Answers
First you have to find the missing angle in the triangle
The angles in a triangle add to 180°
180 - 97 - 30 = 53
The angle we just found and angle x form a straight line meaning they add to 180°
180 - 53 = 127
The answer is 127°
Hope this helps!
X is part of a straight angle, which should add up to 180°.
So first we need to figure out what the other side of this straight angle is. Since the other side is inside the triangle, it should be easy.
All angles in any triangle always add up to exactly 180°. To figure out what the last angle is in the triangle, we just need to add the two existing angle measurements together and subtract the sum from 180.
97 + 30 = 127
180 - 127 = 53
So the last angle in the triangle is 53°.
Now, we can figure out the value of x. Since x is part of a 180° straight angle, and we know that half of said angle is 53°, we can find x by subtracting 53 from 180.
180 - 53 = 127
x = 127
Hope this helps :)
Related Questions
Create a function to model the height of a firework when shot in the air. Explain whether the function will have a maximum or a minimum value.
Answers
0 is the initial height, since you cannot hold one in your hand and safely set it off.
This will have a maximum; the firework will reach a highest point and fall back down. This is also evident from the equation; the negative in front of t² flips the parabola upside down, which gives it a maximum.
Find the determinant of G
Answers
where the letters are: a and b is the top row and c and d is the bottom row
so ad = 12*2 and cb = -6*0
12x2 = 24
-6x0 =0
24-0 = 24
the determinant = 24
can you please help me on this worksheet and try to show the work
Answers
180° = π radians
13. (105°)*(π radians)/180° = 7π/12 radians
14. 5π/6 radians * 180°/(π radians) = 150°
15. (The geometry shown cannot actually exist. The total of subtended arcs on any linked sprockets must be 360°.)
a.
The arc length on the smaller sprocket is (12 in)*(160/360) = 5 1/3 in.
The arc length on the larger sprocket is (20 in)*(185/360) = 10 5/18 in.
The total chain length is 5 6/18 +10 + 10 5/18 + 10 = 35 11/18 in. Rounded to the nearest tenth inch, the chain length is 10.3 inches.
b.
10.3/0.5 ≈ 20.6 ≈ 21
The chain has about 21 links. (There are no "teeth" on a chain.)
1. The radius can be found by looking at the distance between the origin and the line 8x+6y-48=0. That distance is |-48|/√(8²+6²) = 4.8.
Since the radius of the circle is 4.8, so its circumference is ...
9.6π cm, about 30.159 cm.
2. s = rθ
s = (12 cm)(π/3 radians) = 4π cm
The length of arc AB is 4π cm, about 12.566 cm.
Which of these is an example of continuous random variable
A. Number of flights leaving an airport
B. Distance of a javelin throw
C. Pieces of mail in your mailbox
D. Attendance at a sporting event
Answers
A continuous random variable is a variable that can be any number between the integers not just the integers themselves.
For example, the distance for a javelin throw could be 6.70819203 feet.
However, you would not say that you have 6.70819203 flights, pieces of mail or people at a sporting event.
The B. distance of a javelin throw is an example of a continuous random variable.
Explanation:A continuous random variable is a variable that can take any value within a specified range, often associated with measurements and quantities. Unlike discrete variables, it can assume an infinite number of values, typically represented by intervals on the real number line.
Out of the given options, the distance of a javelin throw is an example of a continuous random variable. A continuous random variable is one that can take on any value within a given range. In the case of a javelin throw, the distance can be any real number value within a certain range, such as 0 to infinity. This is in contrast to the other options, which are discrete random variables that can only take on specific whole number values.
A man has 25 25 coins in his pocket, all of which are dimes and quarters. if the total value of his change is $ 4.45 $4.45, how many dimes and how many quarters does he have?
Answers
$1.95/$0.15 = 13
quarters.
The man has 12 dimes and 13 quarters.
_____
Check:
12*$0.10 +13*$0.25 = $1.20 +3.25 = $4.45 . . . (answer checks OK)
does anyone know the surface area for this problem. if so i will mark brainliest.
Answers
From the given volume, we know that the ratio is:
[tex]\text {Ratio of yellow eraser to white eraser = } \sqrt[3]{ \dfrac{16}{2} } [/tex]
From the given area, we know that the ratio is:
[tex]\text {Ratio of yellow eraser to white eraser = } \sqrt{ \dfrac{52}{x} } [/tex]
Equate the two ratio and solve for x:
[tex]\sqrt[3]{ \dfrac{16}{2} } = \sqrt{ \dfrac{52}{x} } [/tex]
Cube both sides:
[tex]\dfrac{16}{2} = \bigg(\sqrt{ \dfrac{52}{x} }\bigg)^3[/tex]
Square both sides:
[tex]\bigg(\dfrac{16}{2} \bigg)^2 = \bigg( \dfrac{52}{x} }\bigg)^3[/tex]
Simplify each term:
[tex]\dfrac{16^2}{2^2} = \dfrac{52^3}{x^3} [/tex]
Cross multiply:
[tex]256x^3 = 562432[/tex]
Divide both sides by 256:
[tex] x^3 = 2197[/tex]
Cube root both sides:
[tex] x = 13[/tex]
A ball is thrown upward from the top of a building. The function below shows the height of the ball in relation to sea level, f(t), in feet, at different times, t, in seconds:f(t) = −16t2 + 48t + 100The average rate of change of f(t) from t = 3 seconds to t = 5 seconds is _____feet per second.
Answers
f (t) = -16t2 + 48t + 100
The average rate of change is:
(f (t2) - f (t1)) / (t2 - t1)
We have then:
For t1 = 3:
f (3) = -16 * (3) ^ 2 + 48 * (3) + 100
f (3) = 100
For t2 = 5:
f (5) = -16 * (5) ^ 2 + 48 * (5) + 100
f (5) = -60
Substituting values:
AVR = (-60 - 100) / (5 - 3)
AVR = -80
Answer:
The average rate of change of f (t) from t = 3 seconds to t = 5 seconds is -80 feet per second.
What is the best estimate for the sum of 3/8 and 1/12?
Answers
describe the steps you would use to solve the following inequality
Answers
[tex] \frac{x + 1}{2x - 3} > 2 \\ x + 1 > 2(2x - 3) \\ x + 1 > 4x - 6 \\ - 3x + 1 > - 6 \\ - 3x > - 7x \\ x < \frac{7}{3} [/tex]
Answer:
Rewrite the inequality so there is a single rational expression on one side, and 0 on the other.
Combine under a common denominator.
Test points in the critical regions.
Construct the solution.
Step-by-step explanation:
Answer Above
Solve system by elimination
y=x^2
y=x+2
SHOW YOUR WORK
Answers
y=x²----> equation 1
y=x+2-----> equation 2
multiply equation 1 by -1
-y=-x²
add equation 1 and equation 2
-y=-x²
y=x+2
------------
0=-x²+x+2-------------> -x²+x+2=0-----> x²-x-2=0
Group terms that contain the same variable, and move the constant to the opposite side of the equation
(x²-x)=2Complete the square. Remember to balance the equation by adding the same constants to each side
(x²-x+0.5²)=2+0.5²
Rewrite as perfect squares
(x-0.5)²=2+0.5²(x-0.5)²=2.25-----> (x-0.5)=(+/-)√2.25-----> (x-0.5)=(+/-)1.5
x1=1.5+0.5-----> x1=2
x2=-1.5+0.5---- > x2=-1
for x=2
y=x²----> y=2²----> y=4
the point is (2,4)
for x=-1
y=x²----> y=(-1)²---> y=1
the point is (-1,1)
the answer is
the solution of the system are the points
(2,4) and (-1,1)
PLEASE HELP! - Caitlyn recorded the height of each plant after she exposed each plant to a set amount of darkness daily. The scatterplot shows her results after 2 weeks of exposing each plant to the amount of darkness.
(Graph Below)
Which statement about the scatterplot is true?
A. The point (17, 12) could cause the description of the data set to be overstated.
B. The point (17, 12) could cause the description of the data set to be understated.
C. The point (17, 12) shows that there is no relationship between the number of hours of darkness and the height of the plant.
D. Although (17, 12) is an extreme value, it should be part of the description of the relationship between number of hours of darkness and the height of the plant.
Answers
Answer:
I think it's b, please correct me if I'm wrong.
Step-by-step explanation:
I'm taking the test rn so I hope its right.
Answer:
its b i just did it
Step-by-step explanation:
The shorter leg of a right triangle is 7 inches shorter than the longer leg. The hypotenuse is 17 inches longer than the longer leg. Find the side lengths of the triangle.
Answers
Or round them all up.
a² + b² = c²
c² - a² = b²
17² - 7² = x² ↓
289 - 49 = √240
√240 = 15.49193338482967 (rounded 15.49 or 15.5)
Hope this helps,
♥A.W.E.S.W.A.N.♥
look at the image then answer the question that goes with it
Answers
[tex]\sin(60) = \dfrac{x}{8} [/tex]
[tex]x = 8\sin(60) [/tex]
[tex]x = \dfrac{8 \sqrt{3} }{2} [/tex]
[tex]x = 4 \sqrt{3} [/tex]
x = 4sqrt3
PLEASE HELP ME
If the maximum density of killer whales per cubic foot is 0.000011142, what is the maximum number of killer whales allowed in the main show tank at any given time? You must explain your answer using words, and you must show all work and calculations
Answers
Given that the the tank has a radius of 70 ft;
The volume of the sphere will be:
V=1/4[4/3*pi*r^3]
v=1/4[4/3*3.14*70^3]
V=359,006.6675
b] The maximum number of whales will be given by:
density=mass/volume
thus
0.000011142=x/(359006.6675)
thus
x=0.000011142*359006.6675
x=4
thus the maximum number of whale would be 4
Rachel enjoys exercising outdoors. Today she walked5 2/3 miles in2 2/3 hours. What is Rachel’s unit walking rate in miles per hour and in hours per mile
Answers
the answer to your question is
Unit walking rate in miles per hour = 5 2/3 / 2 2/3 = 17/3 / 8/3 = 17/3 x 3/8 = 17/8 = 2 1/8 miles per hour.
Unit walking rate in hours per mile = 1/ 17/8 = 8/17 hours per mile
Answer:
Rachel enjoys exercising outdoors. Today she walked 5 2/3 miles or 5.67 miles in 2 2/3 hours or 2.67 hours.
Rachel’s unit walking rate in miles per hour is = [tex]\frac{5.67}{2.67}[/tex] = 2.125 miles per hour or 2 1/8 miles per hour.
Rachel’s unit walking rate in hours per mile = [tex]\frac{2.67}{5.67}[/tex] = 0.47 hours per mile.
Write the inverse function for the function, ƒ(x) = x + 4. Then, find the value of ƒ -1(4). Type your answers in the box.
ƒ -1(x) = _________ x ___________ ______
ƒ -1(4) = _____________
Answers
Answer: The required values are
[tex]f^{-1}(x)=x-4~~~~~~~~\textup{and}~~~~~~~~~f^{-1}(4)=0.[/tex]
Step-by-step explanation: We are given the following function f(x) :
[tex]f(x)=x+4~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)[/tex]
We are to find the values of [tex]f^{-1}(x)[/tex] and [tex]f^{-1}(4).[/tex]
Let us consider that
[tex]y=f(x)~~~~~~~~\Rightarrow x=f^{-1}(y).[/tex]
So, from equation (i), we get
[tex]f(x)=x+4\\\\\Rightarrow y=f^{-1}(y)+4\\\\\Rightarrow f^{-1}(y)=y-4\\\\\Rightarrow f^{-1}(x)=x-4.[/tex]
Substituting x = 4 in the above equation, we get
[tex]f^{-1}(4)=4-4=0\\\\\Rightarrow f^{-1}(4)=0.[/tex]
Thus, the required values are
[tex]f^{-1}(x)=x-4~~~~~~~~\textup{and}~~~~~~~~~f^{-1}(4)=0.[/tex]
Write an algebraic expression for the verbal description. the distance a car travels in t hours at a rate of 50 miles per hour
Answers
d=50t
Where; d is the distance covered in miles
t is the time taken in hours
The algebraic expression is D = 50t where D is the distance covered in miles and t is the time taken in hours.
What is the distance?Distance is defined as the product of speed and time.
What are Arithmetic operations?Arithmetic operations can also be specified by adding, subtracting, dividing, and multiplying built-in functions.
The operator that performs the arithmetic operations is called arithmetic operator.
* Multiplication operation: Multiplies values on either side of the operator
For example 4*2 = 8
/ Division operation: Divides left-hand operand by right-hand operand
For example 4/2 = 2
Given that the distance a car travels in t hours at a rate of 50 miles per hour
Let the distance covered in miles = D
And t is the time taken in hours
Since distance is defined as the multiplication of speed and time.
So an algebraic expression for the verbal description as
⇒ D = 50t
Hence, the algebraic expression is D = 50t where D is the distance covered in miles and t is the time taken in hours.
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Find the area of the shaded region.
Answers
Area of bigger rectangle
A = (x+1)(x+1) = x^2 + 2x + 1
2)
Area of smaller rectangle
A = 5(x - 7) = 5x - 35
3)
The difference:
(x^2 + 2x + 1) - (5x - 35)
= x^2 + 2x + 1 - 5x +35
= x^2 - 3x + 36
area of the shaded region = x^2 - 3x + 36
You first have to find the area of the bigger rectangle
You find this by multiplying the sides
(x + 1)(x + 1) = x^2 + 2x + 1
Next you have to find the area of the smaller rectangle
(x - 7) * 5 = 5x - 35
Subtract the two polynomials
(x^2 + 2x + 1) - (5x - 35) = x^2 - 3x + 36
The answer is [tex] x^{2} -3x + 36[/tex]
Hope this helps!
Rewrite with only sin x and cos x. Sin2x-cos2x
Answers
So sin2x-cos2x = 2sinxcosx - ([tex]cos^{2} x - sin^{2} x[/tex]) in terms of sinx and cosx
Simplify. x + 4.7 = −9.2 A. −13.9 B. 13.9 C. −12.9 D. 12.9
Answers
Isolate the x, subtract 4.7 from both sides
x + 4.7 = - 9.2
x + 4.7 (-4.7) = - 9.2 (-4.7)
x = -9.2 - 4.7
x = -13.9
-13.9, or A, is your answer
hope this helps
Use the given graph to determine the limit, if it exists.
Find limit as x approaches two from the left of f of x. and limit as x approaches two from the right of f of x.
Answers
As the value of x approaches 2 from the positive side i.e. the right hand side, the value of function approaches -4
Since the Right hand limit and the Left hand limit as x approaches 2 are not the same, the limit does not exist at x = 2
find the greatest common factor 9x^2a + 9xa^2
Answers
9x^2a+9xa^2
factor out 9 we get:
9(x^2a+xa^2)
=factoring out x we get
9x(xa+a^2)
factoring out a we get
9ax(x+a)
therefore the GCF is 9ax
if (x+1) is a factor of x^3+2x^2-x-2, find the remaining factors.
Answers
Answer:
The other factors are (x-1) and (x+2).
SOLVE FOR X one question 20 points
Answers
a^2 + b^2 = c^2 (c^2 is the hypertenuse)
The equation will look like this:
14^2 + 14^2 = x^2
196 + 196 = x^2
x^2 = 392
So the answer is the square root of 392:
14[tex] \sqrt{2} [/tex]
If you want a decimal form:
14 * 1.414 = 19.796
Answer:
ANSWER IS
Step-by-step explanation:
a boy flies a kite with a 100 foot long string. the angle of elevation of the string is 48 degrees. how high is the kite from the ground?
Answers
in a right triangle
sin ∅=opposite side angle ∅/hypotenuse
opposite side angle ∅=hypotenuse*sin ∅
in this problem
angle ∅=48°
hypotenuse=100 ft
opposite side angle ∅=?----> height of the kite from the ground
opposite side angle ∅=100*sin 48°------> 74.31 ft
the answer is
74.31 ft
Find the values of b such that the function has the given maximum value. f(x) = −x2 + bx − 14; Maximum value: 86
Answers
[tex]y_{max}\ \text{is in the vertex of the parabola}.[/tex]
[tex]\text{The coordinates of the vertex of the parabola}\ f(x)=ax^2+bx+c:\\\\V\left(\dfrac{-b}{2a};\ \dfrac{-(b^2-4ac)}{4a}\right)[/tex]
[tex]\text{We have:}\\\\f(x)=-x^2+bx-14\to a=-1;\ b=b;\ c=-14[/tex]
[tex]\text{Substitute:}\\\\\dfrac{-\left(b^2-4\cdot(-1)\cdot(-14)\right)}{4\cdot(-1)}=86\\\\\dfrac{b^2-56}{4}=86\ \ \ \ |\cdot4\\\\b^2-56=344\ \ \ \ |+56\\\\b^2=400\to b=\pm\sqrt{400}\\\\b=-20\ \vee\ b=20[/tex]
Answer: b = -20 or b = 20.
To find the value of 'b' that gives the maximum value of 86 for the function f(x) = -x² + bx - 14, substitute the x-coordinate of the vertex (-b/2a) into the function, equate it with 86 and solve for 'b'.
Explanation:To find the values of b for which the quadratic function f(x) = -x² + bx - 14; Maximum value: 86 holds true, we need to utilize the properties of quadratic functions. In a quadratic function in the form f(x) = ax² + bx + c, the maximum value occurs at the vertex of the parabola. The x-coordinate of the vertex is given by -b/2a.
Given that our quadratic is a maximum (opens downwards since a=-1), the y-coordinate of the vertex, which is our maximum value, is 86. Substituting these into our function, we get f(-b/2a) = 86, replacing a=-1, this reduces to -b²/4 -14 = 86. Simplifying this equation will give you the value of b.
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i really need help. please help
Part A: Explain why the x-coordinates of the points where the graphs of the equations y = 4^−x and y = 8^(−x−1) intersect are the solutions of the equation 4^−x = 8^(−x−1). (4 points)
Part B: Make tables to find the solution to 4^−x = 8^(−x−1). Take the integer values of x between −3 and 3. (4 points)
Part C: How can you solve the equation 4^−x = 8^(−x−1) graphically? (2 points)
Answers
The answer to B is forming a table. This is quite easier than you think :D.
Start with x on one side (3) and go all the way down to -3 like this:
(x) 4^-x 8^-x-1
-3 64 (Do the same with this)
-2 16
-1 4
0 1
1 1/4
2 .0625
3 .015625
So basically you just fill in the x variable with every number between 3 and -3. After that, it's just a bit of math and you've done it!
C is a bit misguided. I think it would be quite simple to just use a graphing calculator, but you would have to actually input all the x and y values from the table. After that, you will have a graph.
I hope I helped! Good luck and good day!
If the areas of two rhombi are equal, are the perimeters sometimes, always or never equal. explain your answer – you can use examples with actual numbers to do so.equalif the areas of two rhombi are equal
Answers
The area of a rhombus = length of base * altitude. To keep the area the same we would have to keep the same base and move the opposite side so its parallel to the base. But doing this will lengthen the other 2 sides of the figure so its no longer a rhombus.
The perimeters of two rhombi with equal areas are sometimes equal, but not always.
Let's consider the formula for the area of a rhombus, which is given by [tex]\( A = \frac{1}{2} \times d_1 \times d_2 \), where \( d_1 \) and \( d_2 \)[/tex] are the lengths of the diagonals. For the perimeter P, we have [tex]\( P = 4 \times s \)[/tex], where s is the length of one side of the rhombus.
For two rhombi to have equal areas, the product of their diagonals must be the same. That is, if we have two rhombi with diagonals [tex]\( (d_1, d_2) \) and \( (d_1', d_2') \), then \( d_1 \times d_2 = d_1' \times d_2' \)[/tex].
However, the perimeter depends only on the length of one side, s, and not on the diagonals. Therefore, two rhombi with the same area can have different side lengths and thus different perimeters.
Let's consider an example:
Rhombus 1:
[tex]- Diagonals \( d_1 = 8 \) units and \( d_2 = 4 \) units[/tex]
[tex]- Area \( A = \frac{1}{2} \times 8 \times 4 = 16 \) square units[/tex]
[tex]- Side length \( s \), using the Pythagorean theorem (since the diagonals bisect each other at right angles), is \( s = \sqrt{4^2 + 2^2} = \sqrt{16 + 4} = \sqrt{20} \) units[/tex]
[tex]- Perimeter \( P = 4 \times s = 4 \times \sqrt{20} \) units[/tex]
Rhombus 2:
[tex]- Diagonals \( d_1' = 10 \) units and \( d_2' = 3.2 \) units[/tex]
[tex]- Area \( A' = \frac{1}{2} \times 10 \times 3.2 = 16 \) square units (equal to the area of Rhombus 1)[/tex]
[tex]- Side length \( s' \), using the Pythagorean theorem, is \( s' = \sqrt{1.6^2 + 2.56^2} = \sqrt{2.56 + 6.5536} = \sqrt{9.1136} \) units[/tex]
[tex]- Perimeter \( P' = 4 \times s' = 4 \times \sqrt{9.1136} \) units[/tex]
Comparing the perimeters:
[tex]- \( P = 4 \times \sqrt{20} \approx 4 \times 4.472 = 17.888 \) units[/tex]
[tex]- \( P' = 4 \times \sqrt{9.1136} \approx 4 \times 3.0188 = 12.0752 \) units[/tex]
Clearly, the perimeters are not equal, even though the areas are the same.
However, it is possible for two rhombi with equal areas to have equal perimeters if their side lengths are the same. For instance, if Rhombus 1 and Rhombus 2 both had side lengths of s = s' = 4 units, then their perimeters would both be P = P' = [tex]4 \times 4 = 16 \)[/tex] units, regardless of the lengths of their diagonals, as long as the diagonals satisfy the area condition [tex]\( d_1 \times d_2 = d_1' \times d_2' \)[/tex].
In conclusion, the perimeters of two rhombi with equal areas can sometimes be equal, specifically when the side lengths are the same, but they are not always equal, as demonstrated by the example above.
How many degrees is w?
Answers
w = 180 -38 - 49
w = 93
Answer
w = 93°
The length of a rectangle is 4 feet shorter than its width. the area of the rectangle is 42 square feet. find the length and width. round your answer to the nearest tenth of a foot
Answers
Length is [tex]\boldsymbol{4.4}[/tex] feets and width is equal to [tex]\boldsymbol{8.8}[/tex] feets.
Area of a rectangleThe area of a two-dimensional region, form, or planar lamina in the plane is the quantity that expresses its extent.
A quadrilateral having four right angles is known as a rectangle.
Let [tex]\boldsymbol{w}[/tex] feets denotes width of a rectangle.
Length of a rectangle [tex]=\boldsymbol{w-4}[/tex] feets
Area of a rectangle [tex]=\boldsymbol{42}[/tex] square feet
Length [tex]\times[/tex] Width [tex]=42[/tex]
[tex]w(w-4)=42[/tex]
[tex]w^2-4w-42=0[/tex]
[tex]w=\frac{4\pm \sqrt{16+168}}{2}[/tex]
[tex]=2\pm \sqrt{46}[/tex]
As dimension can not be negative, [tex]2-\sqrt{46}[/tex] is rejected.
So,
[tex]w=2+ \sqrt{46}[/tex]
[tex]=\boldsymbol{8.8}[/tex] feets
Length [tex]=8.8-4[/tex]
[tex]=\boldsymbol{4.4}[/tex] feets
So, length is [tex]4.4[/tex] feets and width is equal to [tex]8.8[/tex] feets.
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