Secondary Maths questions, give me your maths questions

[jinny1]

nothing in the remainder theorem about the simultaneous thing...the info is only a paragraph long.. all it practically states is:

If f(x) is a polynomial in x and is divided by x-a; the remainder is the value of f(x) at x = a i.e., Remainder = f(a).

 

[mdc]

nothing in the remainder theorem about the simultaneous thing...the info is only a paragraph long.. all it practically states is:

If f(x) is a polynomial in x and is divided by x-a; the remainder is the value of f(x) at x = a i.e., Remainder = f(a).

That's all you need. You've been given a polynomial, you've been told what to divide it by and what the remainder is.

You first piece of information is "exactly divisible by x-2" (meaning remainder = 0). So begin with P(2) = 0, subbing in 2 for x in the polynomial. You'll get some equation with a and b in it.

Then do the same with x+5 (remainder -84). You'll get another equation.

Then solve.

 

[BILC]

Ok i have a quick one, i dont know where to start

a tennis ball is thrown back into the court area from a point 2 metres outisde the enclosed area of th court.it just clears the fence surrounding the court. if the base of the fence is taken as the origin. the path of the tennis ball is given by y = -xsquared + 3x + 10

I have to find the
Heght of the fence
The horizontal distance which the ball travels before it lands
maximum height of the ball

Help please??

 

[SJ]

Ok i have a quick one, i dont know where to start

a tennis ball is thrown back into the court area from a point 2 metres outisde the enclosed area of th court.it just clears the fence surrounding the court. if the base of the fence is taken as the origin. the path of the tennis ball is given by y = -xsquared + 3x + 10

I have to find the
Heght of the fence
The horizontal distance which the ball travels before it lands
maximum height of the ball

Help please??

You can do this surely. Draw it.

 

[mdc]

Ok i have a quick one, i dont know where to start

a tennis ball is thrown back into the court area from a point 2 metres outisde the enclosed area of th court.it just clears the fence surrounding the court. if the base of the fence is taken as the origin. the path of the tennis ball is given by y = -xsquared + 3x + 10

I have to find the
Heght of the fence - The base of the fence is "at the origin". When you read origin, that means the fence is at x = 0, right? You're also told that the ball just clears the fence, which means that the height of the fence will be the same as the height of the ball at the origin.

The horizontal distance which the ball travels before it lands - The ball starts 2m behind the fence, so at x = -2. Where does it hit the court?

maximum height of the ball - This is a straight-forward "find the maximum" question using the equation given. Follow the steps given in your textbook.

Help please??

Hints given in bold.

 

[magic_johnson!]

Domain: [45, 45]

p(x)= -x/40 + 2

1) How many times does the graph cross the intercept?

I use the calc and it keeps giving me the answer non, despite the next question being "how could you move the graph so that it didn't cross teh x intercept?"

 

[mdc]

None is the right answer, you've either written the question down wrong, or the book has a typo.

 

[treefingers]

Prove that:
√2 sin (x-π) = sin(x) - cos(x)

I know I have to change around the LHS, but I'm not really sure how to approach this.

 

[mdc]

Prove that:
√2 sin (x-π) = sin(x) - cos(x)

I know I have to change around the LHS, but I'm not really sure how to approach this.

Easiest way is to look up your trig identities.

Best way is to work with the trig circle and the definition of "sin(x)":

Basically, sin(x) is defined as the distance from the horizontal axis to the point where your line meets the circle (as shown by the dotted red line).

You're looking for sin(x-π). To get to x-π, you need to rotate π radians in the negative direction (or in english, clockwise).

So rotate the line clockwise for π radians, and draw in sin(x-π). The relationship between sin(x) and sin(x-π) should now be obvious.

Again, note that you're not expected to do this for these types of questions, and should usually just use the identity. But if you get comfortable working with the trig circle, the whole subject of trig becomes very, very easy.

 

[magic_johnson!]

Sorry, methods exam tomorrow and havn't done matricies since week 1

What does det A etc mean??

 

[Speck]

How do you convert the polar form of a complex number -> 25cis(-106) to the cartesian form -> x + yi?

I know how to go the other way using trig but there is no simple way to simply do the 'opposite'. Can't find anything on the internet either. This is for a project so we weren't shown.

 

[mdc]

How do you convert the polar form of a complex number -> 25cis(-106) to the cartesian form -> x + yi?

I know how to go the other way using trig but there is no simple way to simply do the 'opposite'. Can't find anything on the internet either. This is for a project so we weren't shown.

"cis" is not a function, as much as a convenient notation. Remember what it's short for (or look it up) and the answer falls out.

 

[Speck]

"cis" is not a function, as much as a convenient notation. Remember what it's short for (or look it up) and the answer falls out.

Oh ok hadn't thought about that.

So 25cos(-106) + 25isin(-106)

I assume 25cos(-106) = x and 25sin(-106) = y

Thanks.

 

[worbod]

A farmer goes into his shed and finds some chook wire. When he unrolls it he measures its length and finds he has 50 metres of wire. He wishes to use it to make a chook pen. What is the largest area he can get with 50 metres?

 

[SJ]

A farmer goes into his shed and finds some chook wire. When he unrolls it he measures its length and finds he has 50 metres of wire. He wishes to use it to make a chook pen. What is the largest area he can get with 50 metres?

A = L x W

P = 2L + 2W = 50

L = 25 - W

A = L x W = (25 - W)W = 25W - W^2

To find maximum Area, set dA/dW = 0

dA/dW = 25 - 2W = 0

W = 12.5m

L = 25 - W = 25 - 12.5 = 12.5m

Thus A = L x W = 12.5 x 12.5 = 156.25m

 

[worbod]

A = L x W

P = 2L + 2W = 50

L = 25 - W

A = L x W = (25 - W)W = 25W - W^2

To find maximum Area, set dA/dW = 0

dA/dW = 25 - 2W = 0

W = 12.5m

L = 25 - W = 25 - 12.5 = 12.5m

Thus A = L x W = 12.5 x 12.5 = 156.25m

I must confess that I do know the answer to this. It is always interesting to see how people think in relation to solving this problem. You are right to go to two decimal places but the actual answer is closer to 200 square metres.

 

[SJ]

I must confess that I do know the answer to this. It is always interesting to see how people think in relation to solving this problem. You are right to go to two decimal places but the actual answer is closer to 200 square metres.

P = 2 x pi x R = 50

R = 7.96m

A = pi x R^2 = 198.94m

Poor assumption on my behalf.

 

[worbod]

P = 2 x pi x R = 50

R = 7.96m

A = pi x R^2 = 198.94m

Poor assumption on my behalf.

Correct. Use the chook wire to make a circular chook pen. You learn quickly grasshopper.

 

[magic_johnson!]

The distance d travelled by a car when coming to rest from speed v is proportional to the square of the speed. A car travelling at a speed of 30 m/s along a residential street will stop in a distance of 15m. Find the stopping distance required for a car travelling at 60 m/s.

This is how i did it:

d=kv^2
15=k 30^2
15 = 900 k
k = 60

d=kv^2
d=60x60^2
d=21600

What am i doing wrong??????

 

[mdc]

15 = 900 k
k = 60

In the future, when you're getting an answer that's significantly larger than what you expect, it's likely you've done something like this.

 

[magic_johnson!]

In the future, when you're getting an answer that's significantly larger than what you expect, it's likely you've done something like this.

so d=kv

 

[mdc]

so d=kv

How did you get from line 1 to line 2 in my quote? That's your only mistake.

 

[magic_johnson!]

How did you get from line 1 to line 2 in my quote? That's your only mistake.

by squaring 30 sorry im not following...

 

[mdc]

by squaring 30 sorry im not following...

My fault, quoted the wrong thing. Take a look at the edited quote.

 

[magic_johnson!]

My fault, quoted the wrong thing. Take a look at the edited quote.

OHH i get ya, 15/900 rather than 900/15. cheers.