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shepuryov [24]
3 years ago
9

An isosceles right triangle has legs lengths of 4 centimeters what is the length of the altitude drawn from the right angle to t

he hypotenuse
​
Mathematics
1 answer:
Free_Kalibri [48]3 years ago
8 0

Answer:

answer is 2 squareroot of 2 cm

Step-by-step explanation:

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The diagonal length of a rectangular playing field is 76 feet, and its width is 48 feet. How long is the playing field?
Sauron [17]

The length of playing field is 58.9 feet

<u>Solution:</u>

Given that, the diagonal length of a rectangular playing field is 76 feet,  

And its width is 48 feet.  

<em><u>To find: length of playing field</u></em>

Now, we know that, <em>diagonal, width and length of a rectangle will form an right angle triangle with diagonal as hypotenuse. </em>

So, now, in a right angled triangle  we can use pythagorean theorem to find the length

Pythagorean theorem, states that the square of the length of the hypotenuse is equal to the sum of squares of the lengths of other two sides of the right-angled triangle.

By above definition, In a right angled triangle ABC we get

c^2 = a^2 + b^2

Where "c" is the length of hypotenuse

"a" is the length of one leg of right angled triangle

"b" is the length of other leg of right angled triangle

\begin{array}{l}{\text {Then, diagonal = width }^{2}+\text { length }^{2}} \\\\ {76^{2}=48^{2}+\text {length }^{2}} \\\\ {\text {Length }^{2}=76^{2}-48^{2}} \\\\ {\text {Length }^{2}=5776-2304}\end{array}

\begin{array}{l}{\text { Length }^{2}=3472} \\\\ {\text { Length }=\sqrt{3472}} \\\\ {\text { Length }=58.92}\end{array}

Hence, the length of the rectangular field is 58.9 feet

7 0
3 years ago
Write the first five digits of 1/7 in base 9 expression
kaheart [24]

Compare 1/7 to consecutive multiples of 1/9. This is easily done by converting the fractions to a common denominator of LCM(7, 9) = 63:

1/9 = 7/63

2/9 = 14/63

while

1/7 = 9/63

Then 1/7 falls between 1/9 and 2/9, so 1/7 = 1/9 plus some remainder. In particular,

1/7 = 1/9¹ + 2/63.

We do the same sort of comparison with the remainder 2/63 and multiples of 1/9² = 1/81. We have LCM(63, 9²) = 567, and

1/9² = 7/567

2/9² = 14/567

3/9² = 21/567

while

2/63 = 18/567

Then

2/63 = 2/9² + 4/567

so

1/7 = 1/9¹ + 2/9² + 4/567

Compare 4/567 with multiples of 1/9³ = 1/729. LCM(567, 9³) = 5103, and

1/9³ = 7/5103

2/9³ = 14/5103

3/9³ = 21/5103

4/9³ = 28/5103

5/9³ = 35/5103

6/9³ = 42/5103

while

4/567 = 36/5103

so that

4/567 = 5/9³ + 1/5103

and so

1/7 = 1/9¹ + 2/9² + 5/9³ + 1/5103

Next, LCM(5103, 9⁴) = 45927, and

1/9⁴ = 7/45927

2/9⁴ = 14/45927

while

1/5103 = 9/45927

Then

1/5103 = 1/9⁴ + 2/45927

so

1/7 = 1/9¹ + 2/9² + 5/9³ + 1/9⁴ + 2/45927

One last time: LCM(45927, 9⁵) = 413343, and

1/9⁵ = 7/413343

2/9⁵ = 14/413343

3/9⁵ = 21/413343

while

2/45927 = 18/413343

Then

2/45927 = 2/9⁵ + remainder

so

1/7 = 1/9¹ + 2/9² + 5/9³ + 1/9⁴ + 2/9⁵ + remainder

Then the base 9 expansion of 1/7 is

0.12512..._9

8 0
2 years ago
Calculate the sample mean and sample variance for the following frequency distribution of heart rates for a sample of American a
spin [16.1K]

Answer:

Mean = 68.9

s^2 =18.1 --- Variance

Step-by-step explanation:

Given

\begin{array}{cccccc}{Class} & {51-58} & {59-66} & {67-74} & {75-82} & {83-90} \ \\ {Frequency} & {6} & {3} & {11} & {13} & {4} \ \end{array}

Solving (a): Calculate the mean.

The given data is a grouped data. So, first we calculate the class midpoint (x)

For 51 - 58.

x = \frac{1}{2}(51+58) = \frac{1}{2}(109) = 54.5

For 59 - 66

x = \frac{1}{2}(59+66) = \frac{1}{2}(125) = 62.5

For 67 - 74

x = \frac{1}{2}(67+74) = \frac{1}{2}(141) = 70.5

For 75 - 82

x = \frac{1}{2}(75+82) = \frac{1}{2}(157) = 78.5

For 83 - 90

x = \frac{1}{2}(83+90) = \frac{1}{2}(173) = 86.5

So, the table becomes:

\begin{array}{cccccc}{x} & {54.5} & {62.5} & {70.5} & {78.5} & {86.5} \ \\ {Frequency} & {6} & {3} & {11} & {13} & {4} \ \end{array}

The mean is then calculated as:

Mean = \frac{\sum fx}{\sum f}

Mean = \frac{54.5*4+62.5*3+70.5*11+78.5*13+86.5*4}{6+3+11+13+4}

Mean = \frac{2547.5}{37}

Mean = 68.9 -- approximated

Solving (b): The sample variance:

This is calculated as:

s^2 =\frac{\sum (x - \overline x)^2}{\sum f - 1}

So, we have:

s^2 =\frac{(54.5-68.9)^2+(62.5-68.9)^2+(70.5-68.9)^2+(78.5-68.9)^2+(86.5-68.9)^2}{37 - 1}

s^2 =\frac{652.8}{36}

s^2 =18.1 -- approximated

5 0
2 years ago
Theresa Morgan, a chemist, has a 30% hydrochloric acid solution and a 50% hydrochloric acid solution. How many liters of each sh
Bogdan [553]
Let x represent the number of liters of 50% acid Theresa puts into the mix. The the number of liters of 30% acid will be (420-x). The total amount of acid in the final solution will be ...
  0.50x + 0.30(420-x) = 0.45(420)
  0.20x + 126 = 189 . . . . . . . . . . . . . . . simplify
  0.20x = 63 . . . . . . . . . . . . . . . . . . . . . subtract 126
  x = 63/0.20 = 315 . . . . . . . . . . . . . . . liters of 50% solution
  (420-x) = 420-315 = 105 . . . . . . . . . liters of 30% solution

Theresa should mix ...
  105 liters of 30% solution
  315 liters of 50% solution
7 0
3 years ago
A triangle has a 90° angle. What type of triangle is it?
CaHeK987 [17]

Answer:

Its a right triangle

Step-by-step explanation:

8 0
3 years ago
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