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denis-greek [22]
3 years ago
6

In a classroom, 5/8 of the students are wearing blue shirts, and 1/4 are wearing white shirts. There are 36 students in the clas

s.How many students are wearing a shirt other than a blue or white?
Mathematics
2 answers:
Serhud [2]3 years ago
6 0

Answer:

5 students

Step-by-step explanation:

1/4 x 2 = 2/8

2/8 + 5/8 = 7/8

36 students/ 8= 4.5 students

you divide by 8 because you only have one eighth left

dexar [7]3 years ago
3 0

Answer:

the answer would be 4.5 but there cant be a half a student so im kind of confused on that

Step-by-step explanation:


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Drag the tiles to the correct boxes to complete the pair.
seraphim [82]

Answer:

1. h(x) = x + 5—> f(x) = x² + 4x - 5, and g(x) = x - 1

2. h(x) = x + 3 —> f(x) = x² - 9, and g(x) = x - 3

3. h(x) = x + 4—> f(x) = x² - 16, and g(x) = x - 4

4.h(x) = x - 1—> f(x) = x² - 4x + 3, and g(x)= x - 3

Step-by-step explanation:

A) f(x) = x² - 9, and g(x) = x - 3

We are told that; h(x) = f(x)/g(x)

f(x) = x² - 9 can be factorized to;

f(x) = (x + 3)(x - 3)

Thus; h(x) = (x + 3)(x - 3)/(x - 3)

(x - 3) will cancel out to give;

h(x) = x + 3

B) f(x) = x² - 4x + 3, and g(x)= x - 3

x² - 4x + 3 can be factorized as;

(x - 1)(x - 3)

Thus; f(x) = (x - 1)(x - 3)

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

h(x) = x - 1

C) f(x) = x² + 4x - 5, and g(x) = x - 1

x² + 4x - 5 can be factorized as;

(x - 1)(x + 5)

Thus; f(x) = (x - 1)(x + 5)

h(x) = (x - 1)(x + 5)/(x - 1)

h(x) = x + 5

D) f(x) = x² - 16, and g(x) = x - 4

x² - 16 can be expressed as;

(x + 4)(x - 4)

Thus; f(x) = (x + 4)(x - 4)

h(x) = (x + 4)(x - 4)/(x - 4)

h(x) = x + 4

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f(x)=x^3-3x^2-9x+4 find the intervals on which f is increasing or decreasing b. find the local maximum and minimum values of f.
MArishka [77]

Answer:

Please read the complete answer below!

Step-by-step explanation:

You have the following function:

f(x)=x^3-3x^2-9x+4   (1)

a) To find the interval on which f is increasing or decreasing, you first calculate the critical points of f(x).

You calculate the derivative f(x) respect to x:

\frac{df}{dx}=3x^2-6x-9    (2)

Next, you equal the derivative to zero, and then you find the roots of the polynomial by using the quadratic formula:

3x^2-6x-9=0\\\\x_{1,2}=\frac{-(-6)\pm\sqrt{(-6)^2-4(3)(-9)}}{2(3)}\\\\x_{1,2}=\frac{6\pm12}{6}\\\\x_1=-1\\\\x_2=3

Then, the critical points are x=-1 and x=3

Next, you calculate df/dx for a values of x to the left and to the right of the critical points x1 and x2. If df/dx < 0 the function is decreasing, if df/dx > 0 the function is increasing.

for x = -1.01

\frac{df(-1.01)}{dx}=3(-1.01)^2-6(-1.01)-9=0.12

Then, in the interval (-∞,-1), the function is increasing

for x = -0.99

\frac{df(-0.99)}{dx}=3(-0.99)^2-6(-0.99)-9=-0.11

In the interval (-1,3) the function is decreasing

for x = 3.01

\frac{df(3.01)}{dx}=3(3.01)^2-6(3.01)-9=0.12

In the interval (3,+∞) the function is increasing

b) To find the local minimum and maximum you use the second derivative of the function:

\frac{d^2f}{dx^2}=6x-6     (3)

you evaluate the second derivative for the critical points x1 and x2, if the second derivative is positive, you have a local minimum. If the second derivative is negative, you have a local maximum:

for x1 = -1

6(-1)-6=-12

x=-1 is a local maximum

for x2 = 3

6(3)-6=12>0

x=3 is a local minimum

c) upward concavity: (-1,3)

downward concavity: (-∞,-1)U(3,+∞)

The inflection points are calculated with the second derivative equal to zero:

6x-6=0\\\\x=1

For x = 1 you have an inflection point

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Alexus [3.1K]

Answer:

10

Step-by-step explanation:

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