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labwork [276]
3 years ago
11

Can someone please help me with question 7?

Mathematics
1 answer:
storchak [24]3 years ago
6 0

The goal is to raise AT LEAST $180, so this automatically indicates that the amount of keychains sold with the price must be the same or more than $180.

k = number of keychains

each keychain is $2.25

2.25k ≥ 180

divide both sides by 2.25 to get the k alone

k ≥ 80

The answer is 2.25k ≥ 180, k ≥ 80

(Hope this isn't confusing)


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The diagram shows a right-angled triangle.
Ratling [72]
Where is the diagram?
4 0
3 years ago
Which expressions have the same value as .04 x .005. Select all that apply.
Novay_Z [31]

Answer:

Hello There!!

Step-by-step explanation:

0.04×0.005=0.0002

.0002 ✔

(4/100)=>0.04 × (5/1000)=>0.005=0.0002 ✔

(4 × .01)=>0.04 × (5 × .001)=>0.0005 =0.00002 ❌

.002 ❌

4 × 5 × .00001=0.0002 ✔

hope this helps,have a great day!!

~Pinky~

4 0
3 years ago
An object is translated by (x-2, y-6). If one point in the pre-image has the coordinates (-3, 7,) what would be the coordinates
Katen [24]

Answer:

(- 5, 1 )

Step-by-step explanation:

Given the translation rule

(x, y ) → (x - 2, y - 6 )

This means subtract 2 from the original x- coordinate and subtract 6 from the original y- coordinate, that is

(- 3, 7 ) → (- 3 - 2, 7 - 6 ) → (- 5, 1 )

5 0
4 years ago
Prove that $5^{3^n} + 1$ is divisible by $3^{n + 1}$ for all nonnegative integers $n.$
Viktor [21]

When n=0, we have

5^{3^0} + 1 = 5^1 + 1 = 6

3^{0 + 1} = 3^1 = 3

and of course 3 | 6. ("3 divides 6", in case the notation is unfamiliar.)

Suppose this is true for n=k, that

3^{k + 1} \mid 5^{3^k} + 1

Now for n=k+1, we have

5^{3^{k+1}} + 1 = 5^{3^k \times 3} + 1 \\\\ ~~~~~~~~~~~~~ = \left(5^{3^k}\right)^3 + 1^3 \\\\ ~~~~~~~~~~~~~ = \left(5^{3^k} + 1\right) \left(\left(5^{3^k}\right)^2 - 5^{3^k} + 1\right)

so we know the left side is at least divisible by 3^{k+1} by our assumption.

It remains to show that

3 \mid \left(5^{3^k}\right)^2 - 5^{3^k} + 1

which is easily done with Fermat's little theorem. It says

a^p \equiv a \pmod p

where p is prime and a is any integer. Then for any positive integer x,

5^3 \equiv 5 \pmod 3 \implies (5^3)^x \equiv 5^x \pmod 3

Furthermore,

5^{3^k} \equiv 5^{3\times3^{k-1}} \equiv \left(5^{3^{k-1}}\right)^3 \equiv 5^{3^{k-1}} \pmod 3

which goes all the way down to

5^{3^k} \equiv 5 \pmod 3

So, we find that

\left(5^{3^k}\right)^2 - 5^{3^k} + 1 \equiv 5^2 - 5 + 1 \equiv 21 \equiv 0 \pmod3

QED

5 0
2 years ago
40 POINTS PLEASE HELP!
White raven [17]

Answer:

-1.70588235294

1. X3,    15x - 10y = 80   15x - 7y = 51

-17y = 29

2. 8 = 1/60 + a. 8 - 1/60 = 7.98333333333

  6 = 1/80 + b. 6 - 1/80 =  5.9875

                                         13.9708333333

3. I genuinely dont know sorry.

4. Pencil = $0.05. Pen = $1.05

Step-by-step explanation:

7 0
3 years ago
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