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3 years ago

Is 5 and 13 a right triangle

Mathematics

2 answers:

Margaret [11]3 years ago

6 0

Yes it is I am correct

natima [27]3 years ago

3 0

Answer:

Your question wasn't clear, but if the side lengths are 5, 13, AND 12, then it is a yes.

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Can anyone please explain? Need some help :)

DedPeter [7]

Answer:

93.5 square units

Step-by-step explanation:

Diameter of the Circle = 12 Units

Therefore:

Radius of the Circle = 12/2 =6 Units

Since the hexagon is regular, it consists of 6 equilateral triangles of side length 6 units.

Area of the Hexagon = 6 X Area of one equilateral triangle

Area of an equilateral triangle of side length s = $\dfrac{\sqrt{3} }{4}s^2$

Side Length, s=6 Units

$\text{Therefore, the area of one equilateral triangle =}\dfrac{\sqrt{3} }{4}\times 6^2\\\\=9\sqrt{3} $ square units$

Area of the Hexagon

$= 6 X 9\sqrt{3} \\=93.5$ square units (to the nearest tenth)$

7 0

3 years ago

Which operation should be performed first to solve the inequality below? -3x + 5 = 23

Ne4ueva [31]

Answer:

Pass the number 5 to the other side with te opposite sign in order to leave the x alone

it would be

-3x = 23-5

Then you divide by -3 and thats ur answer

4 0

3 years ago

Read 2 more answers

Wat is the answer for this question -x2 + 5x + 24.

Helen [10]

Answer:

Step-by-step explanation:

5 0

3 years ago

A. 3 B. -3/4 C. -1 D. 2

yawa3891 [41]

B
because rise over run

6 0

3 years ago

Read 2 more answers

Given the equation (x+a)^2(x-2)=x^3+bx^2+12x-72 find a and b

GaryK [48]

Isolate the variable by dividing each side by factors that don't contain the variable.

$a=- \frac{ x^{2}- \sqrt{( x^{3} + x^{2} b+12x-72(x-2)-2x} }{x-2} ,- \frac{ x^{2}+ \sqrt{( x^{3} + x^{2} b+12x-72(x-2)-2x} }{x-2}$

Solve for b by simplifying both sides of the equation then isolating the variabel.

$b= \frac{12}{x}+ \frac{72}{ x^{2} }-2+2a- \frac{4a}{x}+ \frac{ a^{2} }{x}- \frac{2a^{z} }{ x^{2} }$

Hopefully i helped ^.^ Mark brainly if possible

7 0

3 years ago