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8 months ago

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Identify the coefficient of second term in the expression:3x^2 + 7y^3 + 2x + 9y -5y^2.can someone please explain what coefficien

t of second means and how to solve this equation please !

1 answer:

Paladinen [302]8 months ago

4 0

So the coefficient are the numbers when you multiply a number by a varriable

for example the first part of this equation starts with 3x

3 is the number and x is the viable

that means that the first coefficient in this problem is 3

so moving forward the second coeffifient would be 7 since 7y comes after the first coefficient

$\begin{gathered} \\ \end{gathered}$

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The height of the sail on a boat is 7 feet less than 3 times the length of its base. If the The area of the sail is 68 square fe

Dima020 [189]

Step-by-step explanation:

It is given that,

The height of the sail on a boat is 7 feet less than 3 times the length of its base.

Let the length of the base is x.

ATQ,

Height = (3x-7)

Area of the sail is 68 square feet.

Formula for area is given by :

$A=lb\\\\68=x(3x-7)\\\\3x^2-7x=68\\\\3x^2-7x-68=0$

x = 8 feet and x = -3.73 feet

So, length is 8 feet

Height is 3(8)-7 = 17 feet.

So, its height and the length of the base is 17 feet and 8 feet respectively.

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2 years ago

Are they right??? <br><br> Due tomorrow

DedPeter [7]

Yes they are right you did a great job!

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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.

<u>Ben:</u>

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

Thus Eqn. (2) becomes:

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

<u>Josh:</u>

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

Thus Eqn. (2) becomes:

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

<em><u>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 </u></em> $h(t)=0$ <em><u>. Now since both are quadratic equations we will solve one showing the full method which can be repeated for the other one. </u></em>

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:

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

Since time can only be positive we reject the $t_{2}$ solution and we keep that Ben's book took $t=4.3276$ seconds to reach the ground.

Similarly solving for Josh we obtain

$t_{1}=3.6794sec\\t_{2}=-0.6794sec$

Thus again we reject the negative and keep the positive solution, so Josh's book took $t=3.6794$ seconds to reach the ground.

Then we can find the difference between Ben and Josh times as

$t_{Ben}-t_{Josh}= 4.3276 - 3.6794 = 0.6482$

So to answer the original question:

<em>Whose textbook reaches the ground first and by how many seconds?</em>

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

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2 years ago

Product of (8.6×10^2)(3×10^2)in scientific notation

Snowcat [4.5K]

Answer:

in scientific notation = 2.58 × 10⁵

I hope I helped you^_^

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2 years ago

What’s the answer to this?

Elina [12.6K]

Step-by-step explanation:

Line is passing through the points:

$(-2,\:3)=(x_1,\:y_1) \:\&\: (2,\:5)=(x_2 ,\:y_2)$

Equation of line in two point form is given as:

$\frac{y -y_1 }{y_1 -y_2 } = \frac{x -x_1 }{x_1 -x_2 } \\ \\ \therefore \frac{y -3}{ 3 -5} = \frac{x -(-2) }{-2 -2} \\ \\ \therefore \frac{y -3}{ - 2} = \frac{x -(-2) }{-4} \\ \\ \therefore \frac{y - 3}{1} = \frac{x +2}{2} \\ \\ \therefore \: y-3= \frac{x}{2} + \frac{2}{2} \\ \\\therefore \: y= \frac{x}{2} + 1 +3\\ \\ \huge \red{ \boxed{\therefore \: y= \frac{1}{2} \:x+ 4}}\\ is \: the \: required \: equation \: of \: line.$

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2 years ago