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3 years ago

Find the perimeter of the right triangle.

Mathematics

2 answers:

Contact [7]3 years ago

3 0

Answer:

b. 36

Step-by-step explanation:

You have to use Pythagorean theorem. A squared + b squared = c squared.

a=12 12^2 (12 squared)= 144

b=9 9^2 (9 squared)= 81

144+81=225

Then you find the square root of 225 which is 15. Since you have the length for side c.

12+9+15=36

n200080 [17]3 years ago

3 0

Answer:

Step-by-step explanation

You would first have to find the hypotenuse to do that you would follow the formula c = √a²+b², giving you 15.

Next you would add all the sides P = a+b+c, giving you 36.

The perimeter of this triangle would be 36.

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Mrrafil [7]

Answer:

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Step-by-step explanation:

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3 years ago

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vazorg [7]

2x^2y√6x

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2 years ago

The number of letters in Stephanie's full name is sixteen less than twice the number of letters in Amy's full name. If the numbe

Anestetic [448]

Answer:

19 letters

Step-by-step explanation:

Suppose Number of Amy's full name letters = X

Number of Stephanie's full name letters = Y

Now, according to given condition that ''number of letters in Stephanie's full name is sixteen less than twice the number of letters in Amy's full name'',

Equation 1 becomes

Y = 2X-16

Now, according to second condition, product of their names' letter is 418. So,

Equation 2 becomes

XY = 418

Deducing value of X from equation 1,

X = (Y+16)/2

Putting this value in equation 2 we get,

{(Y+16)/2}*Y = 418,

By simplifying this equation we get a quadritic equation

(Y^2)+16Y-836=0

By breaking middle term (You can use quadratic formula here as well)

(Y^2)+38Y-22Y-836=0

Y(Y+38)-22(Y+38)=0

(Y-22)(Y+38)=0

At this stage we have two values for Y,

Y=22, or Y=-38

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Y=22,

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7 0

3 years ago

Required information Skip to question A die (six faces) has the number 1 painted on two of its faces, the number 2 painted on th

grigory [225]

Answer:

The change to the face 3 affects the value of P(Odd Number)

Step-by-step explanation:

Analysing the question one statement at a time.

Before the face with 3 is loaded to be twice likely to come up.

The sample space is:

$S = \{1,1,2,2,2,3\}$

And the probability of each is:

$P(1) = \frac{n(1)}{n(s)}$

$P(1) = \frac{2}{6}$

$P(1) = \frac{1}{3}$

$P(2) = \frac{n(2)}{n(s)}$

$P(2) = \frac{3}{6}$

$P(2) = \frac{1}{2}$

$P(3) = \frac{n(3)}{n(s)}$

$P(3) = \frac{1}{6}$

P(Odd Number) is then calculated as:

$P(Odd\ Number) = P(1) + P(3)$

$P(Odd\ Number) = \frac{1}{3} + \frac{1}{6}$

Take LCM

$P(Odd\ Number) = \frac{2+1}{6}$

$P(Odd\ Number) = \frac{3}{6}$

$P(Odd\ Number) = \frac{1}{2}$

After the face with 3 is loaded to be twice likely to come up.

The sample space becomes:

$S = \{1,1,2,2,2,3,3\}$

The probability of each is:

$P(1) = \frac{n(1)}{n(s)}$

$P(1) = \frac{2}{7}$

$P(2) = \frac{n(2)}{n(s)}$

$P(2) = \frac{3}{7}$

$P(3) = \frac{n(3)}{n(s)}$

$P(3) = \frac{1}{7}$

$P(Odd\ Number) = P(1) + P(3)$

$P(Odd\ Number) = \frac{2}{7} + \frac{1}{7}$

Take LCM

$P(Odd\ Number) = \frac{2+1}{7}$

$P(Odd\ Number) = \frac{3}{7}$

Comparing P(Odd Number) before and after

$P(Odd\ Number) = \frac{1}{2}$ --- Before

$P(Odd\ Number) = \frac{3}{7}$ --- After

<em>We can conclude that the change to the face 3 affects the value of P(Odd Number)</em>

7 0

3 years ago

Please can someone explain how to work out this

wariber [46]

Answer:

T = 4B + 10C

Step-by-step explanation:

Since there are 4 doughnuts in a bag then there are 4B doughnuts in B bags

Since there are 10 doughnuts in a carton then there are 10C doughnuts in C cartons, thus summing the two quantities gives

T = 4B + 10C

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3 years ago