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djverab [1.8K]
3 years ago
14

2

Mathematics
2 answers:
Nadya [2.5K]3 years ago
8 0

Answer:

64, 400, 289 , 0 are perfect square.

follow me!

tankabanditka [31]3 years ago
5 0

Answer:

64, 289, 400, 0

Step-by-step explanation:

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What is the product: (х + 5)(х + 12) =
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The product is x^{2}+17x+ 60

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What are 2 examples of what weighs less than 1 ounce
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A feather and a leaf both weigh less than 1 ounce.
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How many radians are in one degree, accurate to three decimal places?
qaws [65]
Now, if you recall, there are 180° in π radians, now, so how many radians in 1°?

\bf \begin{array}{ccll}
de grees&radians\\
\text{\textemdash\textemdash\textemdash}&\text{\textemdash\textemdash\textemdash}\\
180&\pi \\
1&r
\end{array}\implies \cfrac{180}{1}=\cfrac{\pi }{r}\implies r=\cfrac{1\cdot \pi }{180}
3 0
4 years ago
In△XYZ , XY¯¯¯¯¯¯=7in , YZ¯¯¯¯¯=5in , and XZ¯¯¯¯¯=4in . This triangle is reflected across the x-axis to result in △X'Y'Z' . Whic
ki77a [65]

Answer:

B

Step-by-step explanation:

Reflection is a transformation that preserves lengths.

Triangle XYZ, with side lengths XY = 7 in, YZ = 5 in, XZ = 4 in, is reflected across the x-axis to result in △X'Y'Z'.

This means that corresponding sides have the same lengths:

  • XY = X'Y' = 7 in;
  • XZ = X'Z' = 4 in;
  • YZ = Y'Z' = 5 in.

Find the perimeters of both triangles:

P_{XYZ}=XY+XZ+YZ=7+4+5=16\ in\\ \\P_{X'Y'Z'}=X'Y'+X'Z'+Y'Z'=7+4+5=16\ in

Hence, only option B is true.

8 0
3 years ago
You enter a chess tournament where your probability of winning a game is 0.3 against half the players (call them type 1), 0.4 ag
Doss [256]

The probability of winning is 0.44 .

<u>Step-by-step explanation:</u>

Here it's given that , You enter a chess tournament where your probability of winning a game is 0.3 against half the players (call them type 1), 0.4 against a quarter of the players (call them type 2), and 0.5 against the remaining quarter of the players (call them type 3). You play a game against a randomly chosen opponent. More precisely :

Probability(type1) = 0.3\\Probability(type2) = 0.4\\Probability(type3) = 0.5\\\\Probability(winning) = w

Now, we choose a random opponent and we need to find probability of winning which is possible as :

<u>1. player chosen from type1</u>

<u>2. player chosen from type2</u>

<u>3. player chosen from type3</u>

Combining all cases we get :

⇒ w = type1(1-type2)(1-type3) + (1-type1)type2(1-type3) + (1-type1)(1-type2)type3

⇒ w = 0.3(0.6)(0.5)+0.7(0.4)(0.5)+0.7(0.5)(0.6)

⇒ w = 0.44

∴ The probability of winning is 0.44 .

8 0
3 years ago
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