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serious [3.7K]
2 years ago
7

If the exchange rate of 1 dollar is rs 72.45, how many dollar can be exchanged for rs 1449?

Mathematics
1 answer:
Shalnov [3]2 years ago
4 0

Answer:

20

Step-by-step explanation:

1449÷72.45=20

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Select the correct numbers on the number line. Which points on the number line are 8 units away from 0?
Harman [31]

Answer:

8 and -8

Step-by-step explanation:

0-8= -8

0+8= 8

Both 8 jumps from 0

6 0
2 years ago
What equation do I need to use?
goldenfox [79]
To ease your problem, consider "L" as you x-axis
Then the coordinate become:
A(- 4 , 3)   and B(1 , 2) [you notice that just the y's changed]

This is a reflection problem.
Reflect point B across the river line "L" to get B', symmetric of B about L.
The coordinates of B'(1 , -1) [remember L is our new x-axis]
JOIN A to B' . AB' intersect L, say in H
We have to find the shortest way such  that AH + HB = shortest.
But HB = HB' (symmetry about L) , then I can write instead of 
AH + HB →→ AH + HB'. This is the shortest since the shortest distance between 2 points is the straight line and H is the point requiered 


6 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
Which equation can be used to solve the problem? 750 is 150 percent of what? StartFraction 150 times 1 Over 750 times 1 EndFract
Bezzdna [24]

Answer:

B

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
Maryland had $(62xsquared + x - 4) on mondays she spent $(11x squared - 33) on a pair of gloves how much money did she have left
Anton [14]

Answer:

$( 51x^2 + x + 29).

Step-by-step explanation:

Amount she had left = original amount - amount spent on the gloves

=  62x^2 + x - 4 - (11x^2 - 33)   (note we place the amount spent on gloves in parentheses because we have to subtract the whole amount)

Now we distribute the negative over the parentheses:

= 62x^2 + x - 4 - 11x^2 + 33     ( note - 33 becomes -33*-1 = +33)

Now simplifying like terms:

= 51x^2 + x + 29  (answer).

7 0
3 years ago
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