All categories

3 years ago

A triangle has sides 12cm, 9cm and 15cm. Is the triangle right-angled? Justify your answer

Mathematics

1 answer:

marusya05 [52]3 years ago

3 0

Answer:

yes

Step-by-step explanation:

Yes this triangle is right angled and can be justified using the Pythagorean theorem. The theorem states that: $a^{2} +b^{2} = c^{2}$ . Substitute using the given values:

(9^2)+(12^2)=15^2

81 + 144 = 255

255 = 255.

It satisfies the theorem therefore it is a right angle triangle.

You might be interested in

When all 240 6th grade students were polled, results showed that 96 students were dissatisfied with the music played at the scho

avanturin [10]

Answer:

40 percent of students were dissatisfied

3 0

3 years ago

A ____________________ is a measurement of the change in variables over a specific period of time.

Montano1993 [528]

Answer:

rate of change

Step-by-step explanation:

The rate of change (ROC) is the speed at which a variable changes over a specific period of time.

5 0

3 years ago

Explain how you can write a quadratic function in a factored form that would have a vertex with an x-coordinate of 3 and two dis

Elena-2011 [213]

Answer:

Step-by-step explanation:

(x-3)²-4=0

(x-3)²=4=2²

x-3=±2

x=3±2

x=3-2,3+2

x=1,5

it has two distinct roots and a vertex at x=3

3 0

3 years ago

Are rational numbers equal to integers

Alinara [238K]

Yes rational numbers are equal to integers.

4 0

3 years ago

Read 2 more answers

Find the domain of the function y = 3 tan(23x)

solmaris [256]

Answer:

$\mathbb{R} \backslash \displaystyle \left\lbrace \left. \frac{1}{23}\, \left(k\, \pi + \frac{\pi}{2}\right) \; \right| k \in \mathbb{Z} \right\rbrace$ .

In other words, the $x$ in $f(x) = 3\, \tan(23\, x)$ could be any real number as long as $x \ne \displaystyle \frac{1}{23}\, \left(k\, \pi + \frac{\pi}{2}\right)$ for all integer $k$ (including negative integers.)

Step-by-step explanation:

The tangent function $y = \tan(x)$ has a real value for real inputs $x$ as long as the input $x \ne \displaystyle k\, \pi + \frac{\pi}{2}$ for all integer $k$ .

Hence, the domain of the original tangent function is $\mathbb{R} \backslash \displaystyle \left\lbrace \left. \left(k\, \pi + \frac{\pi}{2}\right) \; \right| k \in \mathbb{Z} \right\rbrace$ .

On the other hand, in the function $f(x) = 3\, \tan(23\, x)$ , the input to the tangent function is replaced with $(23\, x)$ .

The transformed tangent function $\tan(23\, x)$ would have a real value as long as its input $(23\, x)$ ensures that $23\, x\ne \displaystyle k\, \pi + \frac{\pi}{2}$ for all integer $k$ .

In other words, $\tan(23\, x)$ would have a real value as long as $x\ne \displaystyle \frac{1}{23} \, \left(k\, \pi + \frac{\pi}{2}\right)$ .

Accordingly, the domain of $f(x) = 3\, \tan(23\, x)$ would be $\mathbb{R} \backslash \displaystyle \left\lbrace \left. \frac{1}{23}\, \left(k\, \pi + \frac{\pi}{2}\right) \; \right| k \in \mathbb{Z} \right\rbrace$ .

4 0

3 years ago