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3 years ago

30 degree triangles.. photo provided

Mathematics

2 answers:

lutik1710 [3]3 years ago

7 0

Lets x = short leg

30 60 90 triangle
Ratio of short leg: long leg: hypo = x : x√3 : 2x
Given long leg = 555 feet = 185√3 * √3
So
short leg = 185√3 ft
long leg = 555 ft
hypo = 370 ft
hypo is the longest's height

short leg: The distance from the man's feet to the base of the monument = 185√3 ft
hypo: The distance from the man's feet to the top of the monument = 370 ft

Answer:
A. The distance from the man's feet to the base of the monument is 185√3 feet
D. The segment representing the monument's height is the longest segment in the triangle

Vsevolod [243]3 years ago

6 0

The answer is going to be the second one

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Diano4ka-milaya [45]

Answer:

selecting one item? If you mean each color:

40% chance for a red marble / 2/5 chance

60% chance for a blue marble / 3/5 chance

Step-by-step explanation:

Because there are 12 blue marbles and 8 red marbles, we would total them to see how many marbles there are in total. There are 20 in total, but to find the probability, if we put all 20 marbles in a bag (blue and red), and we picked a marble at random, there are 8 red marbles and 12 blue marble in that bag. The probability that I will choose red marbles is 8/20 (or 8 red marbles out of 20 total marbles), and for blue marbles, it would be 12/20 (or 12 blue marbles out of 20 total marbles). Simplify both expression to 2/5, and 3/5 respectively. Finally, if you need percentage, just multiply the denominator by 5 to get it to a hundred, and do the same to the numerator. This way you don't change the value of the expression.

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3 years ago

Find the area of the parallelogram

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Hello from MrBillDoesMath!

Answer:

32 ft^2

Discussion:

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3 years ago

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Solve the equation 4(3y-1)-3(2y)=6

olga nikolaevna [1]

Answer:

$4(3y - 1) - 3(2y) = 6$

Remove backets

$12y - 4 - 6y = 6$

Group and Evaluate like terms

$12y - 6y = 6 + 4$

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$y = 10 \div 6$

$y = 1.67$

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3 years ago

The scatter plot shows the amount of sleep Aria got the night before a test and her test scores. Use the y-intercept, (0, 62), a

kotegsom [21]

Answer:

y = 3x + 62

Step-by-step explanation:

Trend line equation in the form :

y = mx + b

m = slope

b = intercept

Using the points :

(0, 62) ; (6, 80)

Slope, m = Rise / Run = (y2-y1) / (x2 - x1)

y2 = 80 ; y1= 62 ; x2 = 6 ; x1 = 0

m = (80 - 62) / (6 - 0)

m = 18 / 6

m = 3

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(0, 62) ; hence, b = 62

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3 0

3 years ago

For the given term, find the binomial raised to the power, whose expansion it came from: 15(5)^2 (-1/2 x) ^4

Elina [12.6K]

Answer:

C. $(5-\frac{1}{2})^6$

Step-by-step explanation:

Given

$15(5)^2(-\frac{1}{2})^4$

Required

Determine which binomial expansion it came from

The first step is to add the powers of he expression in brackets;

$Sum = 2 + 4$

$Sum = 6$

Each term of a binomial expansion are always of the form:

$(a+b)^n = ......+ ^nC_ra^{n-r}b^r+.......$

Where n = the sum above

$n = 6$

Compare $15(5)^2(-\frac{1}{2})^4$ to the above general form of binomial expansion

$(a+b)^n = ......+15(5)^2(-\frac{1}{2})^4+.......$

Substitute 6 for n

$(a+b)^6 = ......+15(5)^2(-\frac{1}{2})^4+.......$

[Next is to solve for a and b]

From the above expression, the power of (5) is 2

Express 2 as 6 - 4

$(a+b)^6 = ......+15(5)^{6-4}(-\frac{1}{2})^4+.......$

By direct comparison of

$(a+b)^n = ......+ ^nC_ra^{n-r}b^r+.......$

and

$(a+b)^6 = ......+15(5)^{6-4}(-\frac{1}{2})^4+.......$

We have;

$^nC_ra^{n-r}b^r= 15(5)^{6-4}(-\frac{1}{2})^4$

Further comparison gives

$^nC_r = 15$

$a^{n-r} =(5)^{6-4}$

$b^r= (-\frac{1}{2})^4$

[Solving for a]

By direct comparison of $a^{n-r} =(5)^{6-4}$

$a = 5$

$n = 6$

$r = 4$

[Solving for b]

By direct comparison of $b^r= (-\frac{1}{2})^4$

$r = 4$

$b = \frac{-1}{2}$

Substitute values for a, b, n and r in

$(a+b)^n = ......+ ^nC_ra^{n-r}b^r+.......$

$(5+\frac{-1}{2})^6 = ......+ ^6C_4(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+ ^6C_4(5)^{6-4}(\frac{-1}{2})^4+.......$

Solve for $^6C_4$

$(5-\frac{1}{2})^6 = ......+ \frac{6!}{(6-4)!4!)}*(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+ \frac{6!}{2!!4!}*(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+ \frac{6*5*4!}{2*1*!4!}*(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+ \frac{6*5}{2*1}*(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+ \frac{30}{2}*(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+15*(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+15(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+15(5)^2(\frac{-1}{2})^4+.......$

Check the list of options for the expression on the left hand side

The correct answer is $(5-\frac{1}{2})^6$

3 0

3 years ago