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3 years ago

Sarah needs to go to five different stores. How many ways can she go to two of them before lunch?

Mathematics

1 answer:

Zina [86]3 years ago

6 0

Answer:

Step-by-step explanation:

Solution 1: At first, you might think that because there are 5 ways to choose the first store and 4 ways to choose the second store, the answer is 5 * 4 = 20 but this is over-counting by a factor of 2. Say that two of the stores are A and B. If she went to A then B, that's the same as going to B then A since you still go to the same stores, therefore, the answer is 20 / 2 = 10.

Solution 2: We need to find the number of ways to choose 2 stores from 5, we can do this by calculating ₅C₂ which equals:

5! / 2! * 3!

= 5 * 4 * 3 * 2 * 1 / 2 * 1 * 3 * 2 * 1

= 5 * 4 / 2 * 1

= 10

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The concentration C (in mg/dl), of an antibiotic in a patient’s bloodstream per hour, t, is given by: c(t) = (50t)/(t^2+25) In o

NARA [144]

Answer:

Step-by-step explanation:

<u>Functions </u>

The problem describes a function that expresses the concentration of an antibiotic in mg/dl vs time in hours as:

$\displaystyle c(t)=\frac{50\ t}{t^2+25}$

We need to find the first value of t such that

$\displaystyle c(t)\geq 4$

It means that

$\displaystyle \frac{50\ t}{t^2+25}\geq 4$

Operating with the inequality

$\displaystyle 50\ t\geq 4\ t^2+100$

Rearranging and dividing by 2, we have a polynomial inequality:

$\displaystyle 2t^2-25t+50\leq 0$

Factoring

$\displaystyle 2(t-10)\left (t-\frac{5}{2}\right )\leq 0$

There are two possible values for t, both valids because they are positive

$\displaystyle t=\frac{5}{2}=2.5, \ t=10$

We need to find the first value, i.e.

$t=2.5 \ hours$

Now for the graphic method, we plot the graph for the function and a horizontal line at c=4 to find the values of t.

The graph is shown in the image provided below. We can see both values where the funcion and C=4 intersect. Both values coincide with the previously analitically found values

4 0

4 years ago

39y³(50y² - 98) ÷26y²5y + 7)

olga nikolaevna [1]

Answer:

3y5 • (5y + 7)2 • (5y - 7) then solve that which is (3•2y5) • (5y + 7) • (5y - 7) and keep going { 3x−2y=5 & 2x−5y=7

SOLVE FOR X, Y

x=1

y=−1

I hope this helps a little

7 0

2 years ago

The equation of a parabola is y=x2+6x+9. Write the equation in vertex form.

nignag [31]

Answer:

Step-by-step explanation:

Vertex form is accomplished by completing the square on the quadratic. Do this by first setting the parabola equal to 0 then moving the constant over to the other side:

$x^2+6x=-9$