The hypotenuse of a 45°-45°-90° triangle measures 4 miles. What is the measure of the leg?

GEOMETRY PEOPLE PLEASE HELP ME!!!!!!!!!!!!!!!!! I WILL GIVE BRAINLIEST!

IrinaVladis [17]

$\Delta KLJ\sim\DeltaNPM$ (AAA) Corresponding angles are congruent.

Therefore, the sides of the triangles are proportional:

$\dfrac{4x-4}{20}=\dfrac{30}{25}$ cross multiply

$25(4x-4)=(20)(30)$ use distributive property

$(25)(4x)+(25)(4)=600\\\\100x+100=600$ subtract 100 from both sides

$100x=500$ divide both sides by 100

$x=5$

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$\Delta PQR\cong\Delta STU$ (SAS)

If two sides and the included angle of one triangle are equal to the corresponding sides and angle of another triangle, the triangles are congruent.

Therefore we have the equation:

$3y-2=y+4\qquad|+2\\\\3y=y+6\qquad|-y\\\\2y=6\qquad|:2\\\\y=3$

The perimeter of △PQR:

$P=4+6+3y-2=8+3y$

Substitute the value of y to the expression:

$P=8+3(3)=8+9=17\ ft$

7 0

3 years ago

Just write some random stuff and get points

Tanya [424]

I like pizza pizza is good for my health

8 0

3 years ago

A sociologist was investigating the ages of grandparents of high school students. From a random sample of 10 high school student

umka2103 [35]

Answer:

$(69.8-71.6) -2.101 \sqrt{\frac{8.38^2}{10} +\frac{6.65^2}{10}}= -8.908$

$(69.8-71.6) +2.101 \sqrt{\frac{8.38^2}{10} +\frac{6.65^2}{10}}= 5.308$

So then the confidence interval for the difference of means is given by:

$-8.908 \leq \mu_M -\mu_F \leq 5.308$

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X_{M}= 69.8$ represent the mean for the age of grandmother

$\bar X_{F}= 71.6$ represent the mean for the age of grandfather

$s_{M}= 8.38$ represent the sample deviation for the age of grandmother

$s_{F}= 6.65$ represent the sample deviation for the age of grandfather

$n_M = n_F= 10$

Solution to the problem

For this case the confidence interval is given by:

$(\bar X_{M} -\bar X_F) \pm t_{\alpha/2} \sqrt{\frac{s^2_M}{n_M} +\frac{s^2_F}{n_F}}$

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

$df=n_M +n_F-2=10+10-2=18$

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,18)".And we see that $t_{\alpha/2}=2.101$

And replacing we got:

$(69.8-71.6) -2.101 \sqrt{\frac{8.38^2}{10} +\frac{6.65^2}{10}}= -8.908$

$(69.8-71.6) +2.101 \sqrt{\frac{8.38^2}{10} +\frac{6.65^2}{10}}= 5.308$

So then the confidence interval for the difference of means is given by:

$-8.908 \leq \mu_M -\mu_F \leq 5.308$

5 0

3 years ago

2^5×5×11 and 2^3×5^2×7

IceJOKER [234]

Answer:

1. 1760

2. 1400

Step-by-step explanation:

2^5 = 32

32 × 5 = 160

160 × 11 = 1760

2^3 = 8

5^2 = 25

8 × 25 = 200

200 × 7 = 1400

4 0

3 years ago

Factor COMPLETELY:

Elena L [17]

1. Find the greatest common factor (GCF)
What is the largest number that divides evenly into 4x^2, -16x^4, and 10x^5?
It is 2.
What is the highest degree of x that divides evenly into 4x^2, -16x^4, and 10x^5?
It is x^2.
Multiply the results above, the GCF = 2x^2

2. Factor out the GCF (Write the GCF first. Then, in parentheses, divide each term by the GCF.)
2x^2(4x^2/2x^2 + -16x^4/2x^2 + 10x^5/2x^2)

3. Simplify each term in parentheses
2x(2-8x^2+5x^3)

Have a nice day :D

6 0

3 years ago