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0=26sin60t + (1/2)(-9.8)t^2

0=(26sin60)t + (1/2)(-9.8)t^2     It is better with the extra brackets.

0=t [(26sin60)+ (1/2)(-9.8)t ]

t=0       or      26sin60 = (1/2)(9.8)t 

      \(t=0\qquad or \qquad 26*\frac{\sqrt3}{2} = \frac{(9.8)\;t}{2} \\ t=0\qquad or \qquad 26\sqrt3 = (9.8)\;t\\ t=0\qquad or \qquad \frac{26\sqrt3}{9.8} = \;t\\\)

And you can finish it.         

I assume this is

0=26(sin60)t + (1/2)(-9.8)t^2      so we have

[ 26sqrt(3)/2] t - 4.9 t^2   = 0      factor out t

t [ 13sqrt(3) - 4.9t  ] = 0

Set both factors to 0.......  and t = 0   or t =  [ 13sqrt(3)]  / 4.9 = about  4.595

yes she is correct because 42 divided 7 is 7 and 35 divided by 7 is five.  So, she is correct.

0 x infinity = 0

anything times 0 is 0

eg. 3 x 0 = 0

or 3 x 4 x 4 x 5 x 6 x 0 = 0

Simplify the following:
3 x-2 y+x (3 y-4 (y-4)-2 (x y+y)-2 y)+4 x y

Grouping like terms, 3 y-4 (y-4)-2 (x y+y)-2 y = -(2 (y+x y))+(3 y-2 y)-4 (y-4):
3 x-2 y+x (-(2 (y+x y))+(3 y-2 y)-4 (y-4))+4 x y

3 y-2 y = y:
3 x-2 y+x (-(2 (y+x y))+y-4 (y-4))+4 x y

-2 (y+x y) = -2 y-2 x y:
3 x-2 y+x (-2 y-2 x y+y-4 (y-4))+4 x y

-4 (y-4) = 16-4 y:
3 x-2 y+x (-2 y-2 x y+y+16-4 y)+4 x y

Grouping like terms, -2 y-2 x y+y+16-4 y = -2 x y+(-2 y+y-4 y)+16:
3 x-2 y+x (-2 x y+(-2 y+y-4 y)+16)+4 x y

-2 y+y-4 y = -5 y:
3 x-2 y+x (-2 x y+-5 y+16)+4 x y

x (16-5 y-2 x y) = 16 x-5 x y-2 x^2 y:
3 x-2 y+16 x-5 x y-2 x^2 y+4 x y

Grouping like terms, 3 x-2 y+16 x-5 x y-2 x^2 y+4 x y = -2 x^2 y+(4 x y-5 x y)-2 y+(3 x+16 x):
-2 x^2 y+(4 x y-5 x y)-2 y+(3 x+16 x)

4 (x y)-5 (x y) = -(x y):
-2 x^2 y+-x y-2 y+(3 x+16 x)

3 x+16 x = 19 x:
Answer: | 
| -2x^2y - xy - 2y +19x

you can put 0 into 0 one time so the answer to 0 divided by 0 is 1

y=-6

4x+5y=-18        substitute the first equation into the second for y

4x + 5[ -6] = -18

4x - 30  = -18        add 30 to both sides

4x = 12    divide both sides by 4

x = 3    so   (x , y) = (3, -6)

This is an undetermined quantity......to see why....

0 /  0  = any real number  ......   because any real number *  the denominator of 0  = the numerator, 0  

If you look at the unit circle, the angle of 3pi/2 has its terminal point at  (x. y) = (0, -1)

And sin = y/r    and r = 1 on the unit circle.....so .....    sin (3pi/2) = y / r   = -1  1   = -1

(bat + ball, = 110, > ball, = ?) ball = 5 because if the bat is 100 then your gut instinct is to go to 10 but if you do the math it equals 120. via trial and error ball =5

Let p=lb of peanuts        [HERE IS A METOD OF SUBSTITUTION AND ELIMINATION]

Let m=lb of mixed nuts

p+14=m

2.25p + 45.5=2.65m

Solve the following system:
{p+14 = m}
2.25 p+45.5 = 2.65 m

In the first equation, look to solve for m:
{p+14 = m
2.25 p+45.5 = 2.65 m

p+14 = m is equivalent to m = p+14:
{m = p+14
2.25 p+45.5 = 2.65 m

Substitute m = p+14 into the second equation:
{m = p+14
2.25 p+45.5 = 2.65 (p+14)

2.65 (p+14) = +(2.65 p+37.1) = 2.65 p+37.1:
{m = p+14
2.25 p+45.5 = 2.65 p+37.1

In the second equation, look to solve for p:
{m = p+14
2.25 p+45.5 = 2.65 p+37.1

2.25 p+45.5 = (9 p)/4+91/2 and 2.65 p+37.1 = (53 p)/20+371/10:
(9 p)/4+91/2 = (53 p)/20+371/10

Subtract (53 p)/20+91/2 from both sides:
{m = p+14
-(2 p)/5 = -42/5

Multiply both sides by -5/2:
{m = p+14
p = 21

Substitute p = 21 into the first equation:
Answer: | 
| {m = 35 Pounds of mixed nuts.
    p = 21 Pounds of peanuts.

x=5

I'm guessing that the "27"  in your equation was arrived at by cubing rhe sphere's radius......thus....the radius was 3 and the diameter, twice that  =  6

Here you go.......http://www.purplemath.com/modules/exponent3.htm

x+3(x+2(x-6)+6)+11/4  =

x + 3x + 6(x - 6) + 18 + 11/4  =

x + 3x + 6x - 36 + 18   + 11/4 =

10x - 18 + 11/4  =        

10x - 72/4 + 11/4  =

10x - 61/4

Sorry....I don't see this as a substitution or elimination problem......here's how I would solve it......

Let x be the lbs. of peanuts used....and we have

2.25x + 3.25 (14) =  2.65(14 + x)    simplify

2.25x + 45.50   = 37.10 + 2.65x       subtract 2.25x, 37.10 from both sides

8.40  = .40x      divide both sides by .40

21  = x   = 21 lbs. of peanuts

[Maybe someone else on here knows a different method]

Let L1 be the length of one of the legs and L2 be the lentgh of the other.......and by the Pythagorean Theorem ,we have that

[L1]^2 + [(1/2)L2] ^2 = 19^2     and

[L2]^2 + [(1/2)L1]^2  = 13^2   simplify

L1^2 + (1/4)L2^2  = 361     (1)

L2^2 + (1/4)L1^2  = 169     (2)

Rearranging (1),we  have

L1^2  = 361  - (1/4)L2^2   (3)    and subbing this into (2),we have

L2^2 +(1/4)[ 361 - (1/4)L2^2 ] = 169     multiply through by 16

16L2^2 + 4*361 - L2^2  = 2704

15L2^2  = 2704 - 4*361 

15L2^2 = 1260     divide by 15

L2^2 = 84   →  L2 = 2sqrt(21)

And using (3), we have

L1^2 = 361 - (1/4)L2^2

L1^2 = 361 - (1/4)84

L1^2  = 361 - 21

L1^2  = 340  →  L1 = 2sqrt(85)

So....the hypotenuse =  sqrt (84 + 340) =  sqrt (424) = 2sqrt (106)

I need help with homework please help me with this question {35-[49-(52-28)-15]}x4

Is this whole expression X 4?, if so then we have:25 X 4=100

An infinite number of lines go through (3, -1).  You need to know the slope of the straight line as well before you can find the intercept.  

See below:

Remember......a^m * aⁿ  = a ^m+n     ......so.....

3²⁵ * 3²⁶   = 3 ^{25 + 26}  = 3⁵¹

This plausibly looks to represent the  combined volume of two cones, both with a radius of r ...... one with height "a"  and the other with height "b"

None  of them......however, if the equation is supposed to  be

y = 2x + 3  .......then   (1, 5)  and (0.5, 4)    are OK

2 adult tickets and 3 student tickets cost $12.

1 Adult ticket and 2 student tickets costs $7.

How much does each kind of ticket cost

Let an adult ticket be A

Let a student ticket be S, we have:

2A + 3S=12, and

A + 2S=7

A=7 - 2S substitute

2(7-2S) + 3S=12

14 -4S +3S =12

14 -S=12

S=$2 price of stuent ticket, and

A +4=7

A=$3 price of adult ticket