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3 years ago

12

An empty luggage case weighs three point five kilograms. Sarah is allowed to carry a total fifteen kilograms on the plane. How m

uch weight can she fit in her luggage case?

1 answer:

Nookie1986 [14]3 years ago

7 0

11.5 kilograms
3.5kilo - 15kilo = 11.5kilo she can carry in luggage

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Louis, Lucas, Garcelle, and Sheryl go bowling every week. when ordere for highest to lowest, how many ways can their scores be a

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3 years ago

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The entry fee to an Amusement Park is $125 and the cost of each ride is $ 7.25 ( Mon - Thurs) and $ 8.50 ( Fri - Sun). Ava's mom

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3 years ago

25/9 converted into a mixed number equals

blagie [28]

First figure out how many times 9 can go into 25 **do the highest amount of times it can go in-
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4 years ago

In △ABC, m∠A=39°, a=11, and b=13. Find c to the nearest tenth.

Talja [164]

For this problem, we are going to use the <em>law of sines</em>, which states:

$\dfrac{\sin{A}}{a} = \dfrac{\sin{B}}{b} = \dfrac{\sin{C}}{c}$

In this case, we have an angle and two sides, and we are trying to look for the third side. First, we have to find the angle which corresponds with the second side, $B$ . Then, we can find the third side. Using the law of sines, we can find:

$\dfrac{\sin{39^{\circ}}}{11} = \dfrac{\sin{B}}{13}$

We can use this to solve for $B$ :

$13 \cdot \dfrac{\sin{39^{\circ}}}{11} = \sin{B}$

$B = \sin^{-1}{\Big(13 \cdot \dfrac{\sin{39^{\circ}}}{11}\Big)} \approx 48.1$

Now, we can find $C$ :

$C = 180^{\circ} - 48.1^{\circ} - 39^{\circ} = 92.9^{\circ}$

Using this, we can find $c$ :

$\dfrac{\sin{39^{\circ}}}{11} = \dfrac{\sin{92.9^{\circ}}}{c}$

$c = \dfrac{11\sin{92.9^{\circ}}}{\sin{39^{\circ}}} \approx \boxed{17.5}$

c is approximately 17.5.

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3 years ago

What is the algebraic expression for: four less than three times the number of egg orders and six more than two times the number

Ira Lisetskai [31]

3n - 4 + 6 + 2n

I believe that's the answer!

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3 years ago