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Dimas [21]
3 years ago
8

What is the greatest common factor of 32 and 48

Mathematics
2 answers:
TEA [102]3 years ago
4 0
The gcf of 32 & 48 is 16.
Maru [420]3 years ago
3 0
The greatest common factor would be 16
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Can anyone help me with my homework?
Evgesh-ka [11]

I want more points next time....Hehehee!

<h2><u>______________________</u></h2>

<em>Solution in attachments!</em>

6 0
2 years ago
Given that ABDC : EFGH, what is m G?<br> 27°<br> 90°<br> 117°<br> 120°
oksian1 [2.3K]
I think it would be 117

Because H is 63
So 90+90+63=243
360-243=117
I used 360 because angles in quadrilaterals add up to 360 degrees
4 0
3 years ago
F 6. A carpenter bought a table for 3500. He spent 500 on repairing and painting it. He then sold it for 3500. What was his prof
Anon25 [30]

Answer:

the carpenter lost 500

Step-by-step explanation:

if he bought the table for 3500 and spent an additional 500 on repairs but only sold it for 3500 he didn't make a profit and lost 500

8 0
1 year ago
The sum of two integers is 61 and their difference is 1. Find the integers.
musickatia [10]

We can find this integer using a system of equations.  The first thing we look at is the first sentence.

Let's call X The first integer,

Let's call Y the second integer.


So, X+Y=61 and X-Y= 1


Using Substituion on the second equation, we get

X=Y+1

(Now we plug in Y+1 into the second equation for X)

So,

(Y+1)+Y=61

2Y=60

Y=30

So, one of the integers is 30.

Now to find the second, we go back to

X=Y+1

X=30+1

X=31.


So, the integers are 31 and 30.


Hope this helps!

7 0
3 years ago
A company manufactures and sells x television sets per month. The monthly cost and price-demand equations are
skelet666 [1.2K]

A) The maximum revenue is 450000$

B) The maximum profit is 216000$ when 2400 sets are manufactured and sold for 180$ each

C) When each set is taxed at ​$55​, the maximum profit is 99125$ when 1850 sets are manufactured and sold for 207.5$ each.

A)p(x)=300−(x/20​),

revenue R(x)=p*x

revenue R(x)=300x -(x2/20)

for maximum revenue dR/dx =0 ,

=>300-(2x/20)=0

=>x/10=300

=>x=3000

maximum revenue = R(3000)=300*3000 -(30002/20)

maximum revenue = R(3000)=450000$

B) profit =revenue -cost

profit P(x)=300x -(x2/20)-72000-60x

profit P(x)=240x -(x2/20)-72000

for maximum cost dP/dx =0

240 -(2x/20)=0

x=240*10

x=2400

p(2400)=300−(2400/20​)=180

profit P(2400)=240*2400 -(24002/20)-72000 =216000

The maximum profit is 216000$ when 2400 sets are manufactured and sold for 180$ each

c)

profit =revenue -cost -tax

profit P(x)=300x -(x2/20)-72000-60x-55x

profit P(x)=185x -(x2/20)-72000

for maximum cost dP/dx =0

185-(2x/20)=0

x=185*10

x=1850

p(1850)=300−(1850/20​)=207.5

profit P(1850)=185*1850 -(18502/20)-72000

profit P(1850)=99125$

When each set is taxed at ​$55​, the maximum profit is 99125$ when 1850 sets are manufactured and sold for 207.5$ each.

To know more about maximum profit check the below link:

brainly.com/question/4166660

#SPJ4

5 0
1 year ago
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