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1 year ago

The area of the parallelogram is 56 square miles. Find its base and height.

2 answers:

8 0

The equation for the area of a parallelogram is: $A = bh$
A = area, b = length of base, h = height

You are told that the height, h = x - 5, and base, b = x + 5. You are also told that the area is 56.

1) Plug these values into the equation. Multiply (x-5) and (x+5) by foiling:
$A = bh\\
56 = (x+5)(x-5)\\
56 = x^{2} + 5x -5x + 5(-5)\\
56 = x^{2} - 25$

2) Solve for the value of x
$56 = x^{2} - 25\\
56 + 25 = x^{2} \\
 x^{2} = 81\\
x = \sqrt{81} \\
x = 9$
Note that x could also equal -9, but since we're looking for lengths of a shape, the lengths can't be negative!

3) Plug the value of x = 9 miles back into the equations for height and base to find the length of the base and height:
$\text{height} = x-5 = 9 - 5 = 4\\\\
\text{base}= x+5 = 9 + 5 = 14$

--------

Answer: Height = 4 miles, Base = 14 miles

Andru [333]1 year ago

3 0

The area of a parallelogram is A=bh. We are given that the height is x-5, the base is x+5 and that the area of the figure is 56 square miles. So,

56=(x-5)(x+5)
Multiply x-5 and x+5
56=x²-25
Add 25 to both sides
81=x²
Take the square root of both sides
x=9

With the value of x, we can substitute it into the height and base so,

Height:9-5
Base=9+5

So, the height is 4miles and the base is 14 miles.
We can check the answer by using the formula of the area of a parallelogram.
A=bh
56=4*14
56=56

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Answer:
Interest earned = 2713.8

Explanation:
We will solve this problem on two steps:
1- get the final amount after three years
2- get the interest earned by subtracting the initial amount from the final one.

1- getting the final amount after 3 years:
The formula that we will use is as follows:
A = P (1 + r/n)^(nt)
where:
A is the final amount we want to calculate
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r is the interest = 0.12
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Substitute to get the final amount:
A = P (1 + r/n)^(nt)
A = 6300 (1 + 0.12/12) ^ (12*3)
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2- getting the interest earned:
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From the Venn diagram: 15 players like Chemstrand, 17 players like Chemgrass, 13 players like both Chemstrand and Chemgrass while 10 players like neither Chemstrand nor Chemgrass.

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2 years ago

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Answer:

It is proved that $f_x, f_y$ exixts at (0,0) but not differentiable there.

Step-by-step explanation:

Given function is,

$f(x,y)=\frac{9x^2y}{x^4+y^2}; (x,y)\neq (0,0)$

To show exixtance of $f_x(0,0), f_y(0,0)$ we take,

$f_x(0,0)=\lim_{h\to 0}\frac{f(h+0,k+0)-f(0,0)}{h}=\lim_{h\to 0}\frac{\frac{9h^2k}{h^4+k^2}-0}{h}\\\therefore f_x(0,0)=\lim_{h\to 0}\frac{9hk}{h^4+k^2}=\lim_{h\to 0}\frac{9k}{h^3+\frac{k^2}{h}}=0$ exists.

And,

$f_y(0,0)=\lim_{k\to 0}\frac{f(h,k)-f(0,0)}{k}=\lim_{k\to 0}\frac{9h^2k}{k(h^4+k^2)}=\lim_{k\to 0}\frac{9h^2}{h^4+k^2}=\frac{9}{h^2}$ exists.

To show f(x,y) is not differentiable at the origin cheaking continuity at origin be such that,

$\lim_{(x,y)\to (0,0)}\frac{9x^2y}{x^4+y^2}=\lim_{x\to 0\\ y=mx^2}\frac{9x^2y}{x^4+y^2}=\frac{9x^2\times m x^2}{x^4+m^2x^4}=\frac{9m}{1+m^2}$ where m is a variable.

which depends on various values of m, therefore limit does not exists. So f(x,y) is not continuous at (0,0). Hence it is not differentiable at (0,0).

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A random sample of 28 statistics tutorials was selected from the past 5 years and the percentage of students absent from each on

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Answer:

Step-by-step explanation:

Hello!

X: number of absences per tutorial per student over the past 5 years(percentage)

X≈N(μ;σ²)

You have to construct a 90% to estimate the population mean of the percentage of absences per tutorial of the students over the past 5 years.

The formula for the CI is:

X[bar] ± $Z_{1-\alpha /2}$ * $\frac{S}{\sqrt{n} }$

⇒ The population standard deviation is unknown and since the distribution is approximate, I'll use the estimation of the standard deviation in place of the population parameter.

Number of Absences 13.9 16.4 12.3 13.2 8.4 4.4 10.3 8.8 4.8 10.9 15.9 9.7 4.5 11.5 5.7 10.8 9.7 8.2 10.3 12.2 10.6 16.2 15.2 1.7 11.7 11.9 10.0 12.4

X[bar]= 10.41

S= 3.71

$Z_{1-\alpha /2}= Z_{0.95}= 1.645$

[10.41±1.645* $(\frac{3.71}{\sqrt{28} } )$ ]

[9.26; 11.56]

Using a confidence level of 90% you'd expect that the interval [9.26; 11.56]% contains the value of the population mean of the percentage of absences per tutorial of the students over the past 5 years.

I hope this helps!

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