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weqwewe [10]
3 years ago
6

Kohls sells casual shirts for 9$ and polo shirts for 18$. Total sales were 1,080 and customers bought 4 times as many casual shi

rts as polo shirts. How many casua shirts did Kohls sell? Id like to know how to do the math
Mathematics
1 answer:
lukranit [14]3 years ago
8 0
The ratio would be 36 to 18
because 9*4=36
and 18 is 18 lol
so (calculator) 1,080/9=120
120
20*18=180
100*9=900
so

idek man this is hard as hell,I tried tho,try to do the rest of the problem from what I have set up
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Answer:

C. 700 km/hour

Step-by-step explanation:

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3 years ago
Sketch the graph of y = (x + 3)2 – 4 and identify the axis of symmetry.
Sloan [31]

Answer: Thought I’d return the favor and help u with this question! But anyways, the axis of symmetry is at x = -3.

Explanatio: This can be found by looking at the basic form of vertex form:

y = (x - h)^2 + k

In this basic form the vertex is (h, k). By looking at what is plugged into the equation, it is clear that h = -3 and k = -4. This means the vertex is at (-3, -4).

It is a fact that the axis of symmetry is a vertical line of x = (vertex value of x). So we can determine that the axis of symmetry is at x = -3

i hope this helps u

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determine which postulate or theorem can be used to prove that abc is congruent to dcb is it sas, aas, sss, or asa?
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Terry weighs 40kg janice weighs 2 3/4 lg less than terry. what is their combined weight?
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I hope this helps you



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7 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.

4 0
3 years ago
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