry if wrong but I think it’s y=2x
7 0
3 years ago
I really need help! Please, please, PLEASE do not give me a link.
MArishka [77]

Answer:

7 fiction and 23 non- fiction

Step-by-step explanation:

The number of non-fiction read was 5 less than 4 times the number of fiction books.

Assume fiction books are x.

An equation representing this would be:

x + (4x - 5) = 30

5x - 5 = 30

5x = 30 + 5

x = 35 / 5

x = 7 fiction books

Non-fiction books:

= 4x - 5

= 4 * 7 - 5

= 23 non - fiction books

3 0
2 years ago
Read 2 more answers
A water tank is in the shape of a cone.Its diameter is 50 meter and slant edge is also 50 meter.How much water it can store In i
Aneli [31]
To get the most accurate answer possible, we're going to have to go into some unsightly calculation, but bear with me here:

Assessing the situation:

Let's get a feel for the shape of the problem here: what step should we be aiming to get to by the end? We want to find out how long it will take, in minutes, for the tank to drain completely, given a drainage rate of 400 L/s. Let's name a few key variables we'll need to keep track of here:

V - the storage volume of our tank (in liters)
t - the amount of time it will take for the tank to drain (in minutes)

We're about ready to set up an expression using those variables, but first, we should address a subtlety: the question provides us with the drainage rate in liters per second. We want the answer expressed in liters per minute, so we'll have to make that conversion beforehand. Since one second is 1/60 of a minute, a drainage rate of 400 L/s becomes 400 · 60 = 24,000 L/min.

From here, we can set up our expression. We want to find out when the tank is completely drained - when the water volume is equal to 0. If we assume that it starts full with a water volume of V L, and we know that 24,000 L is drained - or subtracted - from that volume every minute, we can model our problem with the equation

V-24000t=0

To isolate t, we can take the following steps:

V-24000t=0\\ V=24000t\\ \frac{V}{24000}=t

So, all we need to do now to find t is find V. As it turns out, this is a pretty tall order. Let's begin:

Solving for V:

About units: all of our measurements for the cone-shaped tank have been provided for us in meters, which means that our calculations will produce a value for the volume in cubic meters. This is a problem, since our drainage rate is given to us in liters per second. To account for this, we should find the conversion rate between cubic meters and liters so we can use it to convert at the end.

It turns out that 1 cubic meter is equal to 1000 liters, which means that we'll need to multiply our result by 1000 to switch them to the correct units.

Down to business: We begin with the formula for the area of a cone,

V= \frac{1}{3}\pi r^2h

which is to say, 1/3 multiplied by the area of the circular base and the height of the cone. We don't know h yet, but we are given the diameter of the base: 50 m. To find the radius r, we divide that diameter in half to obtain r = 50/2 = 25 m. All that's left now is to find the height.

To find that, we'll use another piece of information we've been given: a slant edge of 50 m. Together with the height and the radius of the cone, we have a right triangle, with the slant edge as the hypotenuse and the height and radius as legs. Since we've been given the slant edge (50 m) and the radius (25 m), we can use the Pythagorean Theorem to solve for the height h:

h^2+25^2=50^2\\ h^2+625=2500\\ h^2=1875\\ h=\sqrt{1875}=\sqrt{625\cdot3}=25\sqrt{3}

With h=25\sqrt{3} and r=25, we're ready to solve for V:

V= \frac{1}{3} \pi(25)^2\cdot25\sqrt{3}\\ V= \frac{1}{3} \pi\cdot625\cdot25\sqrt{3}\\ V= \frac{1}{3} \pi\cdot15625\sqrt{3}\\\\ V= \frac{15625\sqrt{3}\pi}{3}

This gives us our volume in cubic meters. To convert it to liters, we multiply this monstrosity by 1000 to obtain:

\frac{15625\sqrt{3}\pi}{3}\cdot1000= \frac{15625000\sqrt{3}\pi}{3}

We're almost there.

Bringing it home:

Remember that formula for t we derived at the beginning? Let's revisit that. The number of minutes t that it will take for this tank to drain completely is:

t= \frac{V}{24000}

We have our V now, so let's do this:

t= \frac{\frac{15625000\sqrt{3}\pi}{3}}{24000} \\ t= \frac{15625000\sqrt{3}\pi}{3}\cdot \frac{1}{24000} \\ t=\frac{15625000\sqrt{3}\pi}{3\cdot24000}\\ t=\frac{15625\sqrt{3}\pi}{3\cdot24}\\ t=\frac{15625\sqrt{3}\pi}{72}\\ t\approx1180.86

So, it will take approximately 1180.86 minutes to completely drain the tank, which can hold approximately V= \frac{15625000\sqrt{3}\pi}{3}\approx 28340615.06 L of fluid.
5 0
3 years ago
This is finding exact values of sin theta/2 and tan theta/2. I’m really confused and now don’t have a clue on how to do this, pl
Lostsunrise [7]

First,

tan(<em>θ</em>) = sin(<em>θ</em>) / cos(<em>θ</em>)

and given that 90° < <em>θ </em>< 180°, meaning <em>θ</em> lies in the second quadrant, we know that cos(<em>θ</em>) < 0. (We also then know the sign of sin(<em>θ</em>), but that won't be important.)

Dividing each part of the inequality by 2 tells us that 45° < <em>θ</em>/2 < 90°, so the half-angle falls in the first quadrant, which means both cos(<em>θ</em>/2) > 0 and sin(<em>θ</em>/2) > 0.

Now recall the half-angle identities,

cos²(<em>θ</em>/2) = (1 + cos(<em>θ</em>)) / 2

sin²(<em>θ</em>/2) = (1 - cos(<em>θ</em>)) / 2

and taking the positive square roots, we have

cos(<em>θ</em>/2) = √[(1 + cos(<em>θ</em>)) / 2]

sin(<em>θ</em>/2) = √[(1 - cos(<em>θ</em>)) / 2]

Then

tan(<em>θ</em>/2) = sin(<em>θ</em>/2) / cos(<em>θ</em>/2) = √[(1 - cos(<em>θ</em>)) / (1 + cos(<em>θ</em>))]

Notice how we don't need sin(<em>θ</em>) ?

Now, recall the Pythagorean identity:

cos²(<em>θ</em>) + sin²(<em>θ</em>) = 1

Dividing both sides by cos²(<em>θ</em>) gives

1 + tan²(<em>θ</em>) = 1/cos²(<em>θ</em>)

We know cos(<em>θ</em>) is negative, so solve for cos²(<em>θ</em>) and take the negative square root.

cos²(<em>θ</em>) = 1/(1 + tan²(<em>θ</em>))

cos(<em>θ</em>) = - 1/√[1 + tan²(<em>θ</em>)]

Plug in tan(<em>θ</em>) = - 12/5 and solve for cos(<em>θ</em>) :

cos(<em>θ</em>) = - 1/√[1 + (-12/5)²] = - 5/13

Finally, solve for sin(<em>θ</em>/2) and tan(<em>θ</em>/2) :

sin(<em>θ</em>/2) = √[(1 - (- 5/13)) / 2] = 3/√(13)

tan(<em>θ</em>/2) = √[(1 - (- 5/13)) / (1 + (- 5/13))] = 3/2

3 0
2 years ago
ROOTS AND CUBES<br> Can someone please help me answer all 4 following?
Alex Ar [27]

NOTES:

  • squared (²) means multiply that number by itself 2 times
  • cubed (³) means multiply that number by itself 3 times
  • square root (√) means 2 numbers multiplied by itself on the inside simplify to 1 of that number on the outside of the radical
  • cubed root (∛) means 3 numbers multiplied by itself on the inside simplify to 1 of that number on the outside of the radical

Answer: (C) 41

<u>Step-by-step explanation:</u>

\quad 6^2+\sqrt[3]{125} \\= 6 \cdot 6+\sqrt[3]{5\cdot 5 \cdot 5}\\= 36 + 5\\= 41

***************************************************************************************

Answer: (C) 10

<u>Step-by-step explanation:</u>

   \bigg(\dfrac{7}{3}\times \sqrt[3]{27}-2\bigg)\times \dfrac{1}{5} + \sqrt{81}

=\bigg(\dfrac{7}{3}\times \sqrt[3]{3\cdot 3 \cdot 3}-2\bigg)\times \dfrac{1}{5} + \sqrt{9\cdot 9}

=\bigg(\dfrac{7}{3}\times3-2\bigg)\times \dfrac{1}{5}+9

=(7 - 2)\times \dfrac{1}{5}+9

=5 \times \dfrac{1}{5}+ 9

= 1 + 9

= 10

***************************************************************************************

8² = 8 · 8 = 64                                     11² = 11 · 11 = 121

5³ = 5 · 5 · 5 = 125                               3³ = 3 · 3 · 3 = 27

\sqrt{1600}=\sqrt{40\cdot 40}=40                       \sqrt[3]{64}=\sqrt[3]{4\cdot 4\cdot 4}=4

\sqrt{144}=\sqrt{12\cdot 12}=12                            \sqrt[3]{8000}=\sqrt[3]{20\cdot 20\cdot 20}=20

3 0
3 years ago
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