I need help with this practice problem It’s from my trig bookI attempted this problem earlier, later, if you can, check if I am correct. I will send a pic of my work.
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Given the right triangle:
Taking the cosine of the angle θ:
[tex]\cos \theta=\frac{2\sqrt[]{6}}{2\sqrt[]{15}}=\frac{\sqrt[]{2}}{\sqrt[]{5}}[/tex]Now, taking the arc cosine to find θ:
[tex]\begin{gathered} \theta=\arccos (\sqrt[]{\frac{2}{5}}) \\ \therefore\theta\approx50.8\degree^{} \end{gathered}[/tex]
Related Questions
two rectangles are similar. The length of small rectangle is 4 and the length of the big rectangle is 12. If the perimeter of the smaller rectangle is 28, and what is the perimeter of the larger rectangle?
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In this case, we'll have to carry out several steps to find the solution.
Step 01:
Data
small rectangle
length = 4
perimeter = 28
big rectangle
length = 12
perimeter = ?
Step 02:
small rectangle
perimeter = 2l + 2w
28 = 2 * 4 + 2 w
28 - 8 = 2w
20 / 2 = w
10 = w
big rectangle
[tex]\frac{4}{12}\text{ = }\frac{10}{w}[/tex]4 w = 10 * 12
w = 120 / 4 = 30
Perimeter = 2*12 + 2*30
= 24 + 60 = 84
The answer is:
The perimeter of the big rectangle is 84.
Jake wanted to buy candy for $4.87 with a 6% sales tax. He has a $5.00 bill. Does he have enough for his candy?Yes or No
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The candy cost $4.87, and the sales tax is 6%, which means the sales tax can be calculated as follows;
[tex]\begin{gathered} \text{Cost}=4.87 \\ \text{Sales tax}=4.87\times\frac{6}{100} \\ \text{Sales tax}=4.87\times0.06 \\ \text{Sales tax}=0.2922 \end{gathered}[/tex]Therefore, the total cost inclusive of sales tax would be;
[tex]\begin{gathered} \text{Cost}+\text{Sales tax}=4.87+0.2922 \\ \text{Cost}+\text{Sales tax}=5.1622 \end{gathered}[/tex]ANSWER:
The total cost would be $5.1622
Hence, Jake does not have enough for his candy
The answer is NO
Just need help with number 3. Please. Thankyou! Been stuck on this one for a while now.
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A cosine function is given in the form:
[tex]y=A\cos(x-h)+k[/tex]where |A| is the amplitude, h is the horizontal shift (also called a phase shift) and k is the vertical shift.
The function is given to be:
[tex]\begin{gathered} y=8\cos(\frac{1}{4}x) \\ A=8 \end{gathered}[/tex]Therefore, the amplitude is 8.
Write the fraction as decimal 182/1000182/1000 written as decimal is ?
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Let's convert the following number into a decimal:
[tex]\text{ }\frac{182}{1000}[/tex]182 has 3 digits
1000 has 3 zeros
For this fraction with a denominator of 10, 100, 1000, 10000 and so on.
Converting its decimal form, we just have to count the number of zeros they have. Once we got the number of zeros, that's the number of places we move to put the decimal point in the numerator from right to left.
Let's now answer this to better understand the rule.
Since 1000 has 3 zeros, we move the decimal point 3 places from right to left of 182.
Therefore, the answer is 0.182
57 es 95% de que número
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60
1) Considerando que 57 es 95% de algun numero, vamos escribir esta ecuacion
0.95x = 57 Escribindo 95% como 0.95, dividir los dos lados por 0.95
x =57/0.95
x=60
2) Asi 57 es 95% de 60.
Find the value of x that makes A || B.AB5423142 3x10 and 23 = x + 30X=[? ]
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∠2 and ∠3 are alternate interior angles. In order to A II B, the alternate interior angles must be equal.
Then,
[tex]\begin{gathered} \angle2=\operatorname{\angle}3 \\ 3x-10=x+30 \end{gathered}[/tex]To find x, subtract x from both sides of the equation:
[tex]\begin{gathered} 3x-10-x=x+30-x \\ 3x-x-10=x-x+30 \\ 2x-10=30 \end{gathered}[/tex]Now, add 10 to both sides of the equation:
[tex]\begin{gathered} 2x-10+10=30+10 \\ 2x=40 \end{gathered}[/tex]Finally, divide both sides by 2:
[tex]\begin{gathered} \frac{2x}{2}=\frac{40}{2} \\ x=20 \end{gathered}[/tex]Answer: x = 20.
2.Solve the inequality for x. Show each step to the solution.12x > 3(5x – 2) – 15
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we have the inequality
12x > 3(5x-2)-15
step 1
apply distributive property right side
12x > 15x-6-15
combine like terms right side
12x > 15x-21
step 2
Adds both sides 21
12x+21 > 15x-21+21
simplify
12x+21 > 15x
step 3
subtract 12x both sides
12x+21-12x > 15x-12x
21 > 3x
step 4
Divide by 3 both sides
21/3 > 3x/3
7 > x
Rewrite
x < 7Exactly 25% of the marbles in a bag are black. If there are 8 marbles in the bag, how many are black?
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Let the total number of marbles in the bag be 'x'.
Given that exactly 25% of the total marbles are black,
[tex]\begin{gathered} \text{ No. of black marbles}=25\text{ percent of total marbles} \\ \text{ No. of black marbles}=25\text{ percent of x} \\ \text{ No. of black marbles}=\frac{25}{100}\cdot x \\ \text{ No. of black marbles}=0.25x \end{gathered}[/tex]Also, given that there are total 8 marbles in the bag,
[tex]x=8[/tex]Then the number of black marbles will be obtained by substituting x=8,
[tex]\begin{gathered} \text{ No. of black marbles}=0.25(8) \\ \text{ No. of black marbles}=2 \end{gathered}[/tex]Thus, there are 2 black marbles in the bag.
Evaluate the function at the given x-value.5. f(x) = -4x + 5 ; f(3)
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Answer: f (3) = -7
so i have to factor x(3x+10)=77
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SOLUTION
Given the equation as seen below, we can use the following steps to get the factors
[tex]x(3x+10)=77[/tex]Step 1: Remove the bracket by multiplying the value outside the bracket with the one inside the bracket using the distributive law. We have:
[tex]\begin{gathered} x(3x)+x(10)=77 \\ 3x^2_{}+10x=77 \\ 3x^2+10x-77=0 \end{gathered}[/tex]Step 2: Now that we have a quadratic equation, we solve for x using the quadratic formula:
[tex]\begin{gathered} 3x^2+10x-77=0 \\ u\sin g\text{ the form }ax^2+bx+c=0 \\ a=3,b=10,c=-77 \\ x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ =\frac{-10\pm\sqrt[]{10^2-4(3)(-77)}}{2(3)} \\ =\frac{-10\pm\sqrt[]{100+924}}{6} \\ =\frac{-10\pm\sqrt[]{1024}}{6} \\ =\frac{-10+32}{6}\text{ or }\frac{-10-32}{6} \\ \frac{22}{6}\text{ or -}\frac{42}{6} \\ =\frac{11}{3}\text{ or }-7 \end{gathered}[/tex]Hence, it can be seen from above that the factors will be -7 or 11/3.
find the absolute extrema for the function on the given inveral
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In order to find the minimum and maximum value in the given interval, first let's find the vertex coordinates:
[tex]\begin{gathered} f(x)=3x^2-24x \\ a=3,b=-24,c=0 \\ \\ x_v=\frac{-b}{2a}=\frac{24}{6}=4 \\ y_v=3\cdot4^2-24\cdot4=3\cdot16-96=-48 \end{gathered}[/tex]Since the coefficient a is positive, so the y-coordinate of the vertex is a minimum point, therefore the absolute minimum is (4,-48).
Then, to find the maximum, we need the x-coordinate that is further away from the vertex.
Since 0 is further away from 4 than 7, let's use x = 0:
[tex]f(0)=3\cdot0-24\cdot0=0[/tex]Therefore the absolute maximum is (0,0).
TRIGONOMETRY if 0 is in the first quadrant and cos 0=3/5 what is sin (1/20)?Where 0 is theta
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Given:
[tex]\cos \text{ }\theta\text{ = }\frac{3}{5}[/tex]Using the trigonometric identity:
[tex]undefined[/tex]Determine if the sequence below is arithmetic or geometric and determine thecommon difference / ratio in simplest form.4,2,1,...
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An arithmetic progression is a progression where the next term is found by multiplying the previous by a constant number called the common ratio, for the given progression:
[tex]4,2,1[/tex]If we use 1/2 as a common ratio we get:
[tex]\begin{gathered} 2=\frac{4}{2} \\ 1=\frac{2}{2} \end{gathered}[/tex]Therefore this is an arithmetic progression and its common ratio is 1/2
The value of a baseball players rookkie card began to increase once the player retired.When he retired in 1995 hid card was worth 9.43.The value has increased by 1.38 each year since then.Yall I really need help I dont get this at all
Answers
Given that,
The value of card starts increasing after 1995. In this question, we have to find the value of card at present (2020).
Initial worth = I = 9.43
Final worth = F = ?
Total years = 2020 - 1995 = 25 years
Increasing rate = r = 1.38
The final worth of a card after 'n' years is calculated as:
F = I * r^n
F = 9.43 * (1.38)^25
F = 9.43 * 3140.34
F = 29613.43
Hence, the value of the card in 2020 would be 29613.43.
Can someone please help me find the valu of X?
Answers
Answer:
x = 10
Explanation:
Because the transverse lines are parralell, the following must be true
[tex]\frac{x+8}{x+2}=\frac{3}{2}[/tex]cross multipication gives
[tex]\begin{gathered} 2(x+8)=3(x+2) \\ \end{gathered}[/tex]which simplifies to give
[tex]\begin{gathered} 2x+16=3x+6 \\ \end{gathered}[/tex]subtracting 2x from both sides gives
[tex]16=x+6[/tex]subtracting 6 from both sides gives
[tex]10=x[/tex]Hence the value of x is 10.
Suppose you are measuring a moving box to see if it has enough room in it. The moving box is a cube, and the length of one side is 2 ft. long. What is the volume of the box?
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Solution:
Suppose you are measuring a moving box to see if it has enough room in it.
The moving box is a cube.
Given that the length of one side is 2 ft. long, i.e.
[tex]l=2\text{ ft}[/tex]To find the volume of a cube, the formula is
[tex]V=l^3[/tex]Substitute for l into the formula above
[tex]\begin{gathered} V=l^3=2^3=8\text{ ft}^3 \\ V=8\text{ ft}^3 \end{gathered}[/tex]Hence, the volume of the box is 8 ft³
Determine a series of transformations that would map Figure 1 onto Figure J. y 11 Figure J NOW ona 00 05 15 1 -12-11-10-9-8-7-6 -5 -4 -3 -2 -1 1 2 3 4 5 6 7 8 9 10 11 12 -2 Šť b bo v och t co is with Figure I -11 -12 A followed by a o
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EXPLANATION
The transformations that would map Figure 1 onto Figure J are:
A rotation followed by a translation
I’m sorry to keep bothering you guys but you’re the third person that I’m trying the last two their answers went partway and then stopped I just need to see how this is worked out
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In this case, we'll have to carry out several steps to find the solution.
Step 01
Data:
Graph:
Height of Golf Ball
Step 02:
functions:
We must analyze the graph to find the solution.
Function:
Non linear
intercepts:
x-intercepts: 0 and 120
y-intercept: 0
symmetry:
x = 50
positive:
domain: (0 , 120)
negative:
there are no negative values
increasing:
interval on x: (0 , 50)
decreasing:
interval on x: (50, 120)
That is the full solution.
The area of A triangle with base b and height h is given by A 1/2bh. Find the area when b=24 m and h=30
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Given:
Base of a triangle 24m and height = 30m
Required:
Find the area of a triangle.
Explanation:
We have formula for area of triangle
[tex]A=\frac{1}{2}\times b(base)\times height(h)[/tex]Now,
[tex]\begin{gathered} A=\frac{1}{2}\times24\times30 \\ A=360m^2 \end{gathered}[/tex]Answer:
The area of triangle is 360 meter square.
An architect designs a rectangular flower garden such that the width is exactly two-thirds of the length. If 240 feet of antique picket fencing are to be used to enclose the garden, find the dimensions of the garden. What is the length of the garden? The length of the garden is What is the width of the garden? The width of the garden is
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STEP 1:
We'll derive an expression for the width and the length
[tex]\begin{gathered} w=\frac{2l}{3}\text{ where} \\ w\text{ = width} \\ l=\text{ length} \end{gathered}[/tex]STEP 2:
Next, We then derive an expression for the perimeter substituting w as a function of l
[tex]\begin{gathered} \text{Perimeter = 2(l+w)} \\ 240=2(l+\frac{2l}{3}) \end{gathered}[/tex]STEP 3:
Solve for l and subsequently w
[tex]\begin{gathered} \text{Perimeter}=\text{ 240 = 2(}\frac{2l+3l}{3})=2(\frac{5l}{3}) \\ 240=\frac{10l}{3} \\ \text{Cross multiplying gives 240}\times3=5l \\ l=\frac{240\times3}{10}=72ft \\ w=\frac{2l}{3}=\frac{2\times72}{3}=48ft \end{gathered}[/tex]Therefore, length = 72 ft and width = 48ft
4.A pet store sells cats for $50 and dogs for $100. If one day it sells a total of 4pets and makes $300, find out how many cats and dogs it sold by writing asystem of equations and graphing to solve it.Representations:Equations:
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Answer:
2 cats and 2 dogs
Explanation:
Representations:
x = number of cats sold
y = number of dogs sold
Equations:
We know that it sells a total of 4 pets, so the sum of the number of cats and dogs is 4. So:
x + y = 4
On the other hand, they make $300, and they make $50 for each cat and $100 for each dog, so:
$50x + $100y = $300
So, the system of equation is:
x + y = 4
50x + 100y = 300
Graph:
Now, we need to graph the equations, so we need to identify two points for each equation:
For x + y = 4
If x = 0, then:
0 + y = 4
y = 4
If x = 4, then:
4 + y = 4
4 + y - 4 = 4 - 4
y = 0
For 50x + 100y = 300
If x = 0, then:
50(0) + 100y = 300
100y = 300
100y/100 = 300/100
y = 3
If x = 4, then:
50(4) + 100y = 300
200 + 100y = 300
200 + 100y - 200 = 300 - 200
100y = 100
100y/100 = 100/100
y = 1
Therefore, we have the points (0, 4) and (4, 0) to graph the line of the first equation and the points (0, 3) and (4, 1) to graph the line of the second equation.
So, the graph of the system is:
Therefore, the solution is the intersection point (2, 2), so they sold 2 cats and 2 dogs that day.
6 points 3 The coordinates of the vertices of the triangle shown are P (2,13), Q (7,1), and R (2, 1). 14 13 12 11 10 9 8 6 5 3 2. 1 R Q 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 2 7 8 What is the length of segment PQ in units?
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We have the following:
[tex]d=\sqrt[]{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]replacing:
P (2,13) = (x1,y1)
Q(7,1) = (x2,y2)
[tex]\begin{gathered} d=\sqrt[]{(7-2)^2+(1-13)^2} \\ d=\sqrt[]{5^2+12^2} \\ d=\sqrt[]{25+144} \\ d=\sqrt[]{169} \\ d=13 \end{gathered}[/tex]The answer is 13 units
The perimeter, P, of a rectangle is the sum of twice the length and twice the width. P= 21+ 2w units P= 2([+w) units P= 2(x+3) units P= 2(5)-2(9) units P= 4 x units
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We can see the problem states that P = 2(x+3) and also states that P=4x
Those equations lead to the expression
2(x+3)=4x
Operating
2x+6=4x
Subtracting 2x
6 = 2x
Solving for x
x = 6/2 = 3
Thus, the perimeter is
P = 2(3+3) = 12 units
These trianglesare congruent bythe trianglecongruencepostulate [?].A. AASB. ASAC. Neither, they are not congruent
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At the point of intersection, the angles are equal because they are vertically opposite. This means that in both triangle, there are two congruent angles and a congruent sides. Recall,
if any two angles and the side included between the angles of one triangle are equivalent to the corresponding two angles and side included between the angles of the second triangle, then the two triangles are congruent by the (Angle side angle) ASA rule
Since the given triangles obey this rule, then the correct option is B
6x - 5y = - 4Direct variationХ5?k=Not direct variation2y = 14xDirect variationk ==Not direct variation
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Direct Variation is of the form (directly proportional);
[tex]\begin{gathered} y\propto x \\ y=kx \end{gathered}[/tex]While indirect variation (inversely proportional) is of the form;
[tex]\begin{gathered} y\propto\frac{1}{x} \\ y=\frac{k}{x} \end{gathered}[/tex]So, for each of the given question we want to determine if they are direct or indirect variation;
1.
[tex]\begin{gathered} 6x-5y=-4 \\ -5y=-4-6x \\ -5y=-6x-4 \\ y=\frac{-6x-4}{-5} \\ y=\frac{6}{5}x+\frac{4}{5} \end{gathered}[/tex]Therefore, this is a Direct variation with proportionality constant;
[tex]k=\frac{6}{5}[/tex]2.
[tex]\begin{gathered} 2y=14x \\ y=\frac{14x}{2} \\ y=7x \end{gathered}[/tex]Therefore, this is a Direct variation with proportionality constant;
[tex]k=7[/tex]Evaluate the following expression.12!
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Explanation
Factorial, in mathematics, the product of all positive integers less than or equal to a given positive integer and denoted by that integer and an exclamation point.
[tex]a![/tex]so, to evaluate the expression we need to apply the definition
hence
[tex]\begin{gathered} 12\text{ ! = 12}\cdot11\cdot10\cdot9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1 \\ 12\text{ ! =}479001600 \end{gathered}[/tex]I hope this helps you
what is the difference between solving literal equations(with only variables)and solving multistep equations(woth numbers and a variables)
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To solve a literal equation means to express one variable with respect to the other variables in the equation. The most important part of a literal equation is to isolate or keep by itself a certain variable on one side of the variable (either left or right) and the rest on the other side
Solving multistep equations takes more time and more operations compared to solving a literal equation.
if you receive a 175.84 cents on 314 invested at a rate of 7% for how long did yo invest the principle
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Answer:
The number of years you should invest the principal is;
[tex]8\text{ years}[/tex]Explanation:
Given;
[tex]\begin{gathered} \text{Interest i = \$175.84} \\ \text{ Principal P = \$}314 \\ \text{Rate r = 7\% =0.07} \end{gathered}[/tex]Recall that the formula for simple interest is;
[tex]\begin{gathered} i=P\times r\times t \\ t=\frac{i}{Pr} \\ \text{where;} \\ t=\text{time of investment} \end{gathered}[/tex]substituting the given values;
[tex]\begin{gathered} t=\frac{i}{Pr} \\ t=\frac{175.84}{314\times0.07} \\ t=\frac{175.84}{21.98} \\ t=8 \end{gathered}[/tex]Therefore, the number of years you should invest the principal is;
[tex]8\text{ years}[/tex]We can also solve as;
[tex]\begin{gathered} i=P\times r\times t \\ 175.84=314\times0.07\times t \\ 175.84=21.98t \end{gathered}[/tex]then we can divide both sides by 21.98;
[tex]\begin{gathered} \frac{175.84}{21.97}=\frac{21.98t}{21.98} \\ 8=t \\ t=8\text{ years} \end{gathered}[/tex]Compute.
\[ \left(\dfrac 8 3\right)^{-2} \cdot \left(\dfrac 3 4\right)^{-3}\]
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[tex]\left(\cfrac{8}{3}\right)^{-2} \left(\cfrac{3}{4}\right)^{-3}\implies \left(\cfrac{3}{8}\right)^{+2} \left(\cfrac{4}{3}\right)^{+3}\implies \cfrac{3^2}{8^2}\cdot \cfrac{4^3}{3^3}\implies \cfrac{4^3}{8^2}\cdot \cfrac{3^2}{3^3} \\\\\\ \cfrac{64}{64}\cdot \cfrac{1}{3}\implies 1\cdot \cfrac{1}{3}\implies \cfrac{1}{3}[/tex]
The following table gives the frequency distribution of the ages of a random sample of 104 Iris student
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Given:
The frequency values are given for class interval of N = 104 IRSC students.
The objective is to find cumulative frequency, cumuative relative frequency and cumulative percentage.
Cumulative frequency is addition of the previous frequency values.
So, the cumulative freqency values can be cclculated as,
The formula to find the cumulative relative frequency is,
[tex]\text{CRF}=\frac{CF}{N}[/tex]Now, the cumulative relative frequency can be calculated as,
Now, the formula to find the Cumulative percentage is,
[tex]\text{Cumulative \% = CRF }\times100[/tex]Then, the table values for Cumulative percentage will be,
Hence, the required cumulative frequency, cumuative relative frequency and cumulative percentage values are obtained.