Answers
Explanation:
First use Pythagorean theorem to solve for the hypotenuse.
3^2 +15^2= c^2
15.3 =c
Next solve for the volume.
Related Questions
Given the following rectangles, identify all combinations of assembling these rectangles for which it is possible to create a rectangle with the length of 15 and the width 11 with no gaps or overlapping. You can't cut any of the rectangles but you may use some of them multiple times. More than one answer may be correct; mark all that apply.
Rectangles you are given:
answer options:
two C rectangles, two D rectangles, and two B rectangles
one each of rectangles A, B, C, and D
one A rectangle and four B rectangles
three E rectangles and two B rectangles
one E rectangle, one C, one D, and three B rectangles
Answers
The combinations of assembling these rectangles for which it is possible to create a rectangle with the length of 15 and the width 11 with no gaps or overlapping are:
One each of rectangles A, B, C, and D.One A rectangle and four B rectangles.What is a rectangle?A rectangle is a plane figure with four straight sides and four right angles, especially one with unequal adjacent sides.
Required
Which group forms a rectangle of
[tex]\text{Length}=15[/tex]
[tex]\text{Width}=11[/tex]
First, calculate the area of the big rectangle
[tex]\text{Area}=\text{Length}\times\text{Width}[/tex]
[tex]\text{A}_{\text{Big}}=15\times11[/tex]
[tex]\text{A}_{\text{Big}}=165[/tex]
Next, calculate the area of each rectangle A to E.
[tex]\text{A}_{\text{A}}=11\times7[/tex]
[tex]\text{A}_{\text{A}}=77[/tex]
[tex]\text{A}_{\text{B}}=2\times11[/tex]
[tex]\text{A}_{\text{B}}=22[/tex]
[tex]\text{A}_{\text{C}}=6\times6[/tex]
[tex]\text{A}_{\text{C}}=36[/tex]
[tex]\text{A}_{\text{D}}=6\times5[/tex]
[tex]\text{A}_{\text{D}}=30[/tex]
[tex]\text{A}_{\text{E}}=13\times4[/tex]
[tex]\text{A}_{\text{E}}=52[/tex]
Then consider each option.
(a) 2C + 2D + 2B
[tex]2\text{C}+2\text{D}+2\text{B}=(2\times36)+(2\times30)+(2\times22)[/tex]
[tex]2\text{C}+2\text{D}+2\text{B}=72+60+44[/tex]
[tex]2\text{C}+2\text{D}+2\text{B}=176[/tex]
(b) A + B + C + D
[tex]\text{A}+\text{B}+\text{C}+\text{D}=77+22+36+30[/tex]
[tex]\text{A}+\text{B}+\text{C}+\text{D}=165[/tex]
(c) A + 4B
[tex]\text{A} + 4\text{B}=77+(4\times22)[/tex]
[tex]\text{A} + 4\text{B}=77+88[/tex]
[tex]\text{A} + 4\text{B}=165[/tex]
(d) 3E + 2B
[tex]3\text{E}+2\text{B}=(3\times52)+(2\times22)[/tex]
[tex]3\text{E}+2\text{B}=156+44[/tex]
[tex]3\text{E}+2\text{B}=200[/tex]
(e) E + C + D + 3B
[tex]\text{E} + \text{C} + \text{D} + 3\text{B}=52+36+30+(3\times22)[/tex]
[tex]\text{E} + \text{C} + \text{D} + 3\text{B}=52+36+30+66[/tex]
[tex]\text{E} + \text{C} + \text{D} + 3\text{B}=184[/tex]
Recall that:
[tex]\text{A}_{\text{Big}}=165[/tex]
Only options (b) and (c) match this value.
[tex]\text{A}+\text{B}+\text{C}+\text{D}=165[/tex]
[tex]\text{A} + 4\text{B}=165[/tex]
Hence, options (b) and (c) are correct.
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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?
Answers
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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2(x+5)-5 x 12 example pls
Answers
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.
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?
Answers
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.
Is the expression quadratic 3x+5y-2
Answers
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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Ram borrowed Rs. 250000 from sit a at the rate of 21%: per annum. At the end of monts, how much should he pay compounde à half yearly ?
Answers
The end of 6 months, Ram should pay Rs. 276250 compounded half-yearly.
Ram borrowed Rs. 250000 from Sit at an interest rate of 21% per annum. To calculate the compound interest, we need to know the compounding period. In this case, the interest is compounded half-yearly, which means it is calculated twice a year.
To find out how much Ram should pay at the end of 6 months, we can use the compound interest formula:
A = P(1 + r/n)^(nt)
Where:
A = the amount to be paid at the end of the time period
P = the principal amount (the initial amount borrowed) = Rs. 250000
r = the interest rate per period (in decimal form) = 21% = 0.21
n = the number of compounding periods per year = 2 (since it's compounded half-yearly)
t = the number of years = 6 months = 6/12 = 0.5 years
Plugging in these values into the formula, we get:
A = 250000(1 + 0.21/2)^(2*0.5)
Simplifying the equation:
A = 250000(1 + 0.105)^(1)
A = 250000(1.105)
A = Rs. 276250
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0.059 and 0.01 which is greater?