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Lana71 [14]
2 years ago
9

A student earned grades of​ C, A,​ B, and A. Those courses had these corresponding numbers of credit​ hours: 44​, 66​, 11​, and

66. The grading system assigns quality points to letter grades as​ follows: A= 4, B=3, C=2,D=1, and F=0. Compute the grade point average​ (GPA) and round the result to two decimal places
Mathematics
1 answer:
asambeis [7]2 years ago
4 0

Answer:

The grade point average of the student is 49.92.

Step-by-step explanation:

A student earned grades :​ C, A,​ B, and A

Corresponding numbers of credit​ hours : 44​, 66​, 11​, and 66

Assigns quality points to letter grades :

A = 4 , B = 3 , C = 2 , D = 1 , F = 0

The grade point average​ (GPA):

\frac{\sum (\text{grade point}\times \text{credit hours})}{\sum (\text{grade point})}

=\frac{2\times 44+4\times 66+3\times 11+4\times 66}{2+4+3+4}

=\frac{649}{13}=49.9230\approx 49.92

The grade point average of the student is 49.92.

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vichka [17]

Answer:

y = -3(x - 4)² - 2

Step-by-step explanation:

Given the vertex, (4, -2), and the point (2, -14):

We can use the vertex form of the quadratic equation:

y = a(x - h)² + k

Where:

(h, k) = vertex

a  =  determines whether the graph opens up or down, and it also makes the parent function <u>wider</u> or <u>narrower</u>.

  • <u>positive</u> value of a = opens <u><em>upward</em></u>
  • <u>negative</u> value of a = opens <u><em>downward</em></u>
  • a is between 0 and 1, (0 < a < 1) the graph is <u><em>wider</em></u> than the parent function.
  • a > 1, the graph is <u><em>narrower</em></u> than the parent function.

<em>h </em>=<em> </em>determines how far left or right the parent function is translated.

  • h = positive, the function is translated <em>h</em> units to the right.
  • h = negative, the function is translated |<em>h</em>| units to the left.

<em>k</em> determines how far up or down the parent function is translated.

  • k = positive: translate <em>k</em> units <u><em>up</em></u>.
  • k = negative, translate <em>k</em> units <u><em>down</em></u>.

Now that I've set up the definitions for each variable of the vertex form, we can determine the quadratic equation using the given vertex and the point:

vertex (h, k): (4, -2)

point (x, y): (2, -14)

Substitute these values into the vertex form to solve for a:

y = a(x - h)² + k

-14 = a(2 - 4)²  -2

-14 = a (-2)² -2

-14 = a4 + -2

Add to to both sides:

-14 + 2 = a4 + -2 + 2

-12 = 4a

Divide both sides by 4 to solve for a:

-12/4 = 4a/4

-3 = a

Therefore, the quadratic equation inI vertex form is:

y = -3(x - 4)² - 2

The parabola is downward-facing, and is vertically compressed by a factor of -3. The graph is also horizontally translated 4 units to the right, and vertically translated 2 units down.

Attached is a screenshot of the graph where it shows the vertex and the given point, using the vertex form that I came up with.

Please mark my answers as the Brainliest, if you find this helpful :)

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No

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8−2x<12

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8-8−2x<12-8

-2x <4

Divide each side by -2, remembering to flip the inequality

-2x/-2 >4/-2

x > -2

Is -2 a solution

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A lighthouse is located on a small island 5 km away from the nearest point P on a straight shoreline and its light makes seven r
Vaselesa [24]

Answer:

Light is moving at the speed = 77.78 km/min

Step-by-step explanation:

Given:

Distance between island and point P = 5 km

Moving along the the shoreline when it is 1 km from P.

So, x = 1

From the above statement light makes seven revolutions per minute.

Therefore,              

\frac{d\theta}{dt}=\frac{7\times 2\pi\ rad}{1\ min}  = 14\pi\ rad/min  

Solution:

From the given figure.

tan\theta=\frac{x}{5}

Substitute x = 1

tan\theta=\frac{1}{5}  -------------------(1)

Now, first we differentiate both side with respect to t.

\frac{d(tan\theta)}{dt} = \frac{d}{dt}.(\frac{x}{3})

Using chain rule.

\frac{d(tan\theta)}{dt}.\frac{d\theta}{dt}  = (\frac{1}{5})\frac{dx}{dt}

sec^{2} \theta.\frac{d\theta}{dt} = \frac{1}{5}.\frac{dx}{dt}

(1+tan^{2} \theta).\frac{d\theta}{dt} = \frac{1}{5}.\frac{dx}{dt}

Substitute tan\theta=\frac{1}{5} from equation 1.

(1+(\frac{1}{3})^{2} ).\frac{d\theta}{dt} = \frac{1}{5}.\frac{dx}{dt}

(1+\frac{1}{9}).\frac{d\theta}{dt} = \frac{1}{5}.\frac{dx}{dt}

Substitute \frac{d\theta}{dt}=14\pi

\frac{10}{9}.14\pi  = \frac{1}{5}.\frac{dx}{dt}

Applying cross multiplication rule.

\frac{dx}{dt}= \frac{700\pi }{9}

\frac{dx}{dt}= 77.78\pi

Therefore, light is moving along the shore at the speed of 77.78 km/min

6 0
3 years ago
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