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arlik [135]
3 years ago
13

How to answer this problem

Mathematics
1 answer:
ddd [48]3 years ago
8 0

the solution is in pic .....base area is l×b and height is h

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Inessa05 [86]
It is 16 square units

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4 years ago
Pls help on #9 ASAP. thank you. I’ll give u as much points as i can ♡
Alinara [238K]
A. X> lobsters 
7 + x <= 18
X <= 11 lovsters. 

B. (7+3+10)
x= additional lobsters 
10 + x <= 24 
x<= 14 lobsters

C. 24/4 = 6 lobsters

Hope this helps.

6 0
3 years ago
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From a group of 6 candidates, a committee of 2 people is selected. In how many different ways can the committee be selected? I k
kolezko [41]
This is a combination problem. 

6 nCr 2

\frac{6!}{2! * (6-2)!} =  \frac{720}{48}=15<span><span>

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3 years ago
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Find the measure of ∠EGC. Circle A with chords EF and CD that intersect at point G, the measure of arc EC is 50 degrees, the mea
miv72 [106K]

Answer:

m∠EGC=70°

Step-by-step explanation:

we know that

The measure of the inner angle is the semi-sum of the arcs comprising it and its opposite

so

m∠EGC=(1/2)[arc EC+arc DF]

<u><em>Find the value of x</em></u>

we have

m∠EGC=(7x+7)°

arc EC=50°

arc DF=10x°

substitute and solve for x

(7x+7)°=(1/2)[50°+10x°]

14x+14=50+10x

14x-10x=50-14

4x=36

x=9

<u><em>Find the measure of angle EGC</em></u>

m∠EGC=(7x+7)°

substitute the value of x

m∠EGC=(7(9)+7)°=70°

8 0
4 years ago
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300 ml of pure alcohol is poured from a bottle containing 2 l of pure alcohol. Then, 300 ml of water is added into the bottle. A
Ede4ka [16]

Answer:

The present percentage of pure alcohol in the solution is 72.25% of pure alcohol

Step-by-step explanation:

The volume of pure alcohol poured from the 2 l bottle of pure alcohol = 300 ml of pure alcohol

The volume of water added into the bottle after pouring out the pure alcohol = 300 ml of water

The volume of diluted alcohol poured out of the bottle = 300 ml of diluted alcohol

The volume of water added into the bottle of diluted alcohol after pouring out the 300 ml of diluted alcohol = 300 ml of water

Step 1

After pouring the 300 ml of pure alcohol and adding 300 ml of water to the bottle, the percentage concentration, C%₁ is given as follows;

C%₁ = (Volume of pure alcohol)/(Total volume of the solution) × 100

The volume of pure alcohol in the bottle = 2 l - 300 ml = 1,700 ml

The total volume of the solution = The volume of pure alcohol in the bottle +  The volume of water added = 1,700 ml + 300 ml = 2,000 ml = 2 l

∴ C%₁ = (1,700 ml)/(2,000 ml) × 100 = 85% percent alcohol

Step 2

After pouring out 300 ml diluted alcohol from the 2,000 ml, 85% alcohol and adding 300 ml of water, we have;

Volume of 85% alcohol = 2,000 ml - 300 ml = 1,700 ml

The volume of pure alcohol in the 1,700 ml, 85% diluted alcohol = 85/100 × 1,700 = 1,445 ml

The total volume of the diluted solution = The volume of the 85% alcohol in the solution + The volume of water added

∴ The total volume of the twice diluted solution = 1,700 ml + 300 ml = 2,000 ml

The present percentage of pure alcohol in the solution, C%₂ = (The volume of pure alcohol in the 1,700 ml, 85% diluted alcohol)/(The total volume of the diluted solution) × 100

∴ C%₂ = (1,445 ml)/(2,000 ml) × 100 = 72.25 %

The present percentage of pure alcohol in the solution, C%₂ = 72.25%

3 0
3 years ago
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