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Strike441 [17]
3 years ago
14

Need help asap 8 points

Mathematics
2 answers:
lubasha [3.4K]3 years ago
7 0
Question 13 is 14 x is 14
cricket20 [7]3 years ago
3 0

I'll solve 11-14 since they are fairly simple problems once you get to understand them :)

  • Remember: what you do to one side of the equation, you do to the other.
  • To get everything on one side of the equation, you perform the opposite operation.
  • Say for example, 4x = 8. Since you are multiplying 4 * x, the opposite operation of multiplication is division, so in this case you would DIVIDE both sides by 4.

11) 15 = 6.5 + x

  • The opposite of addition, 6.5 + x, is subtraction, so subtract 6.5 from both sides.

15 - 6.5 = x

  • 8.5 = x, which is the same as <u>x = 8.5</u>.

12) -3 + x = -17

  • The opposite of subtracting is addition, so add 3 to both sides.

x = -17 + 3

  • <u>x = -14</u>

13) 7x = -98

  • You have 7 * x, which is multiplication, and the opposite of multiplication is division, so divide both sides by 7.

x = -98/7

  • <u>x = -14</u>

14) x/-3 = -7

  • The opposite of division is multiplication, so multiply both sides by -3.

x = -7 * -3

  • <u>x = 21</u>
<h3>Hope this helped, good luck!</h3>
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Find an equation for the perpendicular bisector of the line segment whose endpoints are (5,-9) and (1,3).
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Given coordinates of the endpoints of a line segment  (5,-9) and (1,3).

In order to find the equation of perpendicular line, we need to find the slope between given coordinates.

Slope between (5,-9) and (1,3) is :

\mathrm{Slope\:between\:two\:points}:\quad \mathrm{Slope}=\frac{y_2-y_1}{x_2-x_1}

m=\frac{3-\left(-9\right)}{1-5}

m=-3

Slope of the perpendicular line is reciprocal and opposite in sign.

Therefore, slope of the perpendicular line = 1/3.

Now, we need to find the midpoint of the given coordinates.

\mathrm{Midpoint\:of\:}\left(x_1,\:y_1\right),\:\left(x_2,\:y_2\right):\quad \left(\frac{x_2+x_1}{2},\:\:\frac{y_2+y_1}{2}\right)

=\left(\frac{1+5}{2},\:\frac{3-9}{2}\right)

=\left(3,\:-3\right)

Let us apply point-slope form of the linear equation:

y-y1 = m(x-x1)

y - (-3) = 1/3 (x - 3)

y +3 = 1/3 x - 1

Subtracting 3 from both sides, we get

y +3-3 = 1/3 x - 1 -3

<h3>y = 1/3 x - 4 .</h3>
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If p=-2, find the value of 4p +7
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4(-2)+7=-1

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-8+7=-1

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What is the slope of 9x-3y=18?
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Whats the lcm of 15 and 16
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At 2:00 PM a car's speedometer reads 30 mi/h. At 2:15 PM it reads 50 mi/h. Show that at some time between 2:00 and 2:15 the acce
g100num [7]

Here is the correct format for the question

At 2:00 PM a car's speedometer reads 30 mi/h. At 2:15 PM it reads 50 mi/h. Show that at some time between 2:00 and 2:15 the acceleration is exactly 80 mi/h².Let v(f) be the velocity of the car t hours after 2:00 PM.Then \dfrac{v(1/4)-v(0)}{1/4 -0} = \Box. By the Mean Value Theorem, there is a number c such that 0 < c < \Box  with v'(c) = \Box. Since v'(t)  is the acceleration at time t, the acceleration c hours after 2:00 PM is exactly 80 mi/h^2.

Answer:

Step-by-step explanation:

From the information given :

At  2:00 PM ;

a car's speedometer v(0) = 30 mi/h

At 2:15 PM;

a car's speedometer v(1/4) = 50 mi/h

Given that:

v(f) should be the velocity of the car t hours after 2:00 PM

Then \dfrac{v(1/4)-v(0)}{1/4 -0} = \Box will be:

= \dfrac{50-30}{1/4 -0}

= \dfrac{20}{1/4 }

= 20 × 4/1

= 80 mi/h²

By the Mean value theorem; there is a number c such that :

\mathbf{0 < c< \dfrac{1}{4}}     with \mathbf{v'(c) = \dfrac{v(1/4)-v(0)}{1/4 -0}} \mathbf{ = 80 \ mi/h^2}

6 0
3 years ago
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