The hypotenuse of a 45°-45°-90° triangle measures 128 cm. what is the length of one leg of the triangle? 64 cm cm 128 cm cm

Answer 64*sqrt(2)

Step-by-step explanation:

Givens

c = 128 cm

a = b = ??

formula

a^2 + b^2 = c^2                            combine the two equal legs.

2a^2 = c^2                                    Substitute 128 for c

2a^2 = 128^2                                Square

2a^2 = 16384                               Divide by 2

a^2 = 16384/2                  

a^2 = 8192                                   Factor (8192)

8192 = 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2    count the 2s

8192 = 2^13                                 Break 13 into 2 equal values with 1 left over

sqrt(8192) = sqrt(2^6   *   2^6    *    2^1)

sqrt(8192) = 2^6 * sqrt(2)

The length of one leg is 64*sqrt(2)

Length of one leg of the given triangle will be 64√2 cm.

Step-by-step explanation:

Length of the hypotenuse of a 45°- 45°- 90° triangle has been given as 128 cm.

Since two angles other than 90° are of same measure so other two sides of the triangle will be same in measure.

Therefore, by Pythagoras theorem,

Hypotenuse² = Leg(1)² + Leg(2)²

Let the measure of both the legs is x cm

(128)² = 2x²

16384 = 2x²

x² =

x² = 8192

x = √8192

  = 64√2 cm

Therefore, length of one leg of the given triangle will be 64√2 cm.

(w – 2.5)(w + 2.5)

(–4v – 9)(–4v + 9)

step-by-step explanation:

d. 24+-0.114

step-by-step explanation:

the question requires us to calculate the approximate 95% confidence interval for the mean number of ounces of ketchup per bottle in the sample given the following statistics;

sample mean = 24

standard deviation, σ = 0.4

sample size , n = 49

the confidence interval for a population mean is calculated using the formula;

sample mean ± z-score*(σ/sqrt(n))

the z-score associated with a 95% confidence interval, from the standard normal tables, is;

1.96

substituting these values into the formula;

24 ± 1.96*(0.4/7)

= 24 ± 0.112

from the alternatives given, choice d is closest to the above expression;