Find the length of the legs.(4,4)[](-2,-2)[?]Enter the number thatbelongs in the green boxEnter
[?]Enter The Number Thatbelongs In The Green BoxEnter](https://brainimage.s3.us-east-005.backblazeb2.com/contents/attachments/1S1iCYE2XCuSUAOABrZ21PYKrabtdBsn.jpeg)
Answers
We have a right triangle and we have to find the length of the legs.
We first analyse the hypotenuse.
It is a line that is defined by two points: (4,4) and (-2,-2).
The line passes through the center of coordinates, but we will check it:
The slope can be calculated as:
[tex]m=\frac{y_2-y_1}{x_2-x_1}=\frac{4-(-2)}{4-(-2)}=1[/tex]We can now write the slope-point equation as:
[tex]\begin{gathered} y-y_0=m(x-x_0) \\ y-4=1(x-4) \\ y-4=x-4 \\ y=x \end{gathered}[/tex]Then, we know that the equation of the line that defines the hypotenuse is y=x.
If the legs are parallel to the axis, we can find the angle that is defined by the hypotenuse and the horizontal leg as:
[tex]\begin{gathered} \tan (\theta)=m=1 \\ \theta=\arctan (1)=45\degree \end{gathered}[/tex]As we have an angle of 45 degree for one leg, the other has to have an equal angle, as 180-90-45=45 (Note: we are substracting from the sum of the 3 angles, 180 degrees, the already known angles, 90 and 45 degrees, so we are left with 45 degrees that correspond to the third angle).
If the two angles for the legs are equal, their lengths are equal too.
So we can start by calculating the length of the hypotenuse: it is the distance between (4,4) and (-2,-2).
[tex]\begin{gathered} D=\sqrt[]{(x_2-x_1)^2+(y_2-y_1)^2}=\sqrt[]{(4-(-2))^2+(4-(-2))^2} \\ D=\sqrt[]{6^2+6^2}=\sqrt[]{2\cdot6^2}=6\sqrt[]{2} \end{gathered}[/tex]Then, we can write the Pythagorean theorem as (a and b are the legs, and c is the hypotenuse):
[tex]\begin{gathered} a^2+b^2=c^2 \\ a=b\longrightarrow2a^2=c^2=(6\sqrt[]{2})^2=36\cdot2=72 \\ a^2=\frac{72}{2} \\ a^2=36 \\ a=\sqrt[]{36} \\ a=b=6 \end{gathered}[/tex]Answer: the length of the legs is 6 units.
Related Questions
DAN have coordinates D(-6, -1) the altitude drawn to side DN
Answers
Explanation:
The slope of the altitude drawn to side DN is the reciprocal and opposite to the slope of side DN, because the altitude is perpendicular to the side.
First we have to find the slope of side DN. The formula for the slope of a line with points (x1, y1) and (x2, y2) is:
[tex]m=\frac{y_1-y_2}{x_1-x_2}[/tex]In this problem the points are D(-6, -1) and N(-3, 10). The slope of side DN is:
[tex]m_{DN}=\frac{-1-10}{-6-(-3)}=\frac{-11}{-6+3}=\frac{-11}{-3}=\frac{11}{3}[/tex]Therefore the slope of the altitude is:
[tex]m_{\text{altitude}}=-\frac{1}{m_{DN}}=-\frac{1}{\frac{11}{3}}=-\frac{3}{11}[/tex]Answer:
The slope of the altitude is -3/11
(7x10^1(4x10^-7
(5.55 x 10^4) - ( 3.41 x 10^4)
(9 x 10^7) divided (3 x 10^3)
Work needs to be shows !!!
Answers
Answer:
(5.55 * 10^4) - (3.41 * 10^4)
=21,400
(9 * 10^7) divided (3 * 10^3)
= 30,000
Step-by-step explanation:
(5.55 * 10^4) - (3.41 * 10^4)
= (5.55 * 10,000) - (3.41 * 10,000)
= 55,500 - 34,100
= 21,400
(9 * 10^7) divided (3 * 10^3)
= (9 * 10,000,000) ÷ (3 * 1,000)
= 90,000,000 ÷ 3,000
= 30,000
Sorry but i don't understand the "(7x10^1(4x10^7". Your question is invalid.
I need help solving this practice problem If you can , answer (a) and (b) separately so I can tell which is which
Answers
Step 1:
Write the expression
[tex](3x^5\text{ - }\frac{1}{9}y^3)^4[/tex]Step 2:
a)
[tex]\begin{gathered} (3x^5\text{ - }\frac{1}{9}y^3)^4 \\ =^4C_0(3x^5)^4(-\frac{1}{9}y^3)^0+^4C_1(3x^5)^3(-\frac{1}{9}y^3)^1+^4C_2(3x^5)^2(-\frac{1}{9}y^3)^2+ \\ +^4C_1(3x^5)^1(-\frac{1}{9}y^3)^3+^4C_0(3x^5_{})^0(-\frac{1}{9}y^3)^4 \end{gathered}[/tex]Step 3:
b) simplified terms of the expression
[tex]\begin{gathered} Note\colon \\ ^4C_0\text{ = 1} \\ ^4C_1\text{ = 4} \\ ^4C_2\text{ = 6} \\ ^4C_3\text{ = 4} \\ ^4C_4\text{ = 1} \end{gathered}[/tex]Next, substitute in the expression
[tex]\begin{gathered} =\text{ 1}\times81x^{20}\times1\text{ - 4}\times27x^{15}\text{ }\times\text{ }\frac{y^3}{9}\text{ + 6 }\times9x^{10}\times\frac{y^6}{81}\text{ - 4}\times3x^5\text{ }\times\text{ }\frac{y^9}{729} \\ +\text{ 1 }\times\text{ 1 }\times\frac{y^{12}}{6561}\text{ } \end{gathered}[/tex][tex]=81x^{20}-12x^{15}y^3\text{ + }\frac{2}{3}x^{10}y^6\text{ - }\frac{4}{243}x^5y^9\text{ + }\frac{1}{6561}y^{12}[/tex]A paper is sold for Php60.00, which is 150% of the cost. How much is the store's cost?
Answers
The store's cost is php40
Let's call the store Cost = C
This means that this cost is elevated a 150% in order to get the price of php60
In an mathematical expression, this is:
C · 150% = php60
Then, let's convert the percentage to decimal. To do this, we just divide the percentage by 100:
150% ÷ 100 = 1.5
Now we can solve:
[tex]\begin{gathered} C\cdot1.5=60 \\ C=\frac{60}{1.5}=40 \end{gathered}[/tex]Then the store cost is C = php40
Cris pays a total of $11 for every 6 Gatoraid bottles. Circle the graph models a relationship with the same unit rate?
Answers
The line that describes this relationship goes from (0,0) to the point (6,11),we can draw it like this:
One year there was a total of 44 commercial and noncommercial orbital launches worldwide. In addition, the number number of commercial orbital launches. Determine the number of commercial and noncommercial orbital launches was two more thank twice the number of commercial orbital launches (HURRY I NEED ANSWER)
Answers
The number of commercial orbital is 14, and the number of noncommercial orbital is 30.
What is algebra?
When numbers and quantities are represented in formulas and equations by letters and other universal symbols.
Given that,
The total number of commercial and noncommercial orbital launches worldwide = 44
Also, the number of noncommercial orbital is two more than twice of commercial orbital
Let the number of commercial orbital =x
Then number of noncommercial orbital = 2x+2
Since, total number of commercial and noncommercial orbital = 44
x + 2x +2 = 44
3x = 42
x = 14
The number of commercial orbital = x = 14
The number of noncommercial orbital = 2x+2 = 2×14+2 = 30
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Express sin 285 as a function of the reference angle.Question 19 options:sin75sin105-sin75sin-75
Answers
Given:
[tex]\sin 285^{\circ}[/tex]To express it as a function of reference angle:
Since the angle lies in the fourth quadrant.
So, the reference angle will be,
[tex]\begin{gathered} \theta_{ref}=360-\theta_4 \\ \theta_{ref}=360^{\circ}-285 \\ \theta_{ref}=75^{\circ} \end{gathered}[/tex]Therefore, the function of the reference angle is,
[tex]\sin 75^{\circ}[/tex]Determine the equation of the line that passes through the point (-1, 2) and isperpendicular to the line y = -2.
Answers
1) In this question, let's find the equation, using the point-slope formula:
[tex](y-y_0)=m(x-x_0)[/tex]2) Notice that since we want a perpendicular line we can write a perpendicular line to y=2, as x=-1/2 for -1/2 is the opposite and reciprocal to 2 (the necessary condition to get a perpendicular line).
So, the slope of that perpendicular line is -1/2
3) Let's plug into that Point-Slope formula, the slope m= -1/2 and the point:
[tex]\begin{gathered} (y-2)=-\frac{1}{2}(x+1) \\ y-2=-\frac{1}{2}x-\frac{1}{2} \\ y=-\frac{1}{2}x-\frac{1}{2}+2 \\ y=-\frac{1}{2}x+\frac{3}{2} \end{gathered}[/tex]4) Thus, the answer is:
[tex]y=-\frac{1}{2}x+\frac{3}{2}[/tex]11) -3(1 + 6r) = 14 - r
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Distributing over parentheses,
[tex]\begin{gathered} -3\cdot1+(-3)\cdot6r=14-r \\ -3-18r=14-r \end{gathered}[/tex]Adding r at both sides,
[tex]\begin{gathered} -3-18r+r=14-r+r \\ -3-17r=14 \end{gathered}[/tex]Adding 3 at both sides,
[tex]\begin{gathered} -3-17r+3=14+3 \\ -17r=17 \end{gathered}[/tex]Dividing by -17 at both sides,
[tex]\begin{gathered} \frac{-17r}{-17}=\frac{17}{-17} \\ r=-1 \end{gathered}[/tex]The graph of the quadratic function with vertex (2,3) is shown in the figure belowFind the domain and range
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Solution
Domain
[tex]Domain=(-\infty,\infty)[/tex]Range
[tex]Range=(-\infty,3][/tex]Kristi Yang borrowed $12,000. The term of the loan was 150 days, and the annual simple interest rate was 6.5%. Find the simple interest due on the loan. (Round your answer to the nearest cent.)
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For an initial ammount borrowed I, an term of the loan t, and an annual interest rate r, the simple interest S due on the loan is given by:
[tex]S=I\cdot r\cdot\frac{t}{365}[/tex]For I = $12000, r = 0.065 and t = 150 days, we have:
[tex]\begin{gathered} S=12000\cdot0.065\cdot\frac{150}{365} \\ S=12000\cdot0.065\cdot0.41096 \\ S=\text{ \$320.55} \end{gathered}[/tex]For the following two numbers, find two factors of the first number such that their product is the first number and their sum is the second number,40, 14
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First we need to factorate the number 40:
[tex]40=2\cdot2\cdot2\cdot5[/tex]The possible numbers we can create using these factors are 2, 4, 5, 8, 10 and 20.
So If the product of the two factors (let's call them 'a' and 'b') is 40 and the sum is 14, we have:
[tex]\begin{gathered} a\cdot b=40 \\ a+b=14 \\ \\ \text{From the second equation:} \\ b=14-a \\ \\ \text{Using this value of b in the first equation:} \\ a(14-a)=40 \\ 14a-a^2=40 \\ a^2-14a+40=0 \end{gathered}[/tex]Using the quadratic formula to solve this equation, we have:
[tex]\begin{gathered} a_1=\frac{-b+\sqrt[]{b^2-4ac}}{2a}=\frac{14+\sqrt[]{196-160}}{2}=\frac{14+6}{2}=10 \\ a_2=\frac{-b-\sqrt[]{b^2-4ac}}{2a}=\frac{14-6}{2}=4 \\ \\ a=10\to b=14-10=4 \\ a=4\to b=14-4=10 \end{gathered}[/tex]So the factors which product is 40 and the sum is 14 are 4 and 10.
If the figure below were reflected across the waxis, what would be the new coordinates of point A
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The coordinates of point A are (-2,3). A reflection across the y-axis is given by:
[tex](x,y)\rightarrow(-x,y)[/tex]Applying this rule to point A we have:
[tex](-2,3)\rightarrow(2,3)[/tex]Therefore, the image of point A is (2,3) and the correct option is B.
how do you solve for x in the following problem... 4 (x + 3) -2x + 8 = 28
Answers
Given the expression
[tex]4(x+3)-2x+8=28[/tex]To solve it for x, the first step is to calculate the term in parentheses, for this you have to apply the distributive property of multiplication.
[tex]\begin{gathered} (4\cdot x)+(4\cdot3)-2x+8=28 \\ 4x+12-2x+8=28 \end{gathered}[/tex]Next order the alike terms toghether and calculate:
[tex]\begin{gathered} 4x-2x+12+8=28 \\ 2x+20=28 \end{gathered}[/tex]Subtract 20 to both sides of the equation:
[tex]\begin{gathered} 2x+20-20=28-20 \\ 2x=8 \end{gathered}[/tex]And finally divide by 2 to reach the value of x:
[tex]\begin{gathered} \frac{2x}{2}=\frac{8}{2} \\ x=4 \end{gathered}[/tex]For this equation x=4
Given ABC shown below. Map ABC using the transformations given below. In each case, start with ABC , graph the image and state the Coodinates of the image's vertices.a) a reflection in the line x = 2 to produce A' B' C'b) a reflection in the line y= 1 to produce A" B"C"
Answers
a) A'(8,7), B' (10, -6) and C' (2,-3)
b) A" (-4,-5) B" (-6,8) C" (2,5)
1) Examining the graph, we can locate the following points of ABC
To reflect across line x=2 let's count to the left the same distance from x=2
Pre-image Reflection in the line x=2
A (-4, 7) (x+8, y) A'(8,7)
B (-6,-6) (x+16, y) B' (10, -6)
C (2,-3) (x,y) C' (2,-3) Remains the same since C is on x=2
b) A reflection about the line y=1 similarly we'll count the distances and then write new points over the line y=1.
So the Image of this is going to be
Pre-image Reflection in the line x=2
A (-4, 7) (x, y-12) A" (-4,-5)
B (-6,-6) (x, y) B" (-6, 8)
C (2,-3) (x,y) C' (2,5)
A" (-4,-5)
B" (-6,8)
C" (2,5)
During a snowstorm, Grayson tracked the amount of snow on the ground. When the storm began, there were 4 inches of snow on the ground. For the first 3 hours of the storm, snow fell at a constant rate of 1 inch per hour. The storm then stopped for 5 hours and then started again at a constant rate of 3 inches per hour for the next 2 hours. As soon as the storm stopped again, the sun came out and melted the snow for the next 2 hours at a constant rate of 4 inches per hour. Make a graph showing the inches of snow on the ground over time using the data that Grayson collected.
Answers
We can plot all that happened in the next graph:
This is the graph showing the inches of the snow on the ground over time using the data that Grayson collected.
Doug travels 5 times as fast as Gloria. Traveling in opposite directions, they are 858 miles apart after 6.5 hours. Find their rates or travel.
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Distance = rate x time
Gloria = X mph
Doug = 5X mph
Where X is the rate of speed
And we have that they are 858 miles apart after 6.5 hours
6.5 X + 5( 6.5 X) = 858 mph
6.5X + 32,4X= 858 mph
39X = 858mph
X = (858mph)/39 = 22
5X = 110
So Gloria is traveling at 22 mph and Doug is traveling at 110 mph. This meand that Gloria rate of speed is 22 and Doug is 110
Which expression is equivalent to sin(71(1) cos (72) - cos () sin (77.)?1?O cos (5)O sin (5)COS2012sin
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Let:
[tex]\begin{gathered} A=\frac{\pi}{12} \\ B=\frac{7\pi}{12} \end{gathered}[/tex]Using the sine difference identity:
[tex]\begin{gathered} \sin (A)\cos (B)-\cos (A)\sin (B)=\sin (A-B) \\ so\colon \\ \sin (\frac{\pi}{12})\cos (\frac{7\pi}{12})-\cos (\frac{\pi}{12})\sin (\frac{7\pi}{12})=\sin (\frac{\pi}{12}-\frac{7\pi}{12}) \\ \sin (\frac{\pi}{12}-\frac{7\pi}{12})=\sin (-\frac{6\pi}{12}) \\ \sin (-\frac{\pi}{2}) \end{gathered}[/tex]Answer:
[tex]\sin (-\frac{\pi}{2})[/tex]Find the missing factor. 8x2 - X - 9 = (x + 1)(
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In this case, we'll have to carry out several steps to find the solution.
Step 01:
Data
8x² - x - 9 = (x + 1) ( ? )
Step 02:
We must find the missing root to solve the exercise.
x1 = - 1
x2 :
a = 8
b = -1
c = -9
[tex]x\text{ = }\frac{-(-1)\pm\sqrt[]{(-1)^2-4\cdot8\cdot(-9)}_{}}{2\cdot8}[/tex][tex]x\text{ = }\frac{1\pm17}{16}[/tex][tex]x\text{ = }\frac{1+17}{16}=\frac{18}{16}=\frac{9}{8}[/tex]x2 = 9 / 8
The answer is:
( x - 9/8)
8x² - x - 9 = (x + 1) ( x - 9/8 )
14a) Kevin asked a random group of students what theirfavorite class was, and the results are below.If Kevin were to randomly select a boy to explain whichclass was his favorite, what is the probability that he willpick a boy who likes History?A. 19.6%BoysGirlsMath810B. 29.3%English1517C. 35.3%Science105History 189D. 52.9%
Answers
If Kevin were to randomly select a boy from this table, this means that we need to add up the total number of boys in this sample in order to find the denominator of the fraction we are dealing with.
Let's add the number of boys up:
[tex]8+15+10+18=51[/tex]Now, the probability that a randomly selected boy likes History is the number of boys who said they like History divided by the total number of boys.
We already have the total number of boys, so now we need to find the number of boys who said they like History, which is 18 based off of the table.
The probability that Kevin will pick a boy who likes History is C) 35.3%.
[tex]\frac{18}{51}=.353[/tex]3. A coin is tossed 140 times. The probability of getting tails is p = 0.500. Would a result of 55heads out of the 140 trials be considered usual or unusual? Why?Unusual, because the result is less than the maximum usual value.O Usual, because the result is between the minimum and maximum usual values.Unusual, because the result is less than the minimum usual value.Unusual, because the result is more than the maximum usual value
Answers
In order to calculate the minimum and maximum usual values, first let's calculate the mean and standard deviation of this distribution:
[tex]\begin{gathered} \mu=n\cdot p=140\cdot0.5=70\\ \\ \sigma=\sqrt{np(1-p)}=\sqrt{140\cdot0.5\cdot0.5}=5.92 \end{gathered}[/tex]Now, calculating the minimum and maximum usual values, we have:
[tex]\begin{gathered} minimum=\mu-2\sigma=70-11.84=58.16\\ \\ maximum=\mu+2\sigma=70+11.84=81.84 \end{gathered}[/tex]Since the given result is 55, it is an unusual reslt, because it is less tahan the minimum usual value.
Correct option: third one.
Use the graphing tool to determine the true statementsregarding the represented function. Check all that apply.f(x) > 0 over the interval (1,).Of(x) < 0 over the interval [1,0).Of(x) 0 over the interval (-∞, 1].Of(x) > 0 over the interval (-∞, 1).Of(x) > 0 over the interval (-∞o).Intro2010-202
Answers
The true statements are,
f(x) > 0 over the interval (1, ∞)
f(x) ≤ 0 over the interval (-∞, 1]
Interval of a function:
If the value of the function f (x) rises as the value of x rises, the function interval is said to be positive. Instead, if the value of the function f (x) drops as the value of x increases, the function interval is said to be negative.
If the endpoints are absent from an interval, it is referred to as being open. It's indicated by ( ). Examples are (1, 2), which denotes larger than 1 and less than 2. Any interval that contains all the limit points is said to be closed. The symbol for it is []. For instance, [2, 5] denotes a value greater or equal to 2 and lower or equal to 5. If one of an open interval's endpoints is present, it is referred to as a half-open interval.
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Find the area and perimeter with these points.(-11,-8)(-11,0)(0,0)(0,-8)
Answers
we have the coordinates
(-11,-8)
(-11,0)
(0,0)
(0,-8)
step 1
plot the give points
using a graphing tool
see the attached figure below
The figure is a rectangle
where
L=0-(-11)=11 units (subtract x-coordinates)
W=0-(-8)=8 units (subtract y-coordinates)
step 2
Find out the area
A=L*W
A=11*8=88 unit2
step 3
Find out the perimeter
P=2(L+W)
P=2(11+8)
P=2(19)=38 units
I need help quick with a math question !
Answers
Step-by-step explanation:
Circle describe and correct each error Graph y-5(x-2)Point =(5, 2)M=3
Answers
Solution;
[tex]\begin{gathered} y-5=3(x-2) \\ y-5=3x-6 \\ y=3x-1 \\ \end{gathered}[/tex][tex]\begin{gathered} Slope(m)=3 \\ \end{gathered}[/tex][tex]\begin{gathered} Intercept\text{ on x axis; ie y=o} \\ 3x=1 \\ x=\frac{1}{3} \end{gathered}[/tex][tex]\begin{gathered} Intercept\text{ on y axis; ie x=0} \\ y=-1 \end{gathered}[/tex]To the describe the error on the graph;
The graph is supposed to pass through point(1/3, -1).
Rounding the problem to the nearest tenth if necessary and find the missing length?
Answers
Step 1:
[tex]\text{Triangle PQR is similar to triangle GHP}[/tex]Step 2:
Write the corresponding sides of the similar triangle
[tex]\begin{gathered} \\ PQ\text{ }\cong\text{ PG} \\ RP\text{ }\cong\text{ PH} \\ \frac{PQ}{PG}\text{ = }\frac{RP}{PH} \\ \\ \frac{PQ}{91}=\frac{72}{56}\text{ } \end{gathered}[/tex]Next
Cross multiply
[tex]\begin{gathered} 56PQ\text{ = 72 }\times\text{ 91} \\ PQ\text{ = }\frac{6552}{56} \\ PQ\text{ = 117} \end{gathered}[/tex]Final answer
PQ ? = 117
Find the indicated probability using the standard normal distribution.P(-2.18
Answers
we have
Z1=0
Z2=-2.18
so
using the z-scores table values
P(-2.18Given y=0.5x^2, describe the transformation (x,y) --> (x,4y) and sketch the graph of this image
Answers
We are given the equation y = 0.5x^2. To describe its transformation from (x, y) to (x, 4y), we can start by first graphing the given equation.
To graph, let's use sample points (x- and y-values):
x y
-2 2
-1 0.5
0 0
1 0.5
2 2
So we have the points (-2, 2), (-1, 0.5), (0, 0), (1, 0.5), and (2, 2) to help us graph the equation.
A transformation of (x, y) --> (x, ay) where a > 1 means a vertical stretch equal to |a|. In this case, because (x, y) is transformed to (x, 4y), the graph stretches vertically by a factor of 4.
To graph, let's use sample points (x- and y-values):
x y
-2 4(2) = 8
-1 4(0.5) = 2
0 4(0) = 0
1 4(0.5) = 2
2 4(2) = 8
The new graph would now look like this:
The area of a triangle is 2312 . Two of the side lengths are 93 and 96 and the included angle is obtuse. Find the measure of the included angle, to the nearest tenth of a degree.
Answers
to facilitate the exercise we will draw the triangle
We start using the area
[tex]A=\frac{b\times h}{2}[/tex]where A is the area, b the base and h the height
if we replace A=2312 and b=96 we can calculate the height(h)
[tex]\begin{gathered} 2312=\frac{96\times h}{2} \\ \\ h=\frac{2312\times2}{96} \\ \\ h=\frac{289}{6} \end{gathered}[/tex]now to calculate the measure of the angles we can solve the red triangle
first we find Y using trigonometric ratio of the sine
[tex]\sin (\alpha)=\frac{O}{H}[/tex]where alpha is the reference angle, O the opposite side from the angle and H the hypotenuse of the triangle
using Y like reference angle and replacing
[tex]\sin (y)=\frac{\frac{289}{6}}{93}[/tex]simplify
[tex]\sin (y)=\frac{289}{558}[/tex]and solve for y
[tex]\begin{gathered} y=\sin ^{-1}(\frac{289}{558}) \\ \\ y=31.2 \end{gathered}[/tex]value of angle y is 31.2°
Y and X are complementary because make a right line then if we add both numbers the solution is 180°
[tex]\begin{gathered} y+x=180 \\ 31.2+x=180 \end{gathered}[/tex]and solve for x
[tex]\begin{gathered} x=180-31.2 \\ x=148.8 \end{gathered}[/tex]measure of the included angle is 148.8°
Yolanda has a rectangular poster that is 16 cm long and 10 cm wide what is the area of the poster in square meters do not round your answer is sure to include the correct unit in your answer
Answers
The area of a rectangle can be calculated as the height times the wide.
But be careful, the problem asks it in square meters! So let's use meters instead of centimeters.
Remember that : 1 m = 100 cm ----> 1 cm = 0.01 m
[tex]\begin{gathered} A=b\cdot h \\ \\ A=0.16\cdot0.10 \end{gathered}[/tex]Doing the multiplication
[tex]A=0.016\text{ m}^2[/tex]Therefore the area of the poster is 0.016 square meters