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2 years ago

Pplssssssssss help i will vote brainliest!!!!

Mathematics

1 answer:

Fed [463]2 years ago

7 0

Answer:

x = 38°

Step-by-step explanation:

Let's find the internal angles of the triangle, at the base.

We see that one external angle is 134° and we need to find its supplementary angle (a+b = 180°) so:

134° + α = 180°

α = 46°

Now the other base angle:

β + 84° = 180°

β = 96°

We also know that the sum of all angles in a triangle is always 180°:

α + β + x = 180°

x = 180° - α - β = 38°

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shtirl [24]

Solution:

Step-1: Find the data.

70's: 1
80's: 1
90's: 2
100's: 3
110's: 4
120's: 4
130's: 2
140's: 1

Data obtained (Smallest to biggest):

70, 80, 90, 90, 100, 100, 100, 110, 110, 110, 110, 120, 120, 120, 120, 130, 130, 140

Step-2: Count the number of digits.

18 digits in data

Step-3: Use the formula "Total digits/2".

18/2 = 9th digit

The 9th digit in the data is the median.

Step-4: Revise the data.

70, 80, 90, 90, 100, 100, 100, 110, 110, 110, 110, 120, 120, 120, 120, 130, 130, 140

The median value of the line plot is 110.

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2 years ago

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Find sin3A, cos 3A and tan 3A when sin=1/2

Slav-nsk [51]

Answer:

Sin 3A = 1

cos 3A = 0

tan 3 A = undefined

Step-by-step explanation:

$\sin A = \frac{1}{2} \\ \therefore \: \sin A = \sin \: 30 \degree.. \bigg( \because \: \sin \: 30 \degree = \frac{1}{2} \bigg) \\ \therefore \: A = 30 \degree \\ \\ now \\ \sin 3A =\sin (3 \times 30 \degree)= \sin \: 90 \degree = 1 \\ \\ \cos 3A =\cos (3 \times 30 \degree)= \cos \: 90 \degree = 0\\\\ \tan 3A =\tan (3 \times 30 \degree)= \tan \: 90 \degree = \infty$

4 0

4 years ago

A school bus is moving 26.8 m/s on flat ground when it begins to accelerate at 4.73 m/s^2. How much time does it take to travel

leva [86]

The school bus will take 5.34 seconds to travel 95.1 m if the bus is moving 26.8 m/s on flat ground when it begins to accelerate at 4.73 m/s².

<h3>What is a quadratic equation?</h3>

Any equation of the form $\rm ax^2+bx+c=0$ where x is variable and a, b, and c are any real numbers where a ≠ 0 is called a quadratic equation.

As we know, the formula for the roots of the quadratic equation is given by:

$\rm x = \dfrac{-b \pm\sqrt{b^2-4ac}}{2a}$

We have:

A school bus is moving 26.8 m/s on flat ground when it begins to accelerate at 4.73 m/s².

From the second equation of motion:

$\rm s = u + \dfrac{1}{2}at^2$

We have given:

s = 95.1 m

u = 26.8 m/s

a = 4.73 m/s²

$\rm 95.1 = 26.8 + \dfrac{1}{2}(4.73)t^2$

The above equation is a quadratic equation:

After simplification:

$\rm \dfrac{4.73}{2}t^2=68.3$

$\rm t^2=28.879$

t = 5.373 seconds ≈ 5.34 seconds

Thus, the school bus will take 5.34 seconds to travel 95.1 m if the bus is moving 26.8 m/s on flat ground when it begins to accelerate at 4.73 m/s².

Learn more about quadratic equations here:

brainly.com/question/2263981

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