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Sloan [31]
2 years ago
10

If PQRS is a rhombus,which statement must be true? Check all that apply.

Mathematics
1 answer:
Oxana [17]2 years ago
8 0

Answer: A, B, C, D

Step-by-step explanation:

A. This is true because all rhombi are parallelograms, and diagonals of a parallelogram bisect each other.

B. This is true because the diagonals of a rhombus are perpendicular.

C. This is true because diagonals of a rhombus bisect the angles from which they are drawn,

D. This is true because all sides of a rhombus are congfruent.

E. This is not always true - all rhombi are parallelograms, and adjacent angles of a parallelogram are supplementary, but not always congruent.

F. This is not always true - diagonals of a rhombus are not always congruent.

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The length will be letter C, 50 m

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Oblicz.Pamiętaj o kolejności wykonywania działań
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Answer:

33.275

Step-by-step explanation:

<u>1 step:</u> Difference in brackets

3.5-2\dfrac{1}{3}=3\dfrac{1}{2}-2\dfrac{1}{3}=(3-2)+\left(\dfrac{1}{2}-\dfrac{1}{3}\right)=1+\dfrac{3-2}{3\cdot 2}=1+\dfrac{1}{6}=1\dfrac{1}{6}

<u>2 step:</u> 0.75\div\dfrac{1}{3}

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1.28\div 0.04=128\div 4=32 \ \text{Move point two decimal places}

<u>4 step:</u>

2\dfrac{1}{4}\times 1\dfrac{1}{6}=\dfrac{2\cdot 4+1}{4}\times \dfrac{1\cdot 6+1}{6}=\dfrac{9}{4}\times \dfrac{7}{6}=\dfrac{63}{24}=\dfrac{21}{8}

<u>5 step:</u>

\dfrac{9}{4}+32-\dfrac{21}{8}+1.65=(32+1.65)+\left(\dfrac{9}{4}-\dfrac{21}{8}\right)=33.65+\dfrac{9\cdot 2-21}{8}=33.65-\dfrac{3}{8}=33.65-0.375=33.275

6 0
3 years ago
Find the equation of a circle with a center at (7,2) and a point on the circle at (2,5)?
monitta

Answer:

(x-7)^2+(y-2)^2=34

Step-by-step explanation:

We want to find the equation of a circle with a center at (7, 2) and a point on the circle at (2, 5).

First, recall that the equation of a circle is given by:

(x-h)^2+(y-k)^2=r^2

Where (<em>h, k</em>) is the center and <em>r</em> is the radius.

Since our center is at (7, 2), <em>h</em> = 7 and <em>k</em> = 2. Substitute:

(x-7)^2+(y-2)^2=r^2

Next, the since a point on the circle is (2, 5), <em>y</em> = 5 when <em>x</em> = 2. Substitute:

(2-7)^2+(5-2)^2=r^2

Solve for <em>r: </em>

<em />(-5)^2+(3)^2=r^2<em />

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Therefore, our equation is:

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No student has passed more than one exam.

According to question, exactly three students from a randomly chose group of four students have not passed Exam P/1 or Exam FM/2.

So, Probability will be

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Hence, there is probability of 0.57 chances that exactly three students from a group of four students have not passed Exam P/1 or Exam FM/2.

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