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3 years ago

PLSS HELPPPPPPP IMMA FAIL

Mathematics

2 answers:

Aloiza [94]3 years ago

8 0

3.3333 because it is divided by 3

Leto [7]3 years ago

3 0

Answer:

it should 3

Step-by-step explanation:

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in fireside ski shop, 11/28 of the ski caps have tassals. Of the caps with tassals, 7/11 are blue. What fraction of the caps are

Kisachek [45]

Answer:

One fourth of the caps are blue with tassels.

Step-by-step explanation:

Take 7/11 of the caps with tassels, which boils down to (7/11)(11/28), or 1/4.

One fourth of the caps are blue with tassels.

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3 years ago

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3 years ago

The method of completing the square was used to solve the equation 2x^2 - 12x + 6 =0 which equation is a correct step when using

trapecia [35]

The equation we are looking at would have to be divided by 2 first.
x²-6x+3
When looking at that, I can tell that 1 may be the possible answer since you would have to subtract 6 from both sides. That would leave a 3. Let's check.
(x-3)(x-3)
x²-6x+9=6
Now subtract 6 from both sides.
x²-6x+3=0
So, 1- (x-3)²=6 would be used.

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3 years ago

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Natasha2012 [34]

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3 years ago

Ben and Josh went to the roof of their 40-foot tall high school to throw their math books offthe edge.The initial velocity of Be

Taya2010 [7]

Answer

Josh's textbook reached the ground first

Josh's textbook reached the ground first by a difference of $t=0.6482$

Step-by-step explanation:

Before we proceed let us re write correctly the height equation which in correct form reads:

$h(t)=-16t^2 +v_{o}t+s$ Eqn(1).

Where:

$h(t)$ : is the height range as a function of time

$v_{o}$ : is the initial velocity

$s$ : is the initial heightin feet and is given as 40 feet, thus Eqn(1). becomes:

$h(t)=-16t^2 + v_{o}t + 40$ Eqn(2).

Now let us use the given information and set up our equations for Ben and Josh.

Ben:

We know that $v_{o}=60ft/s$

Thus Eqn. (2) becomes:

$h(t)=-16t^2+60t+40$ Eqn.(3)

Josh:

We know that $v_{o}=48ft/s$

Thus Eqn. (2) becomes:

$h(t)=-16t^2+48t+40$ Eqn. (4).

Now since we want to find whose textbook reaches the ground first and by how many seconds we need to solve each equation (i.e. Eqns. (3) and (4)) at $h(t)=0$ . Now since both are quadratic equations we will solve one showing the full method which can be repeated for the other one.

Thus we have for Ben, Eqn. (3) gives:

$h(t)=0-16t^2+60t+40=0$

Using the quadratic expression to find the roots of the quadratic we have:

$t_{1,2}=\frac{-b+/-\sqrt{b^2-4ac} }{2a} \\t_{1,2}=\frac{-60+/-\sqrt{60^2-4(-16)(40)} }{2(-16)} \\t_{1,2}=\frac{-60+/-\sqrt{6160} }{-32} \\t_{1,2}=\frac{15+/-\sqrt{385} }{8}\\\\t_{1}=4.3276 sec\\t_{2}=-0.5776 sec$