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3 years ago

Assssssssssssssssssssssssssssssssss

Mathematics

1 answer:

uysha [10]3 years ago

7 0

Answer:

15, 9, 3n

Step-by-step explanation:

Just multiply the number of red balls by 3 to get total number, and divide the total number by three to get the number of red balls

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Solve x2 – 64 = 0. Isolate x2: x2 = 64 Apply the square root property of equality: StartRoot x squared EndRoot = plus or minus S

Archy [21]

Answer:

8, -8

Step-by-step explanation:

I took the assignment on edge :)

3 0

3 years ago

Read 2 more answers

In this figure a || b Find the measure of each angle using either of the given angle measures. Identify the reason for each answ

geniusboy [140]

Answer:

$m$

Step-by-step explanation:

Part 1) Find the measure of angle 1

we know that

$m$ -----> by supplementary angles

$m$

Part 2) Find the measure of angle 2

we know that

$m$ -----> by alternate exterior angles

Part 3) Find the measure of angle 3

we know that

$m$ -----> by vertical angles

Part 4) Find the measure of angle 4

we know that

$m$ ------> by corresponding angles

Part 5) Find the measure of angle 5

we know that

$m$ -----> by vertical angles

$m$

Part 6) Find the measure of angle 6

we know that

$m$ -----> by supplementary angles

we have

$m$

substitute

$86\°+m$

$m$

3 0

3 years ago

Simplify if possible <br><br> (1/2)^4

mash [69]

Answer:

See the attachment

Step-by-step explanation:

Hope this helps :)

5 0

3 years ago

Do I add all the GPAs together and then add all the GPAs above Salley and then subtract them from each other?

solong [7]

No you add them all together and then you divide the number of GPA's to get the mean. You go from greatest to lowest to figure out the median and to find the mode excepts my mind. <span />

8 0

3 years ago

colby and jaquan are growing bacteria in an experiment in a laboratory. Colby starts with 50 bacteria in his culture and the num

kaheart [24]

To get started, we will use the general formula for bacteria growth/decay problems:

$A_{f} = A_{i} ( e^{kt} )$

where:
A_{f} = Final amount
A_{i} = Initial amount
k = growth rate constant
t = time

For doubling problems, the general formula can be shortened to:

$kt = ln(2)$

Now, we can use the shortened formula to calculate the growth rate constant of both bacteria:

Colby (1):
$k_{1} = ln(2)/t$
$k_{1} = ln(2)/2 = 0.34657$ per hour

Jaquan (2):
$k_{2} = ln(2)/t$
$k_{2} = ln(2)/3 = 0.23105$ per hour

Using Colby's rate constant, we can use the general formula to calculate for Colby's final amount after 1 day (24 hours).

Note: All units must be constant, so convert day to hours.

$A_{f1} = 50( e^{0.34657(24)})$
$A_{f1} = 204,800$

Remember that the final amount for both bacteria must be the same after 24 hours. Again, using the general formula, we can calculate the initial amount of bacteria that Jaquan needs:

$A_{f2} = 204,800 = A_{i2} ( e^{0.23105(24)} )$
$A_{i2} = 800$

3 0

3 years ago