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3 years ago

Can you construct a triangle that has side lengths 5 cm, 5cm, and 10 cm

Mathematics

2 answers:

Reil [10]3 years ago

6 0

Answer:

Step-by-step explanation:

Ⓗⓘ ⓣⓗⓔⓡⓔ

No, because two side added up have to be greater than the other one. 5+5 is not greater than 10, it is equal.

Hence, it doesn't work.

(っ◔◡◔)っ ♥ Hope this helped! Have a great day! :) ♥

Please, please give brainliest, it would be greatly appreciated, I only need one more before I advance, thanks!

Ugo [173]3 years ago

6 0

Answer:

Step-by-step explanation:

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yawa3891 [41]

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3 years ago

|3x - 9| = 12 <br> Show all steps please and thank you!

bezimeni [28]

Answer:

ayy I gotchuu

Step-by-step explanation:

theres gonna be two solutions

a negative then positive

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|3x - 9| = 12

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2 years ago

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How do I find the radius, diameter and circumference

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2 years ago

What is the standard deviation of the data set given below? 4, 7, 8, 9, 12

astraxan [27]

Answer:

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Step-by-step explanation:

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3 years ago

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The angle \theta_1θ 1 theta, start subscript, 1, end subscript is located in Quadrant \text{II}IIstart text, I, I, end text, a

andrew11 [14]

Answer:

$\cos(\theta_1)=-\dfrac{\sqrt{15}}{4}$

Step-by-step explanation:

The angle $\theta_1$ is located in Quadrant II.

$\sin(\theta_1)=\dfrac{1}{4}$

From trigonometry, we know that:

$\sin(\theta)=\dfrac{Opposite}{Hypotenuse}\\$Therefore:\\Opposite=1\\Hypotenuse=4\\Using Pythagorean theorem:\\Hypotenuse^2=Opposite^2+Adjacent^2\\4^2=1^2+Adjacent^2\\Adjacent^2=16-1\\Adjacent^2=15\\Adjacent=\sqrt{15}$

Now, in Quadrant II,

The x-axis is negative
The y-axis is positive

Therefore, the Adjacent angle to $\theta_1 =-\sqrt{15}$

Therefore:

$\cos(\theta_1)=\dfrac{Adjacent}{Hypotenuse}=\dfrac{-\sqrt{15}}{4}\\\\\cos(\theta_1)=-\dfrac{\sqrt{15}}{4}$

8 0

3 years ago