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3 years ago

Can somebody please help me on c and d please

Mathematics

1 answer:

Shalnov [3]3 years ago

5 0

Answer:

i am not really, sure but i believe it could be this

Step-by-step explanation:

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a triangle with a base of 9 meters has an altitude of 12 meters.which would give the area of the triangle

user100 [1]

Answer:

54 $m^2$

Step-by-step explanation:

To find the area of a triangle use the formula a = 1/2bh

The altitude is the same as height so we can substitute 12 for h and 9 for b

Now the equation will become

a = 1/2 · 9 · 12

When you multiply it, it become 108/2 = a simplified to a = 54

7 0

4 years ago

Find the value of x of the triangle

Hunter-Best [27]

Answer:

Step-by-step explanation:

Subtract the sum of the two angles from 180 degrees. The sum of all the angles of a triangle always equals 180 degrees. Write down the difference you found when subtracting the sum of the two angles from 180 degrees. This is the value of X.

4 0

3 years ago

If a square has an area of 81 square units how long is each side?

Sloan [31]

$A=81 \\ \\ A = s^2 \\ \\ s^2 = 81 \\ \\s=\sqrt{81} \\ \\s= 9 \\ \\ Ansewr : \ each \ side \ is \ 9 \ units \ on \ each \ side$

7 0

3 years ago

Which side of the mountain would protect you from the wind best?

Stolb23 [73]

Answer:

well on the ocean you will feel the wind so no

Step-by-step explanation:

6 0

3 years ago

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If cos(θ)= -2/3 and θ lies in quadrant II, find the value of cotθ

Lera25 [3.4K]

$\text{Given that,}\\\\\cos \theta = - \dfrac 23\\\\\implies \cos^2 \theta = \dfrac 49\\\\\implies 1- \sin^2 \theta = \dfrac 49\\\\\implies \sin^2 \theta =1-\dfrac 49\\\\\implies \sin^2 \theta =\dfrac 59\\\\\\\text{Now,}\\\\\cot^2 \theta = \dfrac{\cos^2 \theta}{\sin^2 \theta} \\\\\\\implies \cot^2 \theta = \dfrac{\tfrac 49}{\tfrac 59}\\\\\\\implies \cot^2 \theta = \dfrac 49 \times \dfrac 95\\\\\\\implies \cot^2 \theta = \dfrac 45\\\\\\$

$\implies \cot \theta = \pm\sqrt{\dfrac 45} = \pm \dfrac 2{\sqrt 5}\\\\\\\text{Since}~ \theta ~ \text{lies in the second quadrant,} \cot \theta ~\text{will be negative.}\\\\\text{Hence,}~ \cot \theta = - \dfrac{2}{\sqrt 5}$

6 0

2 years ago

Read 2 more answers