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sammy [17]
3 years ago
11

In a class in which the final course grade depends entirely on the average of four equally weighted 100-point tests, Brad has sc

ored 87, 89, and 78 on the first three. What range of scores on the fourth test will give Brad a C for the semester (an average between 70 and 79, inclusive)? Assume that all test scores have a non-negative value.
Mathematics
1 answer:
antiseptic1488 [7]3 years ago
7 0

Answer:

26\leq x\leq 62

Step-by-step explanation:

Let x represent Brad's score on 4th test.

We have been given Brad has scored 87, 89, and 78 on the first three.

To find the range of Brad's test score, we will represent our given information in an inequality as:

70\times 4\leq 87+89+78+x\leq 79\times 4

280\leq 254+x\leq 316

280-254\leq 254-254+x\leq 316-254

26\leq x\leq 62

Therefore, the range of score on the fourth test will be 26\leq x\leq 62.

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If line AD is a tangent to circle B at point C, and m ABC = 55°, what is the measure of BAD?
nikdorinn [45]

Answer:

145°

Step-by-step explanation:

There are a couple of ways you can get there:

1. ∠ACB is a right angle, 90°. Hence, ∠BAC is the complement of ∠ABC, so is ...

... ∠BAC = 90° -∠ABC = 90° -55° = 35°

Then, ∠BAC and ∠BAD are a linear pair, so total 180°. That makes ∠BAD the supplement of ∠BAC, so ...

... ∠BAD = 180° -35° = 145°

2. ∠BAD is the exterior angle at A for the triangle ABC. It will have a measure that is the sum of the opposite interior angles: given ∠ABC = 55° and right angle ACB = 90°.

... ∠BAD = 55° +90° = 145°

8 0
4 years ago
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NNADVOKAT [17]

Answer:

-3.52

Step-by-step explanation:

-2.2 x (-2) / (- 1/4) x 5                     PEMDAS

4.4 / -1.25

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5 0
3 years ago
use green's theorem to evaluate the line integral along the given positively oriented curve. c 9y3 dx − 9x3 dy, c is the circle
Rina8888 [55]

The line integral along the given positively oriented curve is -216π. Using green's theorem, the required value is calculated.

<h3>What is green's theorem?</h3>

The theorem states that,

\int_CPdx+Qdy = \int\int_D(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y})dx dy

Where C is the curve.

<h3>Calculation:</h3>

The given line integral is

\int_C9y^3dx-9x^3dy

Where curve C is a circle x² + y² = 4;

Applying green's theorem,

P = 9y³; Q = -9x³

Then,

\frac{\partial P}{\partial y} = \frac{\partial 9y^3}{\partial y} = 27y^2

\frac{\partial Q}{\partial x} = \frac{\partial -9x^3}{\partial x} = 27x^2

\int_C9y^3dx-9x^3dy = \int\int_D(-27x^2 - 27y^2)dx dy

⇒ -27\int\int_D(x^2 + y^2)dx dy

Since it is given that the curve is a circle i.e., x² + y² = 2², then changing the limits as

0 ≤ r ≤ 2; and 0 ≤ θ ≤ 2π

Then the integral becomes

-27\int\limits^{2\pi}_0\int\limits^2_0r^2. r dr d\theta

⇒ -27\int\limits^{2\pi}_0\int\limits^2_0 r^3dr d\theta

⇒ -27\int\limits^{2\pi}_0 (r^4/4)|_0^2 d\theta

⇒ -27\int\limits^{2\pi}_0 (16/4) d\theta

⇒ -108\int\limits^{2\pi}_0 d\theta

⇒ -108[2\pi - 0]

⇒ -216π

Therefore, the required value is -216π.

Learn more about green's theorem here:

brainly.com/question/23265902

#SPJ4

3 0
2 years ago
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