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Mathematics

1 answer:

Harman [31]3 years ago

3 0

Answer:

The required expression is: $\mathbf{\frac{5(x+y)}{-8x}}$

Option C is correct option.

Step-by-step explanation:

We need to find the equation that fits the description.

a) the expression is the quotient of two quantities

The quotient is expressed as $\frac{a}{b}$

where a is numerator and b is denominator

b) The numerator of the expression is product of 5 and the sum of x and y

Sum of x and y = x+y

Product of 5 and Sum of x and y = 5(x+y)

So, Numerator of the expression is 5(x+y)

c) The denominator is the product of negative 8 and x

Product of negative 8 and x = -8x

So, denominator of the expression is -8x

Now, The required expression is: $\mathbf{\frac{5(x+y)}{-8x}}$

Option C is correct option.

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The graph of which function passes through (0,4) and has a minimum value at (StartFraction 3 pi Over 2 EndFraction, 3)? f (x) =

solniwko [45]

Answer:

f(x) = sin(x) + 4 Let me know if you don't see why.

Step-by-step explanation:

As much as I hate it I think it might be best to check each one.

sin(x) has a point at (0,0) so that's not right.

sin(x)+4 has (0,4) as a point and a minimum at (3*pi/2+2*pi*n,3) where n is some integer. if we have n = 0 then it becomes (3*pi/2, 3) So that looks like our answer.

cos(x) + 3 has a point at (0,4) then minimums at (pi+2*pi*n, 2) the y coordinate is wrong so this won't work

-3sin(x) has a point at (0,0) so that's wrong

4cos(x) has a point at (0,4) then minimums at (pi+2*pi*n, -4) which again has the wrong y value so this is wrong.

Let me know if you don't understand how I got the results I did, I would be happy to explain.

7 0

3 years ago

Read 2 more answers

I need help.. i really want to go sleep.. thank you so much...

Effectus [21]

Answer:

1) True 2) False

Step-by-step explanation:

1) Given $\sum\limits_{k=0}^8\frac{1}{k+3}=\sum\limits_{i=3}^{11}\frac{1}{i}$

To verify that the above equality is true or false:

Now find $\sum\limits_{k=0}^8\frac{1}{k+3}$

Expanding the summation we get

$\sum\limits_{k=0}^8\frac{1}{k+3}=\frac{1}{0+3}+\frac{1}{1+3}+\frac{1}{2+3}+\frac{1}{3+3}+\frac{1}{4+3}+\frac{1}{5+3}+\frac{1}{6+3}+\frac{1}{7+3}+\frac{1}{8+3}$ $\sum\limits_{k=0}^8\frac{1}{k+3}=\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10}+\frac{1}{11}$