Let’s start with the first one: 

We can change this into a logarithm (click here if you’re confused):

Using the change of base rule, we can get it into log base 10:

And then find the answer:

Same for your second question: 

For your final question: 

We found the values of y and x, so we just need to plug them in:

I hope this helps :) 

- Kendra

Let’s start with the first one: 

We can change this into a logarithm (click here if you’re confused):

Using the change of base rule, we can get it into log base 10:

And then find the answer:

Same for your second question: 

For your final question: 

We found the values of y and x, so we just need to plug them in:

I hope this helps :) 

- Kendra

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Can you help me with my homework? i have absolutely no idea what im doing and my teacher never answers any questions i have.

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The first equation is in implicit form. This means that on one side (here: left) are all variables with their coefficients while on the other side of the equation you have a constant.

The second equation is in explicit form. This means you define one variable (here: y) as whatever’s on the other side of the equation.

"To graph an equation in implicit form the easiest way is to find the x-intercept and y-intercept.” Let’s look at the first equation: if you set x=0 you have y=2, so this graph cuts the y-axis at 2. If you then set y=0 you have x=4, so the graph cuts the x-axis at 4. You now only need to connect those points and you’re finished.

"Slope-intercept form is [I assume your teacher wants to hear y=mx+n but I’m not sure. Okay, I /am/ sure, but only 90% sure] where m is the slope and n is the y-intercept.” You get the slope by subtracting y- and x-coordinates from 2 different points and by dividing DELTA y / DELTA x. n is where the graph cuts the y-axis because when you set x=0 the first term is 0, thus you have y=n.

"The solution to the system is the interception point of the two lines.” How so? You are looking for a solution of this linear equations system that is x=& and y have the same value in each equation; you are really looking for a point that both graphs have in common, which is their interception point.

You can now find the solution by drawing the graphs and then checking with the equations by first inserting one variable and then checking the value of the second one. The solution is x=4, y=0.

I hope that helped!

-sorrel

Can you help me with my homework? i have absolutely no idea what im doing and my teacher never answers any questions i have.

—————————————

The first equation is in implicit form. This means that on one side (here: left) are all variables with their coefficients while on the other side of the equation you have a constant.

The second equation is in explicit form. This means you define one variable (here: y) as whatever’s on the other side of the equation.

"To graph an equation in implicit form the easiest way is to find the x-intercept and y-intercept.” Let’s look at the first equation: if you set x=0 you have y=2, so this graph cuts the y-axis at 2. If you then set y=0 you have x=4, so the graph cuts the x-axis at 4. You now only need to connect those points and you’re finished.

"Slope-intercept form is [I assume your teacher wants to hear y=mx+n but I’m not sure. Okay, I /am/ sure, but only 90% sure] where m is the slope and n is the y-intercept.” You get the slope by subtracting y- and x-coordinates from 2 different points and by dividing DELTA y / DELTA x. n is where the graph cuts the y-axis because when you set x=0 the first term is 0, thus you have y=n.

"The solution to the system is the interception point of the two lines.” How so? You are looking for a solution of this linear equations system that is x=& and y have the same value in each equation; you are really looking for a point that both graphs have in common, which is their interception point.

You can now find the solution by drawing the graphs and then checking with the equations by first inserting one variable and then checking the value of the second one. The solution is x=4, y=0.

I hope that helped!

-sorrel

So let’s separate this up, first looking at the house and then at the contents, and later combining them. 

His house was valued at £50,000. Per £100 of that, he gets 20p. Thus, we need to find how many of those £100 fit into what it was valued at. Divide 50,000 by 100 = 500. Now we have 500 times 20p for the house = 10,000p.

The contents were valued at £20,000. Per £100, he gets 25p. £20,000 divided by 100 = 200. 200 times 25p = 5,000p.

The contents and house together = 10,000p + 5,000p = 15,000p. We can convert this into pounds by dividing it by 100 (there are 100 pence in a pound). So, Mr. Day’s total premium is £150 for his house and contents.

- Kendra

So let’s separate this up, first looking at the house and then at the contents, and later combining them. 

His house was valued at £50,000. Per £100 of that, he gets 20p. Thus, we need to find how many of those £100 fit into what it was valued at. Divide 50,000 by 100 = 500. Now we have 500 times 20p for the house = 10,000p.

The contents were valued at £20,000. Per £100, he gets 25p. £20,000 divided by 100 = 200. 200 times 25p = 5,000p.

The contents and house together = 10,000p + 5,000p = 15,000p. We can convert this into pounds by dividing it by 100 (there are 100 pence in a pound). So, Mr. Day’s total premium is £150 for his house and contents.

- Kendra

These are pretty big questions you’re asking… But I shall try my best!

A quadratic is a function with an x squared in it. Let’s compare it to a linear function:

Linear -> y = mx + b -> y = 2x + 1 (for example)

Quadratic -> y = x^2 + px + c -> y = x^2 - 2x + 1 (for example)

They can come in various forms, and I believe the one you’re asking about would look like this: y = (ax + b)(cx + d), where a, b, c, and d are some number.

Remember FOIL? First-inner-outer-last? This rule applies to quadratics. Here it is as a visual: 



So when you have a quadratic, you need to multiply it out, or expand it. I’ll show you how it’s done with a few examples. After expanding it, you combine “like terms” (so all x^2, or x, or only numbers) together, so that your equation is easier to read and work with. 

Example 1: y = (x + 2)(x + 2)

(multiply x by x, x by 2, then 2 by x, and lastly 2 by 2)

y = x^2 + 2x + 2x + 4

(now combine “like terms” - we have two numbers with just an “x”, so we can put them together)

y = x^2 + 4x + 4

That’s it.

Example 2: y = (x - 1)(x + 4)

(multiply x by x, x by 4, -1 by x, and -1 by 4)

y = x^2 + 4x - x - 4

(combine the two “x”)

y = x^2 + 3x -4

Example 3: y = (3x - 1)(x + 5)

(multiply 3x by x, 3x by 5, -1 by x, and -1 by 5)

y = 3x^2 + 15x - x - 5

(again combine the “x”)

y = 3x^2 + 14x - 5

Here are some ones if you want to practice: 

y = (x + 1)(x – 3) 

y = (x + 1)(x – 4)

y = (2x - 2)(x - 4)

y = (5x - 3)(3x - 5)

Here are some links:

(1) (2) (3)

I really hope this, at least somewhat, answered your question. Next time, if you’re a little more specific, we can help you solve the ones you’re stuck on! Feel free to submit again if you need more help.

- Kendra

These are pretty big questions you’re asking… But I shall try my best!

A quadratic is a function with an x squared in it. Let’s compare it to a linear function:

Linear -> y = mx + b -> y = 2x + 1 (for example)

Quadratic -> y = x^2 + px + c -> y = x^2 - 2x + 1 (for example)

They can come in various forms, and I believe the one you’re asking about would look like this: y = (ax + b)(cx + d), where a, b, c, and d are some number.

Remember FOIL? First-inner-outer-last? This rule applies to quadratics. Here it is as a visual: 

So when you have a quadratic, you need to multiply it out, or expand it. I’ll show you how it’s done with a few examples. After expanding it, you combine “like terms” (so all x^2, or x, or only numbers) together, so that your equation is easier to read and work with. 

Example 1: y = (x + 2)(x + 2)

(multiply x by x, x by 2, then 2 by x, and lastly 2 by 2)

y = x^2 + 2x + 2x + 4

(now combine “like terms” - we have two numbers with just an “x”, so we can put them together)

y = x^2 + 4x + 4

That’s it.

Example 2: y = (x - 1)(x + 4)

(multiply x by x, x by 4, -1 by x, and -1 by 4)

y = x^2 + 4x - x - 4

(combine the two “x”)

y = x^2 + 3x -4

Example 3: y = (3x - 1)(x + 5)

(multiply 3x by x, 3x by 5, -1 by x, and -1 by 5)

y = 3x^2 + 15x - x - 5

(again combine the “x”)

y = 3x^2 + 14x - 5

Here are some ones if you want to practice: 

y = (x + 1)(x – 3) 

y = (x + 1)(x – 4)

y = (2x - 2)(x - 4)

y = (5x - 3)(3x - 5)

Here are some links:

(1) (2) (3)

I really hope this, at least somewhat, answered your question. Next time, if you’re a little more specific, we can help you solve the ones you’re stuck on! Feel free to submit again if you need more help.

- Kendra

For both, we want to isolate x: 

2 + sqrt(x+1) = 5 (subtract two from both sides)

sqrt(x+1) = 3 (square both sides to get rid of the square root)

x+1 = 9 (subtract 1)

x = 8

5 + cbrt(a+2) = 3 (subtract 5)

cbrt(a+2) = -2 (cube both sides)

a + 2 = -8 (subtract 2)

a = -10 

I hope this helps!

- Kendra

For both, we want to isolate x: 

2 + sqrt(x+1) = 5 (subtract two from both sides)

sqrt(x+1) = 3 (square both sides to get rid of the square root)

x+1 = 9 (subtract 1)

x = 8

5 + cbrt(a+2) = 3 (subtract 5)

cbrt(a+2) = -2 (cube both sides)

a + 2 = -8 (subtract 2)

a = -10 

I hope this helps!

- Kendra

If your hypotenuse is the side you would consider “opposite” to the angle you’re trying to find, and your triangle is a right-angled triangle, then the angle is 90 degrees. It is always opposite the biggest angle. 

If your hypotenuse is the side you would consider “opposite” to the angle you’re trying to find, and your triangle is a right-angled triangle, then the angle is 90 degrees. It is always opposite the biggest angle. 

andtara:

hey themathkid, this is cute! 

Very cute! They’re all so happy! But…
y = log(x) is wrong

andtara:

hey themathkid, this is cute! 

Very cute! They’re all so happy! But…
y = log(x) is wrong

gayweeb:

spooky-flower:

matthen:

Unrolling these circles gives fills a triangle with base 2 π r and height r (where r is the radius of the filled disk). Such a triangle has area π r2. This does not serve as a complete proof for why this is the area of a circle, but can give you some intuition for why it should be. [code]

HOLY FUCK IGNNGGGGG SHIT

naturalogger

gayweeb:

spooky-flower:

matthen:

Unrolling these circles gives fills a triangle with base 2 π r and height r (where r is the radius of the filled disk). Such a triangle has area π r2. This does not serve as a complete proof for why this is the area of a circle, but can give you some intuition for why it should be. [code]

HOLY FUCK IGNNGGGGG SHIT

naturalogger