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3 years ago

Pick the set of numbers which contain ALL perfect squares:

Mathematics

2 answers:

Bumek [7]3 years ago

8 0

The answer is D I believe.

alexandr1967 [171]3 years ago

6 0

The answer is D. {81,16,25,49,1}

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the area of a rectangular volleyball court is 1800 square feet the courts length is twice it width write a system of equations t

Marianna [84]

Answer:

The dimensions of the rectangular volleyball court are 60 ft x 30 ft

Step-by-step explanation:

Let

x ----> the length of rectangular volleyball court

y ---> the width of the rectangular volleyball court

we know that

The area of the rectangular volleyball court is equal to

$A=xy$

$A=1,800\ ft^2$

$1,800=xy$ ----> equation A

$x=2y$ -----> equation B

substitute equation B in equation A

$1,800=(2y)y$

$1,800=2y^2$

Solve for y

Simplify

$900=y^2$

take square root both sides

$y=30\ ft$

<em>Find the value of x</em>

$x=2y$

substitute the value of y

$x=2(30)=60\ ft$

therefore

The dimensions of the rectangular volleyball court are 60 ft x 30 ft

7 0

3 years ago

What is 214729.08 rounded to nearest thousand

nata0808 [166]

Hello!!

214,729.08 rounded to nearest thousand is 215,000.

And, 214,729.08 rounded to nearest thousandths (3 decimals) is 214,729.080.

Good luck :)

4 0

4 years ago

Given:• PQRS is a rectangle.• mZ1 = 50°Р21SRWhat is mZ2?130°85°70°65°

Over [174]

To answer this question, we need to recall that: "the diagonals of a rectangle bisect each other"

Thus, if we assign the point of intersection of the two diagonals in the rectangle as point O, we can say that the triangle OQR is an "isosceles triangle". Note that this is because the lengths OR and OQ are equal since we know that: "the diagonals of a rectangle bisect each other". See the below diagram for clarity.

Now, we have to recall that:

- the base angles of any isosceles triangle are equal. This is a fact, and this means that the angles

- also the sum of all the angles in any triangle is 180 degrees

Now, considering the isosceles triangle OQR, we have that:

$\angle OQR+\angle ORQ+\angle ROQ=180^o$

Now, since the figure already shows that angle $m\angle2+\angle ORQ+50^o=180^o$ Now, since we have established that the base angles $m\angle2+m\angle2+50^o=180^o$ we can now solve the above equation for m<2 as follows:

$\begin{gathered} m\angle2+m\angle2+50^o=180^o \\ \Rightarrow2m\angle2+50^o=180^o \\ \Rightarrow2m\angle2=180^o-50^o \\ \Rightarrow2m\angle2=130^o \\ \Rightarrow m\angle2=\frac{130^o}{2}=65^o \end{gathered}$

Therefore, the correct answer is: option D