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phi (99)  =  [ (99) / (3 * 11) ] * (3-1)(11-1)  = (3 ) (2) (10)  =  60

well if we say the cubic is \(ax^3+bx^2+cx+d\) then we already know that d is 5 since 0+0+0+d is 5 from f(0) = 5.

8a+4b+2c+5 is 8 from f(2) = 8

27a+9b+3c+5 is 13 from f(3) = 13

343a+49b+7c+5 is -5 from f(7) = -5

now its just a system of equations. im too lazy to type it out so im just going to put the final solution. if you need me to show you the system of equation steps tell me.

so the final solution is \(\boxed{\frac{148}{21}}\). hopefully :P 

\( f(x) \cdot g(x) f(x) = -7x^4-3x^2 + 2,\space g(x) = 2x^8 - 3x^7 + 5x^6 + 3x^5 - 11x^4 + 3x^3 - 18x^2 + 17x - 5.\) 

a) the leading term would be \(-7x^4 * 2x^8\) or -\(14x^{12}\)

b) the coefficient of that would be -14

c) degree of 12 since its x^12

d) the constant term would be \(2*-5\) or -10

e) well to get \(x^6\) you would need either \(x^5*x\),  \(x^4*x^2\),or \(x^3*x^3\). g(x) has an x^5 but f(x) has no x term. so no \(x^5*x\). for the second case you have either -7*-18 or -3*-11, adding gets you 159. f(x) has no x^3 term so final answer \(\boxed{159}\). at least i think so :P  ¯\_(ツ)_/¯

just find how many numbers are divisible by 65 then add three and check if all of them work. 1040 is the least multiple of 65 that works and it is 16*65. 1950 is the greatest multiple of 65, and it is 30 *65. so there are 15 multiple of 65 between 1000 and 2000. we still need to check if adding three changes anything.  1953 is still in the range so the final answer would be \(\boxed{15}\). at least i think so  :P  ¯\_(ツ)_/¯

there are 15 people we need two so 15 choose 2 or \(\boxed{105}\). at least i think so :P ¯\_(ツ)_/¯

well there are 14 members excluding mark. so we just need to get two people from the 14 members. so 14 choose 2. or \(\boxed{91}\). at least i think so :P  ¯\_(ツ)_/¯

Its wrong, but tysm for trying!!! I got that in the beginning too, but it was wrong

pweeez use latex

\((x^2-2x-5)*f(x)=x^4-2x^3+kx^2\) well the \(x^4\) has no coeficient and both polynomials have the same amount of terms so that means f(x) is just  \(x^2\). which means k is \(\boxed{-5}\). at least i think so :P  ¯\_(ツ)_/¯

pweaz use latex

so we need only to find the coeficient of x. so we only need x* a constant. we will get a 7 from x*7 and an 8 from 8x*1. so \(\boxed{15}\). at least i think so :P ¯\_(ツ)_/¯

pweezzzz use latex next time cuz my eyess hurrttt...

and also are u sure the \(x^4+3x^3+\boxed{8x}+7x+1\) 8x shouldn't be an \(8x^2\)?

well im gonna solve for whats given and what i think it should be

so \((x^3+x^2+x+1)(x^4+3x^3+15x+1)\) when exanded we only want the x^2 terms so we need to find the cofficents of \(x^2\) * a constant, and x*x.

we will have a coefficent of 1 from the \(x^2*1\), and a coofficent of 15 from the 15x*x. so in total \(\boxed{16}\). at lease i think so :P ¯\_(ツ)_/¯

well if we consider the all the math books as one group then it just becomes arranging the math grou(m), the psychology books(p), and the chem books(c) so mppcc. to find the arrangements of these books we will do \(\frac{5!}{2!*2!}\) which is \(\boxed{30}\). at least i think so ¯\_(ツ)_/¯

im pretty sure its just 5*4*3 then dividing by 6 to get rid of the overcount. so \(\boxed{10}\). at least i think so ¯\_(ツ)_/¯

This question is really just how to arrange the letters aaabbbb, a the first color, b the second, then dividing by 7 to get rid of the overcount. there are \(\frac{7!}{4!*3!}\) ways to do that or 35 ways. then we need to divide by 7 to deal with the overcount. so \(\boxed{5}\) ways to assemble the bracelet. >:)

Find the discriminant of the quadratic 5x^2 - 2x + 8 - 2x^2 + 6x + 2.   

Combine like terms                              3x² + 4x + 10  

When a quadratic is in the form ax² + bx + c   

the discriminant is (b² – 4ac)   

In this case, that's (4²) – (4)(3)(10)  =  16 – 120  =  – 104  

_.

In a rhombus, all the side lengths are $12,$ and one of the angles is equal to $90^\circ.$ Find the area of the rhombus.   

If one of the angles is 90^o then all the angles are 90^o    

Given that all the side lengths are the same, namely 12   

That makes the rhombus the special case of being a square     

Area  = 12²  =  144 units²    

_.

C(14, 2)  = 91 ways

15 people, choose any  2

C(15, 2)  =  105 ways

Simplify  as

5x = 4y^2  + 9y  + 13

x = (4/5)y^2 + (9/5)y + 13/5

The y coordinate of the vertex = -(9/5) / (2 * 4/5)  = -9/8

The x coordinate is

(4/5) (-9/8)^2 + (9/5)(-9/8) + 13/5 =   127 / 80

Starting with the A moving one letter at a time vertically, horizontally, or diagonally, how many different paths spell ARCH?   

A can reach 3 R's  

R can reach 3 C's  

C can reach 3 H's  

So there are 3 • 3 • 3 = 27 possible paths   

_.

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