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2 years ago

Find the missing side of each triangle

Mathematics

1 answer:

m_a_m_a [10]2 years ago

3 0

The missing angle is 13.6, since we got the sides we can find the hypotenuse of 8 and 11 and that is 13.6 ||

Making the answer: 13.6

Here is my math c=a2+b2=8.2+11.2≈13.60147

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simplest form since it cannot be simplified more

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If f(x)=-12x−3, find x for f(x)=89.

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How many two digit numbers are divisible by 3

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3 years ago

A rocket is launched from the ground at an initial velocity of 39.2 meters per second. Which equation can be used to model the h

Naya [18.7K]

The quadratic equation that models the height of the rocket after t seconds is:

s(t) = -4.9t² + 39.2t.

<h3>What is the quadratic function for a projectile's height?</h3>

Considering the height in meters, the equation is given by:

s(t) = -4.9t² + v(0)t + h(0)

In which:

v(0) is the initial velocity, in m/s.

h(0) is the initial height, in m.

In this problem, we have that v(0) = 39.2, h(0) = 0, hence the equation is:

s(t) = -4.9t² + 39.2t.

More can be learned about quadratic functions at brainly.com/question/24737967

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1 year ago

An urn contains 5 red balls and 2 green balls. Two balls are drawn one after the other. What is the probability that the second

AleksAgata [21]

Answer:

<em>The probability that the second ball is red is 71%</em>

Step-by-step explanation:

<u>Probabilities</u>

We know there are 5 red balls and 2 green balls. Let's analyze what can happen when two balls are drawn in sequence (no reposition).

The first ball can be red (R) or green (G). The probability that it's red is computed by

$\displaystyle P(R)=\frac{5}{7}$

The probability is's green is computed by

$\displaystyle P(G)=\frac{2}{7}$

If we have drawn a red ball, there are only 4 of them out of 6 in the urn, so the probability to draw a second red ball is

$\displaystyle P(RR)=\frac{5}{7}\cdot \frac{4}{6}=\frac{10}{21}$

If we have drawn a green ball, there are still 5 red balls out of 6 in the urn, so the probability to draw a red ball now is

$\displaystyle P(GR)=\frac{2}{7}\cdot \frac{5}{6}=\frac{5}{21}$

The total probability of the second ball being red is

$\displaystyle P(XR)=\frac{10}{21}+\frac{5}{21}=\frac{5}{7}=0.71$

The probability that the second ball is red is 71%

7 0

3 years ago