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rosijanka [135]
4 years ago
10

What is the answer to 2 1/9 - 7/6

Mathematics
2 answers:
ryzh [129]4 years ago
6 0

Answer:

17/18

Step-by-step explanation:

2 1/9=19/9

19/9-7/6

17/18

maw [93]4 years ago
5 0
The correct answer is

0.94444444444
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In a class of 30 students, 12 are girls. What percent of the class is girls
dexar [7]

Answer:

40%

Step-by-step explanation:

Total students = 30

No. of girls. = 12

Girls % = ?

Girls % = No. of girls/Total students * 100%

= 0.4 * 100

= 40%

5 0
3 years ago
Ron is tiling a countertop he needs to place 54 square titles in each of 8 rows to cover the counter
DiKsa [7]
He would only be able to make 6 rows with 3/4 of another 
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3 years ago
Factor Quadratics
icang [17]

Answer:

Step-by-step explanation:

x^2 + 17x + 60

(x + 12)(x + 5)

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3 years ago
Find the value of the variable and DF if D is between C and F if CD 4y -9, DF 2y-7, and CF 14.
Yuri [45]
CF = CD + DF (since D is simply a point on the line, and C to that point added onto F to that point is CF), so 4y-9+2y-7=14. Adding it up, we get 6y-16=14 and adding by 16 we get 30=6y. Dividing both sides by 6, we get y=5. Plugging that into DF, we get 10-7=DF=3
4 0
3 years ago
Use triangle ABC drawn below & only the sides labeled. Find the side of length AB in terms of side a, side b & angle C o
Brrunno [24]

Answer:

AB = \sqrt{a^2 + b^2-2abCos\ C}

Step-by-step explanation:

Given:

The above triangle

Required

Solve for AB in terms of a, b and angle C

Considering right angled triangle BOC where O is the point between b-x and x

From BOC, we have that:

Sin\ C = \frac{h}{a}

Make h the subject:

h = aSin\ C

Also, in BOC (Using Pythagoras)

a^2 = h^2 + x^2

Make x^2 the subject

x^2 = a^2 - h^2

Substitute aSin\ C for h

x^2 = a^2 - h^2 becomes

x^2 = a^2 - (aSin\ C)^2

x^2 = a^2 - a^2Sin^2\ C

Factorize

x^2 = a^2 (1 - Sin^2\ C)

In trigonometry:

Cos^2C = 1-Sin^2C

So, we have that:

x^2 = a^2 Cos^2\ C

Take square roots of both sides

x= aCos\ C

In triangle BOA, applying Pythagoras theorem, we have that:

AB^2 = h^2 + (b-x)^2

Open bracket

AB^2 = h^2 + b^2-2bx+x^2

Substitute x= aCos\ C and h = aSin\ C in AB^2 = h^2 + b^2-2bx+x^2

AB^2 = h^2 + b^2-2bx+x^2

AB^2 = (aSin\ C)^2 + b^2-2b(aCos\ C)+(aCos\ C)^2

Open Bracket

AB^2 = a^2Sin^2\ C + b^2-2abCos\ C+a^2Cos^2\ C

Reorder

AB^2 = a^2Sin^2\ C +a^2Cos^2\ C + b^2-2abCos\ C

Factorize:

AB^2 = a^2(Sin^2\ C +Cos^2\ C) + b^2-2abCos\ C

In trigonometry:

Sin^2C + Cos^2 = 1

So, we have that:

AB^2 = a^2 * 1 + b^2-2abCos\ C

AB^2 = a^2 + b^2-2abCos\ C

Take square roots of both sides

AB = \sqrt{a^2 + b^2-2abCos\ C}

6 0
3 years ago
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