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1 month ago

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N the diagram below, points $A,$ $E,$ and $F$ lie on the same line. If $ABCDE$ is a regular pentagon, and $\angle EFD=90^\circ$,

then how many degrees are in the measure of $\angle FDE$?
[asy]
size(5.5cm);
pair cis(real magni, real argu) { return (magni*cos(argu*pi/180),magni*sin(argu*pi/180)); }

pair a=cis(1,144); pair b=cis(1,72); pair c=cis(1,0); pair d=cis(1,288); pair e=cis(1,216);
pair f=e-(0,2*sin(pi/5)*sin(pi/10));
dot(a); dot(b); dot(c); dot(d); dot(e); dot(f);
label("$A$",a,WNW);
label("$B$",b,ENE);
label("$C$",c,E);
label("$D$",d,ESE);
label("$E$",e,W);
label("$F$",f,WSW);
draw(d--f--a--b--c--d--e);
draw(f+(0,0.1)--f+(0.1,0.1)--f+(0.1,0));
[/asy]

1 answer:

Zina [12.3K]1 month ago

7 0

Answer:

:0000000000000000000000

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The distance AB measures 185.77 feet.
Step-by-step explanation:
According to the attached figure, in triangle ABC,
∠B = 103.2°, ∠C = 14.4° and side BC = 661 feet.
We need to calculate AB's measure.
Since ∠A + ∠B + ∠C = 180°
∠A + 103.2 + 14.4 = 180
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∠A = 180 - 117.6
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Utilizing the sine rule in the triangle ABC,
AB =
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A spinner contains four sections: red, blue, green, and yellow. Joaquin spins the spinner twice. The set of outcomes is given as

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Response:

Detailed explanation:

The spinner has four colors: red, blue, green, and yellow. Joaquin spins this spinner two times. The possible outcomes are represented as

S = {RB, RG, RY, RR, BR, BG, BY, BB, GR, GB, GY, GG, YR, YB, YG, YY}.

Considering the variable “yellow (Y),” let’s determine the potential values for Y.

From the aforementioned set, it appears that Y can assume only three values: 0, 1, or 2.

Within the 16 possible outcomes, Y appears 0 times in 9 instances, 1 time in 6 instances, and 2 times in 1 instance.

Therefore, the probability is calculated as the number of favorable outcomes divided by the total outcomes in the sample space.

This means

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It's known that

When two lines are parallel, their slopes are identical.

The slope between any two points can be calculated using the following formula:

$m=\frac{y2-y1}{x2-x1}$

We will calculate the slope for each case to find the solution to the problem.

Case A) Point $B(5,2)\ C(7,-5)$

Determine the slope of BC

Insert the values into the formula:

$m=\frac{-5-2}{7-5}$

$m=\frac{-7}{2}$

$m=-3.5$

Thus,

The equation $y=-3.5x-15$ is parallel to the line that goes through the points $B(5,2)\ C(7,-5)$

Therefore,

the result for Part A) is

$B(5,2)\ C(7,-5)$ ------> $y=-3.5x-15$

Case B) Point $D(11,6)\ E(5,9)$

Calculate the slope of DE

Plug the values into the formula:

$m=\frac{9-6}{5-11}$

$m=\frac{3}{-6}$

$m=-0.5$

Thus,

The equation $y=-0.5x-3$ is parallel to the line that goes through the points $D(11,6)\ E(5,9)$

Therefore,

the result for Part B) is

$D(11,6)\ E(5,9)$ ------> $y=-0.5x-3$

Case C) Point $F(-7,12)\ G(3,-8)$

Determine the slope of FG

Insert the values into the formula:

$m=\frac{-8-12}{3+7}$

$m=\frac{-20}{10}$

$m=-2$

Thus,

Any linear equation with slope $m=-2$ will be parallel to the line through the points $F(-7,12)\ G(3,-8)$

Case D) Point $H(4,4)\ I(8,9)$

Calculate the slope of HI

Substitute the values in the formula:

$m=\frac{9-4}{8-4}$

$m=\frac{5}{4}$

$m=1.25$

Thus,

The equation $y=1.25x+4$ is parallel to the line through the points $H(4,4)\ I(8,9)$

Therefore,

the result for Part D) is

$H(4,4)\ I(8,9)$ ------> $y=1.25x+4$

Case E) Point $J(7,2)\ K(-9,8)$

Determine the slope of JK

Insert the values into the formula:

$m=\frac{8-2}{-9-7}$

$m=\frac{6}{-16}$

$m=-0.375$

Thus,

Any linear equation characterized by slope $m=-0.375$ will be parallel to the line that runs through the points $J(7,2)\ K(-9,8)$

Case F) Point $L(5,-7)\ M(4,-12)$

Find the slope of LM

Substitute the values in the formula:

$m=\frac{-12+7}{4-5}$

$m=\frac{-5}{-1}$

$m=5$

Thus,

The equation $y=5x+19$ is parallel to the line connecting the points $L(5,-7)\ M(4,-12)$

Therefore,

the result for Part F) is

$L(5,-7)\ M(4,-12)$ ------> $y=5x+19$

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