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Luden [163]
3 years ago
7

3/4+1/2= 2. 5/6+1/2= 3. 3/6+1/2= 4.1/6+4/8= 5.7/8+1/4= 6.3/8+1/2=

Mathematics
1 answer:
kaheart [24]3 years ago
6 0

Answer:

1.) \frac{5}{4\\}

2.) \frac{4}{3}

3.) 1

4.) \frac{2}{3}

5.) \frac{9}{8}

6.) \frac{7}{8}

Step-by-step explanation:

The 2 steps for all of these is to get common denominators and then add and simplify.

1.) 3/4 + 1/2 (common denominator) 3/4 + 2/4 = 5/4

2.) 5/6 + 1/2 (common denominator) 5/6 + 3/6 = 4/3

3.) 3/6 + 1/2 (common denominator) 3/6 + 3/6 = 6/6 (simplify) = 1

4.) 1/6 + 4/8 (common denominator) 8/48 + 24/48 =  32/48 (simplify) = 2/3

5.) 7/8 + 1/4 (common denominator) 7/8 + 2/8 = 9/8

6.) 3/8 + 1/2 (common denominator) 3/8 + 4/8 = 7/8

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Answer:

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Step-by-step explanation:

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3 0
3 years ago
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The results of a linear regression are shown below.
IgorLugansk [536]

Hello!

As we can see, our a value, which would be the coefficient of x, which determines our slope, is negative, meaning that this whole line is negative.

Furthermore, the correlation can be determined using the r value of the linear regression, which is around -0.9.

If the r value of the linear regression is close to 1 or -1, let's say around |r| > 0.8, then we can consider the regression a strong correlation, meaning that this is a strong negative correlation, which is answer choice A.

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3 years ago
Consider the equation below. f(x) = 2x3 + 3x2 − 336x (a) Find the interval on which f is increasing. (Enter your answer in inter
Vaselesa [24]

We have been given a function f(x)=2x^3+3x^2-336x. We are asked to find the interval on which function is increasing and decreasing.

(a). First of all, we will find the critical points of function by equating derivative with 0.

f'(x)=2(3)x^{2}+3(2)x^1-336

f'(x)=6x^{2}+6x-336

6x^{2}+6x-336=0

x^{2}+x-56=0

x^{2}+8x-7x-56=0

(x+8)-7(x+8)=0

(x+8)(x-7)=0

x=-8,x=7

So x=-8,7 are critical points and these will divide our function in 3 intervals (-\infty,-8)U(-8,7)U(7,\infty).

Now we will find derivative over each interval as:

f'(x)=(x+8)(x-7)

f'(-9)=(-9+8)(-9-7)=(-1)(-16)=16

Since f'(9)>0, therefore, function is increasing on interval (-\infty,-8).

f'(x)=(x+8)(x-7)

f'(1)=(1+8)(1-7)=(9)(-6)=-54

Since f'(1), therefore, function is decreasing on interval (-8,7).

Let us check for the derivative at x=8.

f'(x)=(x+8)(x-7)

f'(8)=(8+8)(8-7)=(16)(1)=16

Since f'(8)>0, therefore, function is increasing on interval (7,\infty).

(b) Since x=-8,7 are critical points, so these will be either a maximum or minimum.

Let us find values of f(x) on these two points.

f(-8)=2(-8)^3+3(-8)^2-336(-8)

f(-8)=1856

f(7)=2(7)^3+3(7)^2-336(7)

f(7)=-1519

Therefore, (-8,1856) is a local maximum and (7,-1519) is a local minimum.

(c) To find inflection points, we need to check where 2nd derivative is equal to 0.

Let us find 2nd derivative.

f''(x)=6(2)x^{1}+6

f''(x)=12x+6

12x+6=0

12x=-6

\frac{12x}{12}=-\frac{6}{12}

x=-\frac{1}{2}

Therefore, x=-\frac{1}{2} is an inflection point of given function.

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3 years ago
Help!!
7nadin3 [17]
I would choose letter c
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4 years ago
8x + 5 = 29<br> -5 -5<br> 8x/8 =24/8. X=3
forsale [732]

Answer:

12

Step-by-step explanation:

8 0
3 years ago
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