What is the answer to this??

Answers

Answer 1

Answer: The answer is B 2+3=5 change 5 into 5/1 do 1/5×5/1 which equals 5/5 then change it to 1z -1+1 equals 0, -4z equals -3

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Simplify the expression below: 4+2×[(3+1)×5] (show work please need it for review)

Answers

Answer:

Step-by-step explanation:

Parenthesis

Exponents

Multiply

Divide

Add

Subtract

First we need to do the work inside the parenthesis that are in the boxed parenthesis

3 + 1 = 4

4 * 5 = 20

Now we multiply

20 * 2 = 40

Now we add

40 + 4 = 44

Best Of Luck,

- I.A. -

What is the rule for finding the ratio of the areas of two objects if you know the scale?

Answers

Answer:

The square of the scale

Step-by-step explanation:

Scaling

Suppose we know the scale of the distances, for example, r=5. It means that for each centimeter in the drawing, it corresponds to 5 cm in the real object being scaled. If we wanted to find the ratio of the areas, we would need to scale both dimensions and the ratio would be the square of the scale factor. In our example, the ratio of the areas would be 25, i.e. each square centimeter would correspond to 25 square centimeters

The rule for finding the ratio of the areas of two objects: The square of the scale

Telly’s salary increased from $60 to $80 a week. What percent of increase in salary was this?

Answers

The change is $20. You just need to calculate that how percent of the original salary($60) is $20.

20/60*100 = 33.3%

Which equation represents a circle with a center at (2, –8) and a radius of 11?

Answers

(x-h)^2 +(y-k)^2 = r^2 is the equation for a circle centered at (h,k) and radius r

(x-2)^2 +(y- (-8))^2 = 11^2

(x-2)^2 +(y+8)^2 = 11^2

Answer:(x-2)^2 +(y+8)^2 = 11^2

Answer:

(x-2)^2 + (y+8)^2 = 11^2

Step-by-step explanation:

The standard equation of a circle with center at (h,k) and radius r is

(x-h)^2 + (y-k)^2 = r^2.

Filling in the given info, we get:

(x-2)^2 + (y+8)^2 = 11^2

colby and jaquan are growing bacteria in an experiment in a laboratory. Colby starts with 50 bacteria in his culture and the number of bacteria doubles every two hours. Jaquan has a different type of bacteria that doubles every three hours. How many bacteria should Jaquan start with so that they have they have the same amount at the end of the day?

Answers

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:

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

prints in 2 pounds bags he repackage them into bag that hold 5/6 pounds how many 2 pounds bags of peanut should Chad buy so that he can fill the 5/6 pounds bags without having any peanut left over

Answers

5, 2 pound bags

5 × 2 = 10
10 ÷ 5/6 = 12

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