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maxonik [38]
2 years ago
11

Determine the slope of the line that is perpendicular to the line y = -7 - 12. Select one: O A. O B. -3 O C. -7 O D. – 12​

Mathematics
1 answer:
OlgaM077 [116]2 years ago
4 0

Answer: I think it's D

Step-by-step explanation:

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Finish multiplying<br> 10/3 ⋅11/12?
Mazyrski [523]
When multiplying fractions, just multiply straight across

10 x 11 = 110
3 x 12 = 36

110/36 is your answer

3 1/18 is simplified mixed fraction


hope this helps
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3 years ago
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You have a gift card for your favorite clothing store for the amount of $60. You have found a shirt you want for $15. You don't
Zolol [24]

Let

x--------> the amount you have left to spend

y--------> the cost of a shirt you want

z------> the amount of the gift card

we know that

x+y \leq   z --------> equation 1

y=\$ 15 --------> equation 2

z=\$ 60 --------> equation 3

Substitute the equation 2 and equation 3 in the equation [tex] 1 [/tex]

x+y \leq   z

x+15 \leq   60

therefore

<u>the answer is the option</u>

x+15 is less than or equal to 60

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3 years ago
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Question 7 not answered which set of numbers with a difference of 2 is a factor pair of 48? answer choices 2 and 4 4 and 6 6 and
ehidna [41]

Answer:

  6 and 8

Step-by-step explanation:

All you have to do is find the difference between the numbers in each factor pair and make sure the factors combine to give 48.

2 and 4 . . . . . have a difference of 2, but 2·4 ≠ 48

4 and 6 . . . . . have a difference of 2, but 4·6 ≠ 48

6 and 8 . . . . . have a difference of 2, and 6·8 = 48

24 and 2 . . . . do not have a difference of 2

6 0
3 years ago
13. The least common multiple of two non-zero integers a and b is the unique positive integer m such that (i) m is a common mult
Vlad [161]

Answer:

[a,b] divides n

Step-by-step explanation:

Let us denote the least common multiple of a and b [a,b]=m.

We want to prove that m divides n, where n is a multiple of a and b.

We suppose m does not divide n, then by the Division Theorem, there exists q and r integers such that:

(1) ... n=mq+r, where 0<r<m

As n is a multiple of a and b, there exists s and t integers such that:

sa=n and tb=n

Same thing happens to m as it is the least common multiple, there exists u and v such that:

ua=m and vb=m

So (1) has the following form:

n=mq+r ⇒ sa=uaq+r ⇒sa-uaq=r⇒(s-uq)a=r and

n=mq+r ⇒ tb=vbq+r ⇒ tb-vbq=r⇒ (t-vq)b=r

So r is a multiple of a and b, but r<m which is a contradiction as, m is the least common multiple of a and b. So this concludes the proof.

So this means that \frac{ab}{m} is and integer.

As m= vb, then \frac{m}{b} is an integer, lets say \frac{m}{b}=v; and as m=ua, then \frac{m}{a}=u.

So \frac{ab}{m}v=\frac{ab}{m}\frac{m}{b}=a, so \frac{ab}{m} divides a; on the other hand, \frac{ab}{m}u=\frac{ab}{m}\frac{m}{a}=b, so \frac{ab}{m} divides b. From this we can conclude that \frac{ab}{m} is a common divisor of a and b.

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3 years ago
3. A principal orders 2,592 pencils. She gives an equal amount to each of
AfilCa [17]
Each teacher should receive 108 pencils
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