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2 years ago

Suppose a triangle has two sides of length 42 and 35, and that the angle

Mathematics

2 answers:

lys-0071 [83]2 years ago

6 0

yeah he/she is right ..................

maxonik [38]2 years ago

5 0

9514 1404 393

Answer:

c = √(42² +35² -2·42·35·cos(120°))

Step-by-step explanation:

The third side can be found using the Law of Cosines. It will tell you ...

c² = a² +b² -2ab·cos(C)

where C is the angle opposite the third side (c). For the given side lengths, then the third side can be found from ...

c = √(42² +35² -2·42·35·cos(120°))

<em>Additional comment</em>

You have not provided answer choices to this "which ..." problem. So the above equation may be different from any on your list. We expect your answer choices may leave the equation in the form <em>c² = ...</em>.

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Can you Find the solution of the system of equations?

Lunna [17]

Answer:

x = 3

y = 2

Step-by-step explanation:

multiply by -1 the second equation and add the two equations

$7x-7y=7\\4x+7y=26$

$11x=33$

$x=33/11$

$x=3$

"y" is calculated:

$7(3)-7y=7$

$21-7y=7$

$-7y=-14$

$y=\frac{-14}{-7} =2$

Hope this helps

5 0

2 years ago

For each angle θ listed below, find the reference angle α, and then find sin θ. Round sin θ to four decimal places, if necessary

babunello [35]

Answer with explanation:

It is given that

$\rightarrow\Theta =132^{\circ}$

The Angle lies in Second Quadrant.

Formula to find reference Angle

→If Angle lies in first Quadrant

then Reference Angle = The Angle given

→If Angle lies in Second Quadrant

Then Reference Angle = 180° - Angle

→If Angle lies in third Quadrant

then Reference Angle = Angle - 180°

→If Angle lies in fourth Quadrant

Then Reference Angle = 360° -Angle

Reference Angle

$=180^{\circ}-132^{\circ}\\\\=48^{\circ}$

⇒sin 48°=0.7431448

Rounding to four decimal places

Sin 48°=0.7432

4 0

3 years ago

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MariettaO [177]

Your answer is the third one because your word problem states $25 or more, which would be more than or equal to.

8 0

2 years ago

Read 2 more answers

Food allergies affect an estimated 7% of children under age 5 in the US. What is the probability that in a kindergarden class of

bezimeni [28]

Answer:

Probability of atleast one of 12 student has food allergies ≈ 0.58 ( approx)

Step-by-step explanation:

Given: Probability of a children under age 5 has food allergies = 7%

= $\frac{7}{100}$

To find : Probability of atleast one of 12 student has food allergies

Probability of a chindren under age 5 does not have food allergies = $1-\frac{7}{100}$

⇒ prob = $\frac{93}{100}$

now we find Probability of atleast one of 12 student has food allergies this means we have to find prob of 1 student, 2 student, 3 student, till 12 student have allergy out of 12 student of class then add all prob.

But instead of finding all these probability we find probability of student having no allergy.i.e., 0 student then subtract it from 1(total probability)

Probability of 0 student having allergy out of 12 student = $(\frac{93}{100})^{12}$