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2 years ago

Two sides of a triangle measure 5 in. and 12 in. Which could be the length of the third side?

Mathematics

1 answer:

Airida [17]2 years ago

6 0

Answer: 10.9

Step-by-step explanation:

You would use Pythagorean theorem:

5x5+12x12=c

5x5=25

12x12=144

Now, 25+144=169

Now find the square root of 169 which is 13 which isn't an answer choice, which means your prolly after a leg.

I assume 12 is the hypotenuse

A(squared)+5x5=12x12

A+25=144

144-25=119

Square root of 119=10.9

So I'm going to go with answer 10 in or 11 if you round.

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Circle O has a circumference of 36tt cm. what is length of the radius r

allochka39001 [22]

Answer:

5.73cm

Step-by-step explanation:

Given parameters:

Circumference of the circle = 36cm

Unknown

Length of the radius = ?

Lets represent the radius by r

Solution

The circumference of a circle is the defined as the perimeter of a circle. The formula is given as:

Circumference of a circle = 2πr

Since the unknown is r, we make it the subject of the formula:

r = $\frac{circumference of the circle}{2π}$

r = $\frac{36}{2 x 3.142}$ = $\frac{36}{6.284}$ = 5.73cm

3 0

3 years ago

Read 2 more answers

What is the volume of a cone (in cubic inches) of radius 5 inches and height 3

beks73 [17]

Answer:

78.54 cm^3

Step-by-step explanation:

I think thats the answer

6 0

2 years ago

Plz..... Solve this questions for me...

faust18 [17]

$6.\\\text{We know}\ 9=3^2.\ \text{Therefore}\\\\9^2=(3^2)^2\\\\\text{use}\ (a^n)^m=a^{nm}\\\\9^2=(3^2)^2=3^{2\cdot2}=3^4\\\\\boxed{3^4=9^2}\\=================$

$7.\\METHOD\ 1:\\\\\sqrt{7^4}=\sqrt{7^{2\cdot2}}\\\\\text{use}\ (a^n)^m=a^{nm}\\\\=\sqrt{(7^2)^2}\\\\\text{use}\ \sqrt{a^2}=a\ \text{for}\ a\geq0\\\\=\boxed{7^2}\\\\METHOD\ 2:\\\\\text{use}\ a^\frac{m}{n}=\sqrt[n]{a^m}\\\\\sqrt{7^4}=7^\frac{4}{2}=\boxed{7^2}\\=================$

$8.\\5^a\times5^b=5^{11}\\\\a)\\\text{use}\ a^n\times a^m=a^{n+m}\\\\5^a\times5^b=5^{a+b}\\\\5^{a+b}=5^{11}\Rightarrow a+b=11\\\\11\ \text{is odd. The sum of even and odd is odd. Therefore}\ a\ \text{and}\ b\ \text{can't be even.}\\--------------------\\\\b)\\5\ \text{solutions}\\\\11=5+6\\11=4+7\\11=3+8\\11=2+9\\11=1+10$