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choli [55]
2 years ago
13

answer to #10 and #13. can’t figure out how to use interval notation to write domain and range for graphs.

Mathematics
1 answer:
Lyrx [107]2 years ago
8 0
Interval notation is used to write a set of real numbers from one value to another value.
On the left, you start with left parenthesis or left bracket.
Then you follow by two numbers separated by a comma.
You then finish with a right parenthesis or right bracket.
To include a number, use a square bracket.
To exclude a number use parenthesis.
To write the set of numbers, you need to list the smallest number in the set followed by the largest number in the set. An interval is always stated with two numbers, from the smallest in the set to the largest in the set. The numbers are always separated by a comma.

Examples:

1) All numbers from 6 to 10, including 6 and 10.
Algebra: 6 <= x <= 10
Interval: [6, 10]
Notice brackets since both 6 and 10 are included in this interval.

2) All number from 5 to 20, including 5 but not including 20.
Algebra 5 <= x < 20
Interval: [5, 20)
Bracket with 5 means include 5. Parenthesis with 20 means 20 is not included.

3) All numbers greater than or equal to 7.
Algebra: x >= 7
Interval: [7, ∞)
The 7 has a bracket because it is included. Infinity always has parenthesis.
With the infinity symbol, always use parenthesis, not square bracket.

4) All numbers less than -5.
Algebra: x < - 5
Interval: (-∞, 5)

Now for your problems.

10.
This is a line. Both the domain and range all all real numbers.
That means the interval is from negative infinity to positive infinity.
(-∞, ∞)
Both the domain and range are that same interval, all real numbers, from negative infinity to positive infinity.

13.
The domain is all real numbers as you can see the x-coordinates extend left forever and right forever. The domain is the same interval as the domain and range of problem 10.

The range is zero and all positive numbers.
You can think of it a all values of y such that y is greater than or equal to zero. Notice that zero is included in the interval.

[0, ∞)

Since zero is included, we use a left bracket, not left parenthesis.
With infinity, we alyways use parentheses, not brackets.
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Airida [17]

Answer:

89.1° or -1.4°  

Step-by-step explanation:

1. Location:

You are on the Mont-Saint-Jean escarpment, near the Belgian town of Waterloo.

The French troops are about 50 m below you and 1.2 km distant.

2. Finding the firing angle

Data:

R = 1200 m

u = 600 m/s

h = -50 m (the height of the target)

a = 9.8 m/s²

We have two conditions.

Horizontal distance

(1) 1200 = 600t cosθ

Vertical distance

(2) -50 = 600t sinθ - 4.9t²

Divide each side of (1) by 600cosθ.

(3) \, t =\dfrac{2}{\cos \theta}

Substitute (3) into (2)

-50 = 600t \sin \theta - 4.9t^{2} =  600 \left( \dfrac{2}{\cos \theta} \right ) \sin \theta - 4.9 \left( \dfrac{2}{\cos \theta} \right )^{2}\\\\(4) \, -50 = 1200 \tan \theta - \dfrac{19.6}{\cos^{2} \theta}

Recall that

(5) sec²θ = 1/cos²θ = tan²θ + 1

Substitute (5) into (4)

-50 = 1200 \tan \theta - 19.6 \left(\tan^{2} \theta}+ 1\right )

Set up a quadratic equation

\begin{array}{rcl}-50 & = & 1200 \tan \theta - 19.6\tan^{2} \theta -19.6 \\0 & = & 1200 \tan \theta - 19.6\tan^{2} \theta + 30.4\\0 & =&19.6\tan^{2} \theta - 1200 \tan \theta - 30.4\\0 & =&\tan^{2} \theta - 61.224 \tan \theta - 1.551\\\end{array}

Solve for θ

Use the quadratic formula.

tanθ = 61.249 or -0.025

θ = arctan(61.249) = 89.1° or

θ = arctan(-0.025) = -1.4°

3 0
3 years ago
[-1/4/5] × [-7/10+0.95] - [-2/1/4]
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Answer:

-2.5625

Step-by-step explanation:

<u>1)[-1/4/5] × [-7/10+0.95] - [-2/1/4] turn that into improper fractions</u>

2 1/4=9/4

which is 1 4/5(-7/10+0.95)9/4

<u>2) 1 4/5(-7/10+0.95)= 0.3125</u>

which is 0.3125-9/4

<u>3)Make into decimal form</u>

9/4 equals 2.25

which is 0.3125-2.25

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3 years ago
What is the leading coefficient of the following expression: 5x^4-7x^3+2x+1​
never [62]

Answer:

5

Step-by-step explanation:

The leading coefficient belongs to the term with the largest exponent

Given

5x^{4} - 7x³ + 2x + 1 ← in standard form

The leading term is 5x^{4} ← with leading coefficient 5

5 0
3 years ago
Chloe is painting on a stretched canvas that has an area of 108 square inches. The length of the canvas is 12 inches. The width
Stolb23 [73]

Answer: Width of canvas = 9 inches

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Explanation :

Since we have given that

Area of painting = 108 square inches

Length of canvas = 12 inches

As we know that

\text{Area of rectangle} = length\times breadth

108=12\times breadth\\\\\frac{108}{12}=breadth\\\\9=breadth

So, breadth of canvas = 9 inches .

Now, Chloe paints a 1 inch wide blue border on the canvas.

\text{So, remaining length left for painting after painting the border} = 9-1-1=9-2=7\text{ inches}

\text{ Remaining breadth left for painting after painting the border }= 12-1-1=12-2= 10\text{ inches}

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Hence, remaining area = 70 square inches .

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2 years ago
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