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spayn [35]
3 years ago
12

What is 3/5 divided by 2 1/2 by the lowest quotient

Mathematics
1 answer:
Brilliant_brown [7]3 years ago
7 0

Answer:

<h2>\frac{6}{25}</h2>

Step-by-step explanation:

<h3>\frac{3}{5} ÷ \frac{5}{2} = \frac{3}{5} ÷ \frac{2}{5} = \frac{6}{25}</h3>
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In two or more complete sentences, explain how to solve the cube root equation, ^3√x+1 -2=0
IrinaVladis [17]

Explanation:

The first step is to figure out what the equation is. When unconventional math symbols are used, and when there are no grouping symbols identifying operands, that can be the most difficult step. Here, we think the equation is supposed to be ...

  \sqrt[3]{x+1}-2=0

It usually works well in radical equations to isolate the radical. Here that would mean adding 2 to both sides of the equation, to undo the subtraction of 2.

  \sqrt[3]{x+1}=2

Now, it is convenient to raise both sides of the equation to the 3rd power.

  x+1=2^3=8

Finally, we can isolate the variable by undoing the addition of 1. We accomplish that by adding -1 to both sides of the equation.

  x=7

The equation is solved. The solution is x = 7.

5 0
3 years ago
The population of a city is modeled by P(t)=0.5t2 - 9.65t + 100,where P(t) is the population in thousands and t=0 corresponds to
ira [324]

Answer:

Step-by-step explanation:

This equation is a positive parabola, opening upwards.  Parabolas of this type have a vertex that is a minimum value.  In order to find the year where the population was the lowest, we have to complete the square to find the vertex.  The rule for completing the square is to first set the parabola equal to 0, then next move the constant over to the other side of the equals sign.  The leading coefficient on the x-squared term HAS to be a positive 1.  Ours is a .5, so will factor it out.  Doing those few steps looks like this:

.5(t^2-19.3t)=-100

Next we take half the linear term, square it, and add it to both sides.  Don't forget the .5 sitting out front there as a multiplier.  Our linear term is 19.3.  Taking half of that gives us 9.65, and 9.65 squared is 93.1225

.5(t^2-19.3t+93.1225)=-100+46.56125

In this process, we have created a perfect square binomial on the left.  Stating that binomial and doing the addition on the right looks like this:

.5(t-9.65)^2=-106.8775

Now finally we will divide both sides by .5 then move over the constant again to get the final vertex form of this quadratic:

(t-9.65)^2+106.8775=y

From this we can see that the vertex is (9.65, 106.8775) which translates to the year 2009 and 107,000 approximately.

In our situation, that means that the population was at its lowest, 107,000 in the year 2009.

For part b. we will replace the y in the original quadratic with a 200,000 and then factor to find the t values.  Setting the quadratic equal to 0 allows us to factor to find t:

0=.5t^2-9.65t-199900

If you plug this into the quadratic formula you will get t values of

642.02 and -622.72

The two things in math that will never EVER be negative are distances/measurements and time, so we can safely disregard the negative value of t.  Since the year 2000 is our t = 0 value, then we will add 642 years to the year 2000 to get that

In the year 2642, the population in this town will reach 200,000 (as long as it grows according to the model).

7 0
4 years ago
Help me out please its with triangles and stuff
OverLord2011 [107]

Answer:

11 in = 33 in.

Step-by-step explanation:

6 in. = 18 in.

the second triangle is 3 times bigger than the smaller one.

7 0
3 years ago
Read 2 more answers
Jody has 3 red marbles, 6 blue marbles, and 8 yellow marbles. What is the ratio of red marbles to blue marbles?
Arte-miy333 [17]

Answer:

D) 3:6

Step-by-step explanation:

8 0
3 years ago
Suppose that a function​ f(x) is defined for all real values of x except at xequals=c. can anything be said about the existence
Margaret [11]

we are given that

f(x) is defined for all values of x except at x=c

Limit may or may not exist

case-1:

If there is hole at x=c , then limit exist

case-2:

If there is vertical asymptote at x=c , then limit does not exist

Examples:

case-1:

\lim_{x \to c} \frac{x^2-cx}{(x-c)}

We can simplify it

\lim_{x \to c} \frac{x(x-c)}{(x-c)}

=\lim_{x \to c} x

=c

so, we can see that limit exist and it's value defined

case-2:

\lim_{x \to c} \frac{1}{(x-c)}

Left limit is

\lim_{x \to c-} \frac{1}{(x-c)}

=-\infty

Right Limit is

\lim_{x \to c+} \frac{1}{(x-c)}

=+\infty

so, we can see that left limit is not equal to right limit

so, limit does not exist

4 0
3 years ago
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