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strojnjashka [21]

3 years ago

6

Find the difference. 6rs – (-2rs)

1 answer:

lilavasa [31]3 years ago

8 0

There is no difference, 6rs-(-2rs) if you get rid of the parenthesis you are left with 6rs- -2rs and two negatives make a positive therefore the answer will become 8rs... I hope this helps :)

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What is the equation of the line that is the perpendicular bisector of the segment

uysha [10]

The answer is y=x-4

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8 0

2 years ago

Is anyone willing to do my ixl? I have loads of work to do and I need my ixl due tonight. It’s 8th grade math and I’d appreciate

ch4aika [34]

I failed 8th grade math :/

7 0

2 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.

8 0

2 years ago

Complete the proof by providing the missing statement and reasons

Fittoniya [83]

Answer:

2. angle bisector splits into even halves

Step-by-step explanation:

that's all i can assist with sorry, proofs still confuse me!

3 0

2 years ago

Suppose that you are holding your toy submarine under the water. You release it and it begins to ascend. The graph models the de

Likurg_2 [28]

Answer:

Domain : 0 ≤ t ≤ 3

Range : -4 ≤ d ≤ 0

Step-by-step explanation:

The graph attached models the depth of submarine as a function of time.

Points on x-axis represent the time and points on y-axis represent increase in height of the submarine.

Domain of a function is represented by the points on x-axis.

Therefore, Domain : 0 ≤ t ≤ 3

Range of function is represented by te points on y-axis.

Therefore, Range : -4 ≤ d ≤ 0

3 0

2 years ago