A town had about 2120 acres of pine trees 40 years ago. Only about 13% of the pine trees remain. How many acres of pine trees remain? Round your answer to the nearest tenth.
Answers
There will be 275.6 acres of pine trees are left out of 2120 acres
Data;
- Number of acres = 2120
- percentage left = 13%
Percentage
To calculate the percentage of a value, we find the value of that value to it's total value multiplied by 100.
To find 13% of 2120,
From the calculations above, only 275.6 acres of pine trees are left out of 2120 acres.
Learn more on percentage here;
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a random sample of 130 students is chosen from a population of 4,500 students. if the mean iq in the sample is 120 with a standard deviation of 5, what is the 99% confidence interval for the students' mean iq score? answers given below: 118.87−121.13 107.12−132.88 125−135 115−125
Help me with these questions please
"I am 3 times as old as Sue is," Frank said to Ann. "On the other hand, I am 15 years older than John while Sue is 1 year younger than John. How old am I?"If J represents John's age, then which of the following equations could be used to solve the problem?A. 3(J + 15) = J - 1B. J + 15 = 3J - 1C. J + 15 = 3(J - 1)
2. A highway makes an angle of 6 with the horizontal. This angle is maintained for a horizontal distance of 5 miles. To the nearest hundredth of a mile, how high does the highway rise in this 5-mile section? Show the steps you use to find the distance.
Answers
Answer:
The cannonball lands at approximately 5.093 unit distance from the point of fire
Step-by-step explanation:
The given parameters are;
The arc denoting the equation of motion of the cannon is y₁ = -0.5·x² + 2.5·x + 1
The slope of the field where in the direction the cannon is fired is y₂ = 1.5·x
The points where the cannonball land on the slopping field is given as rightly pointed by equating the two equations, the cannonball path path and the field path as follows;
At the point of contact of the cannonball and the field, the y-values of both equation will be equal
y₁ = y₂
∴ -0.5·x² + 2.5·x + 1 = 0.15·x
Which gives;
-0.5·x² + 2.5·x - 0.15·x + 1 = 0
-0.5·x² + 2.35·x + 1 = 0
-(-0.5·x² + 2.35·x + 1) = 0.5·x² - 2.35·x - 1 = 0
0.5·x² - 2.35·x - 1 = 0
The above equation is in the general form of a quadratic equation, which is given as follows;
a·x² + b·x + c = 0
By the quadratic equation, we have;
Plugging in the values, gives;
∴ x ≈ 5.093 or x ≈ -0.393
Therefore, the cannonball will takeoff at x ≈ -0.393 and land at x ≈ 5.093
The height from which they fire the cannon is given by the substituting the value of x ≈ -0.393 into the equation for the path of the cannonball, to give;
= -0.5·(-0.393)² + 2.5·(-0.393) + 1 = -0.0597
≈ -0.0597.
However, the actual initial height from which the cannonball is fired given by placing x = 0, which gives y = 1, which is the reason for the other (negative) value for x. Please see the attached graph created with Microsoft Excel.
Answers
Answer:-2
Step-by-step explanation:
slope = rise / run
y2-y1/x2-x1
(3,6) is the first set of coordinates
(5,2) is the second set of coordinates
2-6/5-3= -4/2= -2
Answer: the slope is -2/4
Step-by-step explanation: m is the variable for slope.
I think so. Not 100% sure because I just learned this content in pre-algebra class.
3/5
2/3
4/5
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Answer: The scale is 5/4
Step-by-step explanation:
To determine the scale factor, you have to locate which sides are similar to each other .
the side length with 20 is congruent to 16
15 is similar to 12 and 10 is similar to 8.
And to get from triangle A to triangle B of any side length, we need to multiply by a certain number which is the scale factor.
And to find that number we will divide the side length of any side in triangle A by the similar side to it in triangle B.
20/ 16 = 5/4
15/ 12 = 5 / 4
10 / 8 = 5/4
please help me <3
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Answers
After we know that, the sum of 3 odd integers is just 3 times the middle number. ex. 3+5+7 = 3 times 5 = 15
Then we know the some number times three = 225. we find out that the middle number is 75, so the other two are 73 and 77
Answers
Area of prism = base area × altitude
1. (2x²-10)(x+4)
= 3x³-2x - 40
2. Base area = 2πr
Volume = (2πr)(r²+ 5r)
=2πr³ + 10πr²
3. Base area=½(6)(x-4)(x+3)=3(x-4)(x+3)
Volume= 3(x-4)(x+3)(⅓)
=(x-4)(x+3)= x² - x -12
4. Base circumference= 10π
Base radius = 10π/(2π) = 5
Base area = πr² = 25π
Volume = 25π(3x²-2x)
=125πx²-50πx
5. Volume = 3π√50
=15π√2
6. Base diameter = 16
Base radius = 16/2 = 8
Base area = 2πr = 16π
Volume = 16π(23a²)
=368πa²