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morpeh [17]
3 years ago
5

PLS ASAP! help sort the angle pairs!

Mathematics
1 answer:
ollegr [7]3 years ago
5 0

Answer:

complamentary:20 and 70    

47 and 43

34 and 56

supplamentary:

150 and 30

72 and 108

85 and 95

for something to be comp it has to equal 90 when you add it up for something to be supp it has to equal 180 when you add it up

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How to do this question..?
Leokris [45]

uh idk um...

Okay well to answer this you have to work the problem out and enable to do that you have to look at the information provided, also STOP CHEATING AND FIGURE IT OUT

um i mean.....GOODLUCK!!!! I know you can do it :)

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3 years ago
Solve: |2 + 4x| 11
exis [7]
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3 years ago
Two people apply for loans of the same amount. Due to differences in their credit scores, their payments differ by $72
kati45 [8]

Answer:

\large \boxed{\text{d) \$2592 }}

Step-by -step explanation:

The person with the lower credit score pays $72 more each month.

Over 36 mo, they will pay an extra

\text{Total amount} = \text{36 mo} \times \dfrac{\text{\$72}}{\text{1 mo}} = \textbf{\$2592}\\\\\text{The person with the lower credit score will have paid an extra $\large \boxed{\mathbf{\$2592 }}$}

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3 years ago
What is the value of 2/3− (3/4) * (2)÷( -4/3 ) ?
blsea [12.9K]
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3 years ago
Seorang pemilik toko sepatu ingin mengisi tokonya dengan sepatu laki-laki paling sedikit 100 pasang dan sepatu wanita paling sed
taurus [48]

Answer:

<em>The biggest profit that can be obtained by the shop owner= </em>IDR 2,750,000.

Step-by-step explanation:

We need to form a LPP( Linear Programming Problem) for the following problem and solve it to get the maximize the profit.

Let x denote the number of men's shoe and y denote the number of female shoes.

Maximize z= 10000x+5000y----------(1)

             x≥100 ( since a shoe store owner wants to fill his shop with at least    100 pairs of men's shoes )

            y≥150  (at least 150 pairs of women's shoes).

also x+y≤400   (The store can only accommodate 400 pairs of shoes).

         x≤150    (male shoes should not exceed 150 pairs).

⇒         100≤x≤150

Hence the optimal solution of an LPP always lie on the end points.

We have end points as:

(100,300), (150,250), (100,150),(150,150).

by putting these value in equation (1) we see which give the maximum solution.

for (100,300) i.e. x=100 and y=300:   z=2,500,000

for (150,250) i.e. x=150 and y=250:  z=2,750,000

for (100,150) i.e. x=100 and y=150: z=1,750,000

for (150,150) i.e. x=150 and y=150: z=2,250,000

Hence maximum profit is obtained at x=100 and y=300.

i.e. to maximize the profit:

<em>Number of men's shoes to be kept in store=100</em>

<em>and number of women's shoes to be kept in store=300</em>

<em>The biggest profit that can be obtained by the shop owner= </em>IDR 2,750,000.

5 0
3 years ago
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