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Evgesh-ka [11]
2 years ago
15

Find the perimeter. Simplify your answer.

Mathematics
1 answer:
Alexxx [7]2 years ago
6 0

Answer:

wild guess, 50

Step-by-step explanation:

10s+3+10s+3+3s+9+3s+9= 50

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Divide. enter your solution as a mixed number.
Yakvenalex [24]

Answer: 30 6/29

Step-by-step explanation: 6132 \ 203 is 30 and 42/203. 42/203 simplified is 6/29

7 0
2 years ago
What was the temperature
Alenkasestr [34]
It was 65 degrees by lunch time.
6 0
3 years ago
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he data represents the daily rainfall​ (in inches) for one month. Construct a frequency distribution beginning with a lower clas
Licemer1 [7]

Answer:

It is not normally distributed as it has it main concentration in only one side.

Step-by-step explanation:

So, we are given that the class width is equal to 0.2. Thus we will have that the first class is 0.00 - 0.20, second class is 0.20 - 0.40 and so on(that is 0.2 difference).

So, let us begin the groupings into their different classes, shall we?

Data given:

0.31 0.31 0 0 0 0.19 0.19 0 0.150.15 0 0.01 0.01 0.19 0.19 0.53 0.53 0 0.

(1). 0.00 - 0.20: there are 15 values that falls into this category. That is 0 0 0 0.19 0.19 0 0.15 0.15 0 0.01 0.01 0.19 0.19 0 0.

(2). 0.20 - 0.40: there are 2 values that falls into this category. That is 0.31 0.31

(3). 0.4 - 0.6 : there are 2 values that falls into this category.

(4). 0.6 - 0.8: there 0 values that falls into this category. That is 0.53 0.53.

Class interval            frequency.

0.00 - 0.20.                   15.

0.20 - 0.40.                    2.

0.4 - 0.6.                        2.

4 0
3 years ago
Find the area. The figure is not drawn to scale.
ELEN [110]

area of a triangle is 1/2 x base x height

 base = 19

 height = 3

19 *3 = 57

1/2 x 57 = 28.5 square ft.

3 0
3 years ago
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Which of the following is a solution to 2cos2x − cos x − 1 = 0?
Pavel [41]

Answer:

Option A is correct.

Solution for the given equation is, x = 0^{\circ}

Step-by-step explanation:

Given that : 2\cos^2x -\cos x -1 =0

Let \cos x =y

then our equation become;

2y^2-y-1= 0           .....[1]

A quadratic equation is of the form:

ax^2+bx+c =0.....[2] where a, b and c are coefficient and the solution is given by;

x = \frac{-b\pm \sqrt{b^2-4ac}}{2a}

Comparing equation [1] and [2] we get;

a = 2 b = -1 and c =-1

then;

y = \frac{-(-1)\pm \sqrt{(-1)^2-4(2)(-1)}}{2(2)}

Simplify:

y = \frac{ 1 \pm \sqrt{1+8}}{4}

or

y = \frac{ 1 \pm \sqrt{9}}{4}

y = \frac{ 1 \pm 3}{4}

or

y = \frac{1+3}{4} and y = \frac{1 -3}{4}

Simplify:

y = 1 and y = -\frac{1}{2}

Substitute y = cos x we have;

\cos x = 1

⇒x = 0^{\circ}

and

\cos x = -\frac{1}{2}

⇒ x = 120^{\circ} \text{and} x = 240^{\circ}

The solution set:  \{0^{\circ}, 120^{\circ} , 240^{\circ}\}

Therefore, the solution for the given equation  2\cos^2x -\cos x -1 =0 is, 0^{\circ}





8 0
3 years ago
Read 2 more answers
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