The calculated percentage error with the assumed answer of 10%
To find the percentage error in actual weights, we can use the formula:
Percentage Error = [(|Measured Value - Expected Value|) / Expected Value] * 100%
In this case, the expected weight is 4 ounces. Let's assume the measured value is 10% off from the expected value. So the measured value would be:
Measured Value = Expected Value + (10% of Expected Value)
= 4 ounces + (10/100) * 4 ounces
= 4 ounces + 0.4 ounces
= 4.4 ounces
Now we can calculate the percentage error:
Percentage Error = [(|4.4 ounces - 4 ounces|) / 4 ounces] * 100%
= [(0.4 ounces) / 4 ounces] * 100%
= (0.4/4) * 100%
= 0.1 * 100%
= 10%
Comparing the calculated percentage error with the assumed answer of 10%, we can see that they are the same.
The percentage error represents the deviation from the expected value as a percentage of the expected value itself. In this case, it indicates that the actual weights deviate by 10% from the expected weight of 4 ounces. The calculated percentage error with the assumed answer of 10%
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2(x+5)-5 x 12 example pls
When x = 3, the expression 2x - 50 evaluates to -44.
To demonstrate an example using the expression 2(x + 5) - 5 × 12, let's simplify it step by step:
Start with the given expression.
2(x + 5) - 5 × 12
Apply the distributive property.
2x + 2(5) - 5 × 12
Simplify within parentheses and perform multiplication.
2x + 10 - 60
Combine like terms.
2x - 50
The simplified form of the expression 2(x + 5) - 5 × 12 is 2x - 50.
Let's consider an example for substituting a value for the variable x:
Suppose we want to evaluate the expression when x = 3. We substitute x = 3 into the simplified expression:
2(3) - 50
Now, perform the calculations:
6 - 50
The result is -44.
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Question
evaluate the expression 2(x+5)-5 x 12.
Is the expression quadratic 3x+5y-2
No, the expression 3x + 5y - 2200 is not a quadratic expression.
A quadratic expression is an expression of the form ax² + bx + c, where a, b, and c are constants and x is a variable raised to the power of 2.
It is a second-degree polynomial, meaning that the highest power of the variable is 2.Quadratic expressions often have a graph that is a parabola.
"3x + 5y - 2" is a linear expression, not a quadratic expression.
In a quadratic expression, the highest power of the variable(s) is 2, whereas in this expression, the highest power is 1.
The expression 3x + 5y - 2200 is a linear expression since it does not contain a term with a variable raised to the power of 2.
It is a first-degree polynomial, meaning that the highest power of the variable is 1.
Linear expressions often have a graph that is a straight line.
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No. The expression, 3x + 5y - 2, is not quadratic.
What are quadratic expressions?The expression "3x+5y-2" is a linear expression, not quadratic.
Quadratic expressions contain a squared term, like "[tex]ax^2 + bx + c[/tex]." In the given expression, there are no squared terms, only linear terms with variables "x" and "y" raised to the power of 1.
The coefficients for "x" and "y" are 3 and 5, respectively, and there is a constant term of -2. Therefore, it represents a linear relationship between "x" and "y" rather than a quadratic one.
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Select the correct answer.
Omar has a gift card for $40.00 at a gift shop. Omar wants to buy a hat for himself for $13.50. For his friends, he would like to buy souvenir bracelets, which are $3.25 each. All prices include taxes.
Which inequality can be used to solve for how many bracelets Omar can buy?
A.
3.25x + 13.50 ≤ 40
B.
3.25x + 13.50 ≥ 40
C.
13.50x + 3.25 ≤ 40
D.
13.50x + 3.25 ≥ 40
Answer:
A.
3.25x + 13.50 ≤ 40
Step-by-step explanation:
What is the volume of the triangular prism?
3 in.
15 in.
13 in.
Two cyclists, 54 miles apart, start riding toward each other at the same time. One cycles 2 times as fast as the other. If they meet 2 hours later, what is the speed (in mi/h) of the faster cyclist?
Answer:
In summary, the faster cyclist cycles at a speed of 18 mi/h since they travel 36 of the 54 miles in 2 hours while cycling twice as fast as the slower cyclist.
Explanationn:
The two cyclists are 54 miles apart and heading toward each other.
One cyclist cycles 2 times as fast as the other. We will call the faster cyclist A and the slower cyclist B.
They meet 2 hours after starting. This means they travel a total distance of 54 miles in 2 hours.
Since cyclist, A cycles 2 times as fast as cyclist B, cyclist A travels 2/3 of the total distance, and cyclist B travels 1/3 of the total distance.
In two hours, cyclist A travels (2/3) * 54 miles = 36 miles.
We need to find the speed of cyclist A in miles per hour.
Speed = Distance / Time
So the speed of cyclist A is:
36 miles / 2 hours = 18 miles per hour
Therefore, the speed of the faster cyclist is 18 mi/h.
Find y" by implicit differentiation.
cos(y) + sin(x) = 1
y" = cos(y) * dy/dx - sin(x) + sin(y) by implicit differentiation.
To find the second derivative (y") by implicit differentiation, we will differentiate the equation with respect to x twice.
Equation: cos(y) + sin(x) = 1
Differentiating once with respect to x using the chain rule:
-sin(y) * dy/dx + cos(x) = 0
Now, differentiating again with respect to x:
Differentiating the first term:
-d/dx(sin(y)) * dy/dx - sin(y) * d^2y/dx^2
Differentiating the second term:
-d/dx(cos(x)) = -(-sin(x)) = sin(x)
The equation becomes:
-d/dx(sin(y)) * dy/dx - sin(y) * d^2y/dx^2 + sin(x) = 0
Now, let's isolate the second derivative, d^2y/dx^2:
-d^2y/dx^2 = d/dx(sin(y)) * dy/dx - sin(x) + sin(y)
Substituting the previously obtained expression for d/dx(sin(y)) = cos(y):
-d^2y/dx^2 = cos(y) * dy/dx - sin(x) + sin(y)
Thus, the second derivative (y") by the equation:
y" = cos(y) * dy/dx - sin(x) + sin(y)
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The value v of a tractor purchased for $13,000 and depreciated linearly at the rate of $1,300 per year is given by v= -1,300t+13,000, where t represents the number of years since the
purchase. Find the value of the tractor after (a) two years and (b) six years. When will the tractor have no value?
a) the value of the tractor after two years is $10,400.
b) the value of the tractor after six years is $5,200.
To find the value of the tractor after a certain number of years, we can substitute the value of t into the equation v = -1,300t + 13,000.
a) After two years:
Substituting t = 2 into the equation, we get:
v = -1,300(2) + 13,000
v = -2,600 + 13,000
v = 10,400
Therefore, the value of the tractor after two years is $10,400.
b) After six years:
Substituting t = 6 into the equation, we get:
v = -1,300(6) + 13,000
v = -7,800 + 13,000
v = 5,200
Therefore, the value of the tractor after six years is $5,200.
To find when the tractor will have no value, we need to find the value of t when v = 0. We can set the equation v = -1,300t + 13,000 equal to 0 and solve for t:
-1,300t + 13,000 = 0
-1,300t = -13,000
t = -13,000 / -1,300
t = 10
Therefore, the tractor will have no value after 10 years.
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0.059 and 0.01 which is greater?