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agasfer [191]
1 year ago
5

The area of a square is stored in a double variable named area. write an expression whose value is length of the diagonal of the

square java
Mathematics
1 answer:
lbvjy [14]1 year ago
3 0

First of all, a bit of theory: since the area of a square is given by

A = s^2

where s is the length of the square. So, if we invert this function we have

s = \sqrt{A}.

Moreover, the diagonal of a square cuts the square in two isosceles right triangles, whose legs are the sides, so the diagonal is the hypothenuse and it can be found by

d = \sqrt{s^2+s^2} = \sqrt{2s^2} = s\sqrt{2}

So, the diagonal is the side length, multiplied by the square root of 2.

With that being said, your function could be something like this:

double diagonalFromArea(double area) {

double side = Math.sqrt(area);

double diagonal = side * Math.sqrt(2);

return diagonal;

}

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Central school is hosting a “central has talent” show. They will award various prizes for the best 3 acts in the show. First pla
Serjik [45]
Well you didn't post the expressions but let X be the amount of money for first place. This means X-50 would be the amount of second place because <span>each subsequent place after 1st wins $50 less than the previous place. Thirds, would be X-50-50 or X-100

Thus the total amount of money would be all of the values combined which is X+X-50+X-100=3X-150
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8 0
2 years ago
Ming used the calculations shown to find out how much he would spend on 6 pounds of ground beef if 10 pounds of ground beef cost
Angelina_Jolie [31]

Answer:

He determined pounds per dollar by dividing 10 by 25 but wrote the unit rate as a dollar value.

Step-by-step explanation:

Given

\frac{10\ pounds}{\$25} = \frac{10\ pounds}{25} / \frac{\$25}{25} = \frac{0.4}{1}

Unit\ Price = 40\ cents

40\ cents * 6 = \$2.40

Required

Determine Ming's error

Ming's error is from here

\frac{10\ pounds}{\$25} = \frac{10\ pounds}{25} / \frac{\$25}{25} = \frac{0.4}{1}

He calculated the unit rate as pound per dollar.

So, after calculating the unit rate, the unit should be:

Unit\ Rate = 0.4\ pound/\$

But instead, he solved as:

Unit\ Rate = \$0.4/ pound

<em>Hence, (a) is correct</em>

5 0
1 year ago
Read 2 more answers
Yen paid $10.50 for 2.5 pounds of pretzels. What price did she pay for each pound of pretzels?
Furkat [3]

Answer:

The first step is to multiply by a power of 10, so the divisor is a whole number.

Step-by-step explanation:

8 0
2 years ago
Read 2 more answers
Find the sum of the first 63 terms of –19, -13, -7 …
boyakko [2]

The Given Sequence is an Arithmetic Sequence with First term = -19

⇒ a = -19

Second term is -13

We know that Common difference is Difference of second term and first term.

⇒ Common Difference (d) = -13 + 19 = 6

We know that Sum of n terms is given by : S_n = \frac{n}{2}(2a + (n - 1)d)

Given n = 63 and we found a = -19 and d = 6

\implies S_6_3 = \frac{63}{2}(2(-19) + (63 - 1)6)

\implies S_6_3 = \frac{63}{2}(-38 + (62)6)

\implies S_6_3 = \frac{63}{2}(-38 + 372)

\implies S_6_3 = \frac{63}{2}(-38 + 372)

\implies S_6_3 = \frac{63}{2}(334)

\implies S_6_3 = {63}(167) = 10521

The Sum of First 63 terms is 10521

4 0
1 year ago
A manufacturer of banana chips would like to know whether its bag filling machine works correctly at the 449 gram setting. It is
stealth61 [152]

Answer:

We accetp  H₀

Step-by-step explanation:

Information:

Normal distribution  

Population mean      =   μ₀  = 449

Population standard deviation  σ   unknown

Sample size   n  =  23        n < 30    we use t-student test

so   n  =  23    degree of fredom   df = n  - 1  df  = 23- 1   df = 22

Sample mean    μ =  448

Sample standard deviation   s  =  20

Significance level  α  =  0,05  

1.-Hypothesis Test

Null hypothesis                               H₀     μ₀  =  449

Alternative hypothesis                    Hₐ     μ₀  ≠  449

Problem statement ask for determine decision rule for rejecting the null hypothesis. For rejecting the null hypothesis we have to  get an statistic parameter wich implies  that μ is bigger or smaller than μ₀

2.-Significance level   α  =  0,05  ;  as we have a two tail test

α/2    =  0,025

Then from t - student table for  df =  22   and 0,025 (two tail-test)

t(c)  =  ±  2.074

3.- Compute  t(s)

t(s)   =  (  μ  -  μ₀ )  /  s /√n

plugging in values

t(s)   =  (448  -  449) /  20 /√23    ⇒   t(s)   =  -  1*√23 /20

t(s)   =  - 0.2398

4.-Compare t(c)   and  t(s)

t(s)  <  t(c)         - 0.2398  <  - 2.074

Therefore  t(s)  in inside acceptance region.  We accept  H₀

7 0
2 years ago
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