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mihalych1998 [28]

3 years ago

6

A 0.25 mL sample of water drawn from a 5 liter flask contains 1.25 x 10^8 bacteria. Give the approximate number of bacteria in t

he flask, expressing your answer in scientific notation.
(Scientific Notation: find a real number a between 1 and 10, and an integer n, such that x = a x 10^n.)

1 answer:

vazorg [7]3 years ago

7 0

Answer:

The number of bacteria in 0.25 mL is 6.25 x 10^3.

Step-by-step explanation:

5 liter flask contains 1.25 x 10^8 bacteria.

So, 1liter flask contains = $\frac{1.25 \times 10^8}{5} =2.5\times 10^7$

Now

0.25 mL contains = $2.5\times 10^7 \times0.25\times 10^{-3} = 6.25\times 10^3$

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A line passes through (1, –5) and (–3, 7).

son4ous [18]

A. First, find the slope, which is -0.5, and then insert it into point-slope form, which is y-y1=m(x-x1), or y-1=-0.5(x+5). That's point-slope form.

B. First, distribute -0.5 to both terms in the parenthesis, to get y-1=-0.5x-2.5, and lastly add 1 to both sides to get the slope-intercept form of y=-0.5x-1.5

4 0

4 years ago

in the school cafeteria students chose their lunch from 3 sandwiches, 3 soups, 4 salads, and 2 drinks. how many different lunche

Vesna [10]

Answer:

Answer:

3

×

3

×

4

×

2

=

72

Explanation:

Let's look at the 3 sandwiches and 3 soups first and then expand the calculation. There are 9 ways I can have one of the sandwiches and 1 of the soups:

⎛

⎜

⎜

⎜

⎜

⎝

0

Soup 1

Soup 2

Soup 3

Sandwich 1

1

2

3

Sandwich 2

4

5

6

Sandwich 3

7

8

9

⎞

⎟

⎟

⎟

⎟

⎠

And so we can see that we multiply the number of sandwiches and the number of soups to get the total number of ways to get one of each.

The same works for more categories of choices, and so we multiply the 3 sandwiches, the 3 soups, 4 salads, and 2 drinks to get:

3

×

3

×

4

×

2

=

72

7 0

4 years ago

the company also discovered that it costs 30 to 2 widgets, 118 to produce 4 and 766 to produce 10 widgets. Find the total cost o

miv72 [106K]

We will first write cost function. As we are given three points here so we will use them to write quadratic function

$y = ax^{2} +bx+c$ ----------------------------------------(1)

So we will plug point (2,30) in it first ( for 2 widgets cost is 30)

So simply plug 30 in y place and 2 in x place in equation (1) as shown

$30 = a(2)^{2} +b(2)+c$

$30 = 4a +2b +c$ ----------------------------------------------(2)

Similarly plug next point (4,118) now in equation (1)

$118 = a(4)^{2} +b(4)+c$

$118 = 16a +4b +c$ ----------------------------------------------(3)

Similarly plug third point (10,766) now in equation (1)

$766 = a(10)^{2} +b(10)+c$

$766 = 100a +10b +c$ ----------------------------------------------(4)

Now we have to solve system of linear equations (2),(3) and (4) for a,b,c

We can use method of elimination

Subtract equation (3) - equation (2)

118 = 16a +4b +c

-(30 = 4a +2b +c)

----------------------------------------------------

88 = 12a +2b -----------------------------------(5)

Similarly subtract equation (4)-equation (3)

766 =100a +10b +c

-( 118 = 16a +4b +c)

--------------------------------------

648 = 84a + 6b ------------------------(6)

Now to eliminate b in equations (5) and (6) multiply equation (5) by -3 and then add to equation (6)

-3×(88 = 12a +2b)

-264 = -36a - 6b

648 = 84a + 6b

---------------------------------

384 = 48a

Now solve for a

$\frac{384}{48} = \frac{48a}{48}$

8 = a

Plug 8 in a place in equation (5) or (6) and solve for b as shown

88 = 12(8) +2b

88 = 96 + 2b

88 -96 = 96 +2b -96

-8 = 2b

$\frac{-8}{2} = \frac{2b}{2}$

-4 = b

now plug 8 in a place, -4 in b place in either of equations (2),(3) or (4) and solve for c

30 = 4(8) +2(-4) +c

30 = 32 -8 + c

30 = 24 + c

30 - 24 = 24 + c - 24

6 = c

So plugging values of a as 8, b as -4 and c as 6 in equation (1) we get

$y = 8x^{2} -4x+ 6$

So thats the cost function. So now for 6 widgets, plug 6 in x place in this equation and find y value

$y = 8(6)^{2} -4(6)+ 6$

y = 8(36) - 24 +6

y = 288 -24 +6

y = 270

So $270 is the cost for producing 6 widgets and thats the final answer

4 0

3 years ago

One of the roots of the quadratic equation dx^2+cx+p=0 is twice the other, find the relationship between d, c and p

scZoUnD [109]

Answer:

$c^2 = 9dp$

Step-by-step explanation:

Given

$dx^2 + cx + p = 0$

Let the roots be $\alpha$ and $\beta$

So:

$\alpha = 2\beta$

Required

Determine the relationship between d, c and p

$dx^2 + cx + p = 0$

Divide through by d

$\frac{dx^2}{d} + \frac{cx}{d} + \frac{p}{d} = 0$

$x^2 + \frac{c}{d}x + \frac{p}{d} = 0$

A quadratic equation has the form:

$x^2 - (\alpha + \beta)x + \alpha \beta = 0$

So:

$x^2 - (2\beta+ \beta)x + \beta*\beta = 0$

$x^2 - (3\beta)x + \beta^2 = 0$

So, we have:

$\frac{c}{d} = -3\beta$ -- (1)

and

$\frac{p}{d} = \beta^2$ -- (2)

Make $\beta$ the subject in (1)

$\frac{c}{d} = -3\beta$

$\beta = -\frac{c}{3d}$

Substitute $\beta = -\frac{c}{3d}$ in (2)

$\frac{p}{d} = (-\frac{c}{3d})^2$

$\frac{p}{d} = \frac{c^2}{9d^2}$

Multiply both sides by d

$d * \frac{p}{d} = \frac{c^2}{9d^2}*d$

$p = \frac{c^2}{9d}$

Cross Multiply

$9dp = c^2$

or

$c^2 = 9dp$

Hence, the relationship between d, c and p is: $c^2 = 9dp$

8 0

3 years ago

Pls write in comments and write it clearly

Annette [7]

Answer:

(3X+4)^2

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers