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NeX [460]
3 years ago
14

Solve this question using standard algorithm for multiplying a fraction by the whole number.Simplify the product.3x3/4

Mathematics
1 answer:
Eva8 [605]3 years ago
8 0

Answer:

\frac{9}{4}

Step-by-step explanation:

Hi there!

3*\frac{3}{4}

Rewrite 3 as a fraction

= \frac{3}{1} *\frac{3}{4}

Multiply the numerators and the denominators

3*3=9\\1*4=4

= \frac{9}{4}

I hope this helps!

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Please help me!!!!!​
denpristay [2]

Answer:  see proof below

<u>Step-by-step explanation:</u>

Given: A + B + C = π               → A = π - (B + C)

                                               → B = π - (A + C)

                                               → C = π - (A + B)

Use Sum to Product Identity: sin A - sin B = 2 cos [(A + B)/2] · sin [(A - B)/2]

Use the following Cofunction Identity: cos (π/2 - A) = sin A

<u>Proof LHS → RHS:</u>

LHS:                        sin A - sin B + sin C

                             = (sin A - sin B) + sin C

\text{Sum to Product:}\quad 2\cos \bigg(\dfrac{A+B}{2}\bigg)\cdot \sin \bigg(\dfrac{A-B}{2}\bigg)+2\cos \bigg(\dfrac{C}2{}\bigg)\cdot \sin \bigg(\dfrac{C}{2}\bigg)

\text{Given:}\qquad 2\cos \bigg(\dfrac{\pi -(B+C)}{2}+\dfrac{B}{2}}\bigg)\cdot \sin \bigg(\dfrac{A-B}{2}\bigg)+2\cos \bigg(\dfrac{C}2{}\bigg)\cdot \sin \bigg(\dfrac{C}{2}\bigg)\\\\\\.\qquad \qquad =2\cos \bigg(\dfrac{\pi -C}{2}\bigg)\cdot \sin \bigg(\dfrac{A-B}{2}\bigg)+2\cos \bigg(\dfrac{C}2{}\bigg)\cdot \sin \bigg(\dfrac{C}{2}\bigg)

.\qquad \qquad =2\cos \bigg(\dfrac{\pi}{2} -\dfrac{C}{2}\bigg)\cdot \sin \bigg(\dfrac{A-B}{2}\bigg)+2\cos \bigg(\dfrac{C}2{}\bigg)\cdot \sin \bigg(\dfrac{C}{2}\bigg)

\text{Cofunction:} \qquad 2\sin \bigg(\dfrac{C}{2}\bigg)\cdot \sin \bigg(\dfrac{A-B}{2}\bigg)+2\cos \bigg(\dfrac{C}2{}\bigg)\cdot \sin \bigg(\dfrac{C}{2}\bigg)

\text{Factor:}\qquad 2\sin \bigg(\dfrac{C}{2}\bigg)\bigg[ \sin \bigg(\dfrac{A-B}{2}\bigg)+\cos \bigg(\dfrac{C}{2}\bigg)\bigg]

\text{Given:}\qquad 2\sin \bigg(\dfrac{C}{2}\bigg)\bigg[ \sin \bigg(\dfrac{A-B}{2}\bigg)+\cos \bigg(\dfrac{\pi -(A+B)}{2}\bigg)\bigg]\\\\\\.\qquad \qquad =2\sin \bigg(\dfrac{C}{2}\bigg)\bigg[ \sin \bigg(\dfrac{A-B}{2}\bigg)+\cos \bigg(\dfrac{\pi}{2} -\dfrac{(A+B)}{2}\bigg)\bigg]

\text{Cofunction:}\qquad 2\sin \bigg(\dfrac{C}{2}\bigg)\bigg[ \sin \bigg(\dfrac{A-B}{2}\bigg)+\sin \bigg(\dfrac{A+B}{2}\bigg)\bigg]

\text{Sum to Product:}\qquad 2\sin \bigg(\dfrac{C}{2}\bigg)\bigg[ 2\sin \bigg(\dfrac{A}{2}\bigg)\cdot \cos \bigg(\dfrac{B}{2}\bigg)\bigg]\\\\\\.\qquad \qquad \qquad \qquad =4\sin \bigg(\dfrac{A}{2}\bigg)\cdot \cos \bigg(\dfrac{B}{2}\bigg)\cdot \sin \bigg(\dfrac{C}{2}\bigg)

\text{LHS = RHS:}\quad 4\sin \bigg(\dfrac{A}{2}\bigg)\cdot \cos \bigg(\dfrac{B}{2}\bigg)\cdot \sin \bigg(\dfrac{C}{2}\bigg)=4\sin \bigg(\dfrac{A}{2}\bigg)\cdot \cos \bigg(\dfrac{B}{2}\bigg)\cdot \sin \bigg(\dfrac{C}{2}\bigg)\quad \checkmark

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-7 2/3 +(5 1/2) + 8 3/4 =
drek231 [11]

Answer:

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Step-by-step explanation:

-7 2/3 +(5 1/2) + 8 3/4

Get a common denominator of 12

-7 2/3 * 4/4 +(5 1/2 +6/6) + 8 3/4*3/3

-7 8/12    + 5 6/12     + 8 9/12

Subtract the  first two terms first

They have opposite signs so subtract and take the sign of the larger

-7 8/12    + 5 6/12    = -2 2/12

-2  2/12 + 8 9/12

They have opposite signs so subtract and take the sign of the larger

-2  2/12 + 8 9/12 = 6 7/12

7 0
3 years ago
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Korolek [52]

Answer:

180 square inches

Step-by-step explanation:

Area, A is given by A=0.5bh where b is base and h is height. Since both of these dimensions are provided then we substitute the given figures into the formula. Substituting 12 inches for base, b and 30 inches for height, h then the area will be equivalent to

A=0.5*12*30=180 square inches.

Therefore, the area of the described triangle is 180 square inches.

6 0
3 years ago
If $5,500 is invested in an account that pays 3% interest compounded annually, how much money will be in the account at the end
Citrus2011 [14]
He would have 7150 in his account
3 0
3 years ago
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