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3 years ago

15

A. 4 packs of pencils and 3 packs of erasers B. 4 packs of pencils and 5 packs of erasers C. 5 packs of pencils and 3 packs of e

rasers D. 10 packs of pencils and 6 packs of erasers

1 answer:

amid [387]3 years ago

3 0

C. 5 packs of pencils and 3 packs of erasers. it is c because no matter what 20(the pack of erasers) is multiplied by it will always end in a zero, therefore you would have to multiply 12 by a number that would make it a multiple of 20 as well which happens to be 5. 5×12=60 and 3×20=60

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Pleade help and show work

Jet001 [13]

9514 1404 393

Answer:

11) x = 9

12) x = 9

Step-by-step explanation:

11) Corresponding segments are proportional, so ...

x/6 = 12/8

x = 6(12)/8 . . . multiply by 6

x = 9

__

12) Similar triangle relationships show you the geometric mean relationship that applies in this case.

long side/short side = x/3 = 27/x

x² = 3·27 . . . . . . cross multiply

x = √(3·27) . . . . geometric mean relationship

x = 9

8 0

2 years ago

Jhon is making brownies for 48 of his family members. he can make 16 brownies from one package, and he expects everyone yo eat 2

likoan [24]

He will need 48×2=96 brownies so 96÷16= 6 packages of brownie mix

3 0

3 years ago

Find all of the equilibrium solutions. Enter your answer as a list of ordered pairs (R,W), where R is the number of rabbits and

zloy xaker [14]

Answer:

$(0,0)$ $(4000,0)$ and $(500,79)$

Step-by-step explanation:

Given

See attachment for complete question

Required

Determine the equilibrium solutions

We have:

$\frac{dR}{dt} = 0.09R(1 - 0.00025R) - 0.001RW$

$\frac{dW}{dt} = -0.02W + 0.00004RW$

To solve this, we first equate $\frac{dR}{dt}$ and $\frac{dW}{dt}$ to 0.

So, we have:

$0.09R(1 - 0.00025R) - 0.001RW = 0$

$-0.02W + 0.00004RW = 0$

Factor out R in $0.09R(1 - 0.00025R) - 0.001RW = 0$

$R(0.09(1 - 0.00025R) - 0.001W) = 0$

Split

$R = 0$ or $0.09(1 - 0.00025R) - 0.001W = 0$

$R = 0$ or $0.09 - 2.25 * 10^{-5}R - 0.001W = 0$

Factor out W in $-0.02W + 0.00004RW = 0$

$W(-0.02 + 0.00004R) = 0$

Split

$W = 0$ or $-0.02 + 0.00004R = 0$

Solve for R

$-0.02 + 0.00004R = 0$

$0.00004R = 0.02$

Make R the subject

$R = \frac{0.02}{0.00004}$

$R = 500$

When $R = 500$ , we have:

$0.09 - 2.25 * 10^{-5}R - 0.001W = 0$

$0.09 -2.25 * 10^{-5} * 500 - 0.001W = 0$

$0.09 -0.01125 - 0.001W = 0$

$0.07875 - 0.001W = 0$

Collect like terms

$- 0.001W = -0.07875$

Solve for W

$W = \frac{-0.07875}{ - 0.001}$

$W = 78.75$

$W \approx 79$

$(R,W) \to (500,79)$

When $W = 0$ , we have:

$0.09 - 2.25 * 10^{-5}R - 0.001W = 0$

$0.09 - 2.25 * 10^{-5}R - 0.001*0 = 0$

$0.09 - 2.25 * 10^{-5}R = 0$

Collect like terms

$- 2.25 * 10^{-5}R = -0.09$

Solve for R

$R = \frac{-0.09}{- 2.25 * 10^{-5}}$

$R = 4000$

So, we have:

$(R,W) \to (4000,0)$

When $R =0$ , we have:

$-0.02W + 0.00004RW = 0$

$-0.02W + 0.00004W*0 = 0$

$-0.02W + 0 = 0$

$-0.02W = 0$

$W=0$

So, we have:

$(R,W) \to (0,0)$

Hence, the points of equilibrium are:

$(0,0)$ $(4000,0)$ and $(500,79)$

4 0

3 years ago

identify the amplitude and period of the function then graph the function and describe the graph of G as a transformation of the

mrs_skeptik [129]

Given the function:

$g(x)=cos4x$

Let's find the amplitude and period of the function.

Apply the general cosine function:

$f(x)=Acos(bx+c)+d$

Where A is the amplitude.

Comparing both functions, we have:

A = 1

b = 4

Hence, we have:

Amplitude, A = 1

To find the period, we have:

$\frac{2\pi}{b}=\frac{2\pi}{4}=\frac{\pi}{2}$

Therefore, the period is = π/2

The graph of the function is shown below:

The parent function of the given function is:

$f(x)=cosx$

Let's describe the transformation..

Apply the transformation rules for function.

We have:

The transformation that occured from f(x) = cosx to g(x) = cos4x using the rules of transformation can be said to be a horizontal compression.

ANSWER:

Amplitude = 1

Period = π/2

Transformation = horizontal compression.

8 0

1 year ago

Of the cars sold during the month of July, 89 had air conditioning, 99 had an automatic transmission, and 74 had power steering.

DedPeter [7]

Answer:

There are a total of 23 cars with air conditioning and automatic transmission but not power steering

Step-by-step explanation:

Let A be the cars that have Air conditioning, B the cars that have Automatic transmission and C the cars that have pwoer Steering. Lets denote |D| the cardinality of a set D.

Remember that for 2 sets E and F, we have that

$|E \cup F| = |E| + |F| - |E \cap F|$

Also,

|E| = |E ∩F| + |E∩F^c|

We now alredy the following:

|A| = 89

|B| = 99

|C| = 74

$|A \cap B \cap C| = 5$

|(A \cup B \cup C)^c| = 24

|A \ (B U C)| = 24 (This is A minus B and C, in other words, cars that only have Air conditioning).

|B \ (AUC)| = 65

|C \ (AUB)| = 26

$|B \cap C| = 11$

We want to know |(A∩B) \ C|. Lets calculate it by taking the information given and deducting more things

For example:

99 = |B| = |B ∩ C| + |B∩C^c| = 11 + |B∩C^c|

Therefore, |B∩C^c| = 99-11 = 88

And |A ∩ B ∩ C^c| = |B∩C^c| - |B∩C^c∩A^c| = |B∩C^c| - |B \ (AUC)| = 88-65 = 23.

This means that the amount of cars that have both transmission and air conditioning but now power steering is 23.

3 0

3 years ago