The two triangular prisms shown are similar. The dimensions of the larger prism were multiplied by a scale factor of to create the smaller prism.
When the large prism was reduced, the surface area changed by a factor of
A.64/125
B.16/25
C.4/5
D.10/8
Answers
Answer: B. [tex]\frac{16}{25}[/tex]
Step-by-step explanation:
We know that if a figure or solid shape is dilated to create a new shape then the new shape is similar to the original shape.
Also, the scale factor of dilation is the ratio of the sides of new shape to the corresponding sides of the original shape .
Let k be the scale factor of dilation, then
Given : Original shape - larger prism
New shape - smaller prism
Surface area of prism with all sides equal (a)=[tex]4a^2[/tex]
Surface area of original prism =[tex]4(10)^2=400[/tex]
Surface area of new prism =[tex]4(8)^2=256[/tex]
[tex]\Rightarrow k=\frac{256}{400}=\frac{16}{25}[/tex]
Hence, the scale factor = [tex]\frac{16}{25}[/tex]
Related Questions
What is the correct evaluation of 15-x, when x is equal to -5?
Answers
A climber is standing at the top of mount kazbek, approximately 3.1 miles above sea level. the radius of the earth is 3959 miles. what is the climber's distance to the horizon? enter your answer as a decimal in the box. round only your final answer to the nearest tenth. mi
Answers
AC=3959+3.1=3962.1 mi
To evaluate for x we use the Pythagorean theorem, this is given by:
c^2=a^2+b^2
c=AC is the hypotenuse
a and b are the legs
plugging in the values we shall have:
3962.1^2=x^2+3959^2
x^2=3961.1^2-3959^2
x^2=24555.41
hence
x=156.7 mi
Answer: 156.7 mi
The nearest tenth, the climber's distance to the horizon is approximately 156.6 miles.
The climber's distance to the horizon can be calculated using the Pythagorean theorem, where the Earth's radius and the climber's height above sea level form a right-angled triangle.
[tex]\[ d^2 + R^2 = (R + h)^2 \][/tex]
Expanding the right side of the equation gives us:
[tex]\[ d^2 + R^2 = R^2 + 2Rh + h^2 \][/tex]
Subtracting [tex]\( R^2 \)[/tex] from both sides, we get:
[tex]\[ d^2 = 2Rh + h^2 \][/tex]
Since [tex]\( h^2 \)[/tex] is very small compared to [tex]\( 2Rh \)[/tex] (because [tex]\( h \)[/tex] is much smaller than [tex]\( R \))[/tex], we can neglect [tex]\( R^2 \)[/tex] in our calculation for a more practical approximation. This leaves us with:
[tex]\[ d^2 \approx 2Rh \][/tex]
Taking the square root of both sides to solve for [tex]\( d \)[/tex], we have:
[tex]\[ d \approx \sqrt{2Rh} \][/tex]
Now, plugging in the values for [tex]\( R \) and \( h \):[/tex]
[tex]\[ d \approx \sqrt{2 \times 3959 \times 3.1} \][/tex]
[tex]\[ d \approx \sqrt{2 \times 3959 \times 3.1} \][/tex]
[tex]\[ d \approx \sqrt{24516.8} \][/tex]
[tex]\[ d \approx 156.6 \text{ miles} \][/tex]
Rounding to the nearest tenth, the climber's distance to the horizon is approximately 156.6 miles.
On a digital clock, the colon (two dots between the hour and minutes) blinks every second. In 2 hours and 20 minutes, how many times will it blink?
Answers
120+20=140
140·60=8400
8400
See Picture
The colon will blink 8400 times in 2 hours and 20 minutes.
How many times will the colon blink?Since the colon blinks every second. And we know that there are 60 minutes in an hour and there are 60 seconds in a minute.
Thus, we can say:
1 minutes = 60 seconds
1 hour = 60 minutes = 3600 seconds
Note: 60 minute = (60 * 60) seconds = 3600 seconds
Thus, in 2 hours and 20 minutes, we have:
2 hours + 20 minutes = (2 * 3600) + (20 * 60)
= 7200 + 1200
= 8400 second
Therefore, the colon will blink 8400 times in 2 hours and 20 minutes.
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n a study of 250 adults, the mean heart rate was 70 beats per minute. Assume the population of heart rates is known to be approximately normal with a standard deviation of 12 beats per minute. What is the 99% confidence interval for the mean beats per minute?
Answers
The CI bounds are calculated as:
mean +/- z*SD/sqrt(n) = 70 +/- 2.58*12/sqrt(250) = 68.04 to 71.96 beats per minute
. (06.02)
The table below shows data for a class's mid-term and final exams:
Mid-Term Final
96 100
95 85
92 85
90 83
87 83
86 82
82 81
81 78
80 78
78 78
73 75
Which data set has the smallest IQR? (1 point)
They have the same IQR
Mid-term exams
Final exams
There is not enough information
2. (06.02)
The box plots below show student grades on the most recent exam compared to overall grades in the class:
two box plots shown. The top one is labeled Class. Minimum at 74, Q1 at 78, median at 85, Q3 at 93, maximum at 98. The bottom b
Which of the following best describes the information about the medians? (1 point)
The exam median is only 1–2 points higher than the class median.
The exam median is much higher than the class median.
The additional scores in the second quartile for the exam data make the median higher.
The narrower range for the exam data causes the median to be higher.
3. (06.02)
The box plots below show attendance at a local movie theater and high school basketball games:
two box plots shown. The top one is labeled Movies. Minimum at 60, Q1 at 65, median at 95, Q3 at 125, maximum at 150. The botto
Which of the following best describes how to measure the spread of the data? (1 point)
The IQR is a better measure of spread for movies than it is for basketball games.
The standard deviation is a better measure of spread for movies than it is for basketball games.
The IQR is the best measurement of spread for games and movies.
The standard deviation is the best measurement of spread for games and movies.
4. (06.02)
The box plots below show the average daily temperatures in April and October for a U.S. city:
two box plots shown. The top one is labeled April. Minimum at 50, Q1 at 60, median at 67, Q3 at 71, maximum at 75. The bottom b
What can you tell about the means for these two months? (1 point)
The mean for April is higher than October's mean.
There is no way of telling what the means are.
The low median for October pulls its mean below April's mean.
The high range for October pulls its mean above April's mean.
5. (06.02)
The table below shows data from a survey about the amount of time high school students spent reading and the amount of time spent watching videos each week (without reading):
Reading Video
5 1
5 4
7 7
7 10
7 12
12 15
12 15
12 18
14 21
15 26
Which response best describes outliers in these data sets? (2 points)
Neither data set has suspected outliers.
The range of data is too small to identify outliers.
Video has a suspected outlier in the 26-hour value.
Due to the narrow range of reading compared to video, the video values of 18, 21, and 26 are all possible outliers.
6. (06.02)
Male and female high school students reported how many hours they worked each week in summer jobs. The data is represented in the following box plots:
two box plots shown. The top one is labeled Males. Minimum at 0, Q1 at 1, median at 20, Q3 at 25, maximum at 50. The bottom box
Identify any values of data that might affect the statistical measures of spread and center. (2 points)
The females worked less than the males, and the female median is close to Q1.
There is a high data value that causes the data set to be asymmetrical for the males.
There are significant outliers at the high ends of both the males and the females.
Both graphs have the required quartiles.
7. (06.02)
The table below shows data from a survey about the amount of time students spend doing homework each week. The students were either in college or in high school:
High Low Q1 Q3 IQR Median Mean σ
College 50 6 8.5 17 8.5 12 15.4 11.7
High School 28 3 4.5 15 10.5 11 10.5 5.8
Which of the choices below best describes how to measure the spread of this data? (2 points)
Both spreads are best described with the IQR.
Both spreads are best described with the standard deviation.
The college spread is best described by the IQR. The high school spread is best described by the standard deviation.
The college spread is best described by the standard deviation. The high school spread is best described by the IQR.
Answers
1. The dataset with the smallest IQR is the Mid-term exams. 2. The exam median is only 1–2 points higher than the class median. 3. The standard deviation is the best measurement of spread for games and movies.
Explanation:1. The dataset with the smallest IQR is the Mid-term exams. To find the IQR, first, calculate the first quartile (Q1) and the third quartile (Q3). Then, find the difference between Q3 and Q1. By comparing the IQR values for the Mid-term and Final exams, it can be determined that the Mid-term exams have the smallest IQR.
2. The exam median is only 1–2 points higher than the class median. The box plot's median represents the middle value of the dataset. By comparing the medians of the exam and class data, it can be determined that the exam median is only 1–2 points higher than the class median.
3. The standard deviation is the best measurement of spread for games and movies. While the IQR can measure spread, the standard deviation is a more precise measurement. Comparing the spread of the data, the standard deviation is the best measurement for both games and movies.
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Various statistical measures such as IQR, Standard Deviation, Median and Mean were used to interpret the given data in numerous scenarios. Outliers were identified and impacts on datasets were evaluated.
Explanation:To answer these questions, we need to understand few key statistical terms: 'Mean' is the simple average of data, 'Median' is the middle score of data, 'IQR' (Inter Quartile Range) is the difference between the upper quartile (Q3) and the lower quartile (Q1), which helps in understanding the spread and '.'Standard Deviation' measures the absolute variability of a dataset.
Question 1: IQR for the mid-term exams is Q3 (92) - Q1 (82) = 10. IQR for the final exams is Q3 (85) - Q1 (78) = 7. So, the final exams have the smallest IQR.
Question 2: The boxes showing median indicates that the exam median is only 1–2 points higher than the class median.
Question 3: Since the spread of the data at basketball games and local movie theaters demonstrates varied distributions, both the IQR and the standard deviation should be used to evaluate the data spread.
Question 4: Box plots do not provide direct information on the mean. So, there is no way of telling what the means are.
Question 5: For the video hours, we see that values 18, 21, and 26 lie far from the main part of the data, thus they can be considered as possible outliers.
Question 6: The male data set shows a high data value, which causes the data set to be asymmetrical. This could affect statistical measures like the mean and standard deviation.
Question 7: The spread of the college data, having a large standard deviation and IQR, is best described by the standard deviation. The high school data, with smaller numbers, is best described by the IQR.
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Find the mean, median, and mode
15, 3, 11, 15, 1, 14, 7, 2, 1, 1, 2
A. mean = 6.5, median = 8, mode =1
B. mean = 6, median = 3, mode = 1
C. mean = 6, median = 3, mode = 8
D. mean = 6.5, median = 3, mode = 1
Answers
median=3
mode=1
so the answer is d
I know its right for sure :)
Find the fifth term of the arithmetic sequence in which
t1 = 3 and tn = tn-1 + 4.
A) 5
B) 7
C) 19
D) 23
Answers
Someone please help me?
Answers
remove the ^2's from the left side:
(x-3) +(y+5) = sqrt(25)
(x-3) +(y+5) = 5
radius = 5
now solve for zero 's on the left side:
x-3 = 0 x = 3 (3-3=0)
y+5 = 0 y = -5 (-5 +5=0)
so center = (3,-5) and radius is 5
Answer is last one.
Factor 2x2 - 11x - 21.
A) (2x + 3)(x - 7)
Eliminate
B) (x + 3)(2x - 7)
C) (2x - 3)(x + 7)
D) (2x + 7)(x - 3)
Answers
2x² + 3x - 14x - 21
Factor out x from 2x² + 3x.
x × (2x + 3) - 14x - 21
Now factor out -7 from -7(2x + 3)
x × (2x + 3) - 7(2x + 3)
Finally,, factor out 2x + 3 from the expression.
(2x + 3) × (x - 7)
This means the correct answer to your question is (2x + 3) × (x - 7),, or option A.
Let me know if you have any further questions.
:)
Mrs. Jones Algebra 2 class scored very well on yesterday's quiz. With one exception, everyone received an A. Within how many standard deviations from the mean do all the quiz grades fall?
91, 92, 94, 88, 96, 99, 91, 93, 94, 97, 95, 97
A. 1
B. 2
C. 4
D. 3
Answers
Explanation:
1) Find the mean:
mean = sum of the values / number of data
The number of data is 12, so:
mean = (91 + 92 + 94 + 88 + 96 + 99 + 91 + 93 + 94 + 97 + 95 + 97) / 12
mean = 1127 /12 = 93.917
2) Find the variance:
variance = sum of squares of the differences between each data and the media, divided by thenumber of data - 1
variance = (106.97)/(12-1) = 106.97 / 11 = 9.720
3) Find the standard deviation
standard deviation = square root of variance = √(9.720) = 3.118
4) Find the difference of the maximum and the minimun grades with the media:
maximum grade: 99 - 93.917 = 5.083
Find how many standard deviations that is: 5.083 / 3.118 = 1.63
minimum grade: |88 - 93.917| = 5.917
Find how many standard deviations that is: 5.917 / 3.118 = 1.9
5) Conclusion: all the quiz grades falls inside 2 standard deviations form the mean.
Find the length and width (in feet) of a rectangle that has the given area and a minimum perimeter. area: 36 square feet
Answers
A perimeter of 24 feet is achieved in this optimal scenario.
For an area of 36 square feet, we can factor this into pairs: (1, 36), (2, 18), (3, 12), (4, 9), and (6, 6). Since we are looking for the minimum perimeter, the closest factors will yield the smallest perimeter, which in this case is the pair (6, 6). Therefore, a rectangle with the dimensions 6 feet by 6 feet will not only satisfy the area requirement of 36 square feet but will also have the minimum perimeter, which is 24 feet.
The perimeter (P) of a rectangle is calculated by the formula P = 2(l + w), where 'l' is the length and 'w' is the width. Given our dimensions, P = 2(6 + 6) = 24 feet. The square shape of the rectangle, which is a special case where the length and width are equal, is what minimizes the perimeter for a given area.
The rectangle with an area of 36 square feet that has minimal perimeter is actually a square with dimensions of 6 feet by 6 feet.
To find the length and width of a rectangle with an area of 36 square feet that has the minimum perimeter, we need to follow these steps:
Assume the length is l and the width is w. The area equation becomes: l × w = 36 sq ftExpress one variable in terms of the other: l = 36 / wUse the perimeter formula: P = 2l + 2w. Substitute l = 36 / w into this formula: P = 2(36 / w) + 2wTo find the minimum perimeter, take the derivative of P with respect to w and set it to zero: P'(w) = -72/w² + 2 = 0Solve for w: 2 = 72/w² gives w² = 36, so w = 6 ftThen, find l using l = 36 / w, giving l = 6 ftThus, the dimensions that yield the minimum perimeter are 6 feet by 6 feet. This results in a square having the smallest possible perimeter for the given area.
A bathtub can be filled in 8 min. it takes 12 min for the bathtub to drain. if the faucet is turned on but the drain is also left open, how long will it take to fill the tub?
Answers
time taken to drain the bathtub is 12 min
fraction of bathtub filled in 1 hour=1/8
fraction of bathtub drained in 1 hour=1/12
fraction that will be filled in a minute given that both taps are opened at he same time:
1/8-1/12
=1/24
thus time taken for the bathtub to be filled is 24 min
It will take 24 minutes to fill the tub when both the faucet is running and the drain is open. The calculation is done using rates and subtraction.
Explanation:This problem can be approached with the concept of rates. The rate at which the bathtub fills is 1 tub per 8 minutes or 1/8 tubs/min. Similarly, the rate at which it drains is 1 tub per 12 minutes or 1/12 tubs/min. When the faucet is running and the drain is open, the net rate is the fill rate minus the drain rate.
So, (1/8 - 1/12) tubs/minute = 1/24 tubs/minute. Thus, it will take 24 minutes to fill the bathtub with both the faucet running and the drain open.
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What is the standard form for the quadratic function?
g(x)=(x−6)2−5
g(x)=x2−12x+31
g(x)=x2−31
g(x)=x2+12x−41
g(x)=x2−41
Answers
[tex] a{x}^{2} + bx + c = 0[/tex]
multiple answers above fit this description.
Answer:
A) g(x)=x2−12x+31
The domain of f(x)=2logx+3 is x > 3.
true.
false.
Answers
x > 0
So, there is not other function for this problem. Thus, the answer is false. See the figure below that shows f(x)
What is the gcf of 3x^2 and 7y
Answers
Unless we have a relationship defined between x and y there is no common factor apart from 1.
hope it helps
~Reema
Factor 1/2 out of 1/2z+9
Answers
1/2(z+18) is the expression of 1/2z+9 when 1/2 is factored out.
What is Fraction?A fraction represents a part of a whole.
Given,
The expression is 1/2 z+9
A factor is a number that divides another number, leaving no remainder.
The given expression is one by two times of z plus nine.
1/2 z+9
Now we need to take 1/2 as common from the expression
1/2(z+18)
Hence, 1/2(z+18) is the expression of 1/2z+9 when 1/2 is factored out.
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Please help! Math question
Answers
[tex]a_{n}=a_{1}+d(n-1)\\a_{n}=2+3(n-1)\\a_{n}=3n-1[/tex]
The graph shows the first 8 terms.
The length of a rectangle is four times its width. if the width is 15, what is the area?
Answers
You multiply the width by four to get the answer.
60*15 = 900
The area is 900.
Adriana's water bottle contains 2 quarts of water she wants to add a drink to mix into it but the directions for the treatments gives a amount of water fluid ounces how many fluid ounces are in her bottle
Answers
The cost of parking in a garage, in dollars, can be modeled by a step function whose graph is shown. How much does it cost to park for 3 hours and 45 minutes?
$3
$4
$6
$10
Answers
3 hours 45 minutes=3 hours 0.75 hours
3 hours 45 minutes = 3.75 hours
3 hours < 3.75 hours < 4 hours →Cost=$6
Answer: Third option $6
Answer:
the cost is [tex]\$6[/tex]
Step-by-step explanation:
observing the graph
we know that
For the interval of x------> [tex](0,1][/tex] -----> the cost is [tex]\$0[/tex]
For the interval of x------> [tex](1,3][/tex] -----> the cost is [tex]\$4[/tex]
For the interval of x------> [tex](3,4][/tex] -----> the cost is [tex]\$6[/tex]
For the interval of x------> [tex](4,infinite)[/tex] -----> the cost is [tex]\$10[/tex]
In this problem we have
[tex]3[/tex] hours and [tex]45[/tex] minutes
therefore
the value of x belong to the interval [tex](3,4][/tex]
the cost is [tex]\$6[/tex]
Math help please!!!!! If AO = 21 and BC = 14, what is AB?
Answers
Given is a circle O with tangent AB and secant OB.
Given is OA = 21 units and BC = 14 units.
From the diagram, OB = OC + BC.
OC and OA, both are radius, so OC = OA = 21 units.
Now OB = 21 + 14 = 35 units.
In right triangle ΔOAB, using Pythagorean theorem;
OA² + AB² = OB²
⇒ (21)² + AB² = (35)²
⇒ 441 + AB² = 1225
⇒ AB² = 1225 - 441 = 784 square units
⇒ AB = [tex]\sqrt{784} =28[/tex]
⇒ AB = 28 units.
Hence, final answer is AB = 28 units.
Someone who knows math more than I do, can you please answer this question for me, I'd appreciate it so much <3
A right triangle has one angle that measures 28o. The adjacent leg measures 32.6 cm and the hypotenuse measures 35 cm.
What is the approximate area of the triangle? Round to the nearest tenth.
Area of a triangle = 1/2 bh
Answers
x = sqrt(35^2 - 32.6^2) = 12.74 cm
Since this is a right triangle, the base and height are simply the two sides that are perpendicular to each other.
Then the approximate area is (1/2)(leg 1)(leg 2) = (1/2)(32.6)(12.74) = 207.6 cm^2.
2. The height of one square pyramid is 24 m. A similar pyramid has a height of 8 m. The volume of the larger pyramid is 648 m3. The surface area of the smaller pyramid is 124 m^2. Determine each of the following, showing all your work and reasoning: find the surface area of the larger pyramid
Answers
Surface area of square pyramid = a² +2a √((a²/4) + h²)
a = base edge
h = height
for the large pyramid;
volume = 648 m³
height = 24 m
Volume = a²h / 3
648 m³ = a² x 24 m / 3
a² = 81 m²
a = 9 m
by applying surface area equation to the large pyramid,
surface area = a² +2a √((a²/4) + h²)
= 9² + 2 x 9 √((9² / 4 ) + 24²
= 520.53 m²
hence surface area of large pyramid = 520.53 m²
V = a²h / 3648 * 3 / 24= a² a² = 81 a = 9
Area of square is 9 sq.m.
Area of the lateral sides of the larger pyramid:
a = 1/2 b*h
b = 9
h = sqrt (24^2 + 4.5^2)
h = 24.42
a = 1/2 (9 * 24.42) = 109.88
SURFACE AREA = 4 lateral sides area + base area
SURFACE AREA = 4 * 109.88 + 9*9 = 520.53
A model of the is 2.9 ft tall and 1.2 ft wide. The Eiffel Tower is actually 410 ft wide. What is the actual height of the Eiffel Tower?
Answers
Answer:
its B)990.8 ftStep-by-step explanation:
Factor this expression.
3x2 – 6x
A. 3(x – 2)
B. 3(x2 – 2)
C. 3x(x2 – 2)
D. 3x(x – 2)
Answers
Answer:
D.3x(x-2)
Step-by-step explanation:
i already did this
Need help with these 2 questions please and thanks
Answers
1) Problem # 15.
(i) [tex] \lim_{x \to \ 60^{-} } f(x) [/tex]
You have to approach the value of x by the left. That is the horizontal segment at y = 56 (with the open circle to the left and the solid circle to the right).
So, the value of the limit es 56.
(ii) [tex] \lim_{x \to \ 60^{+} } f(x)[/tex]
You have to approach the x value by the right. That is the horizontal segment at y = 68.
So the value of the limit is 68.
(iii) Since, the existence of the limit requires that both limits from the left and the right be equal, the conclusion is that the limit does not exist.
That is shown in the graph because the way the function is defined is different if the value of x is greater or less than 60.
(iv) What causes the graph to jump vertically by the same amount at discontinuites is that the rates are defined in intervals and not in a continuous way.
2) Problem 16. Mathematical induction.
Mathematical induction requires 3 steps: 1) Initial hypothesis, 2) assume the equation is valid for n = k, and 3) prove the equation is valid for n = k+1.
This is the solution step by step.
1) Initial hypothesis: first term ⇒ n = 1
left side: 8
rigth side: 4n(n+1) = 4(1+1) = 4(2) = 8
8 = 8 ⇒ check
2) Assume n = k
letf side: 8 + 16 + 24 + ... + 8k
right side: 4k(k + 1)
3) Proove for n = k + 1
left side: 8 + 16 + 24 + ... + 8k + 8(k+1)
right side: 4(k+1) (k + 1 + 1) = 4(k+1)(k+2)
Since 8 + 16 + 24 + ... 8k is assumed to be equal to 4k(k+1), then the left side ends as:
left side: 4k(k+1) + 8(k+1)
take common factor k + 1 → (k+1)(4k + 8)
take common factor 4 for the second parenthesis → (k+1) 4(k + 2) = 4(k+1)(k+2)
So, we have shown that the left side is equal fo the right side, which completes the proof by induction.
Karl and his dad are building a playhouse for karl's younger sister. the floor of the playhouse will be a rectangle that is 6 by 8 1/2 feet. how much carpeting do karl and his dad need to cover the floor.
Answers
[tex]l \times w[/tex]
Area of playhouse"
[tex]6 \times 8 \frac{1}{2} \\ = 51 {ft}^{2} [/tex]
The playhouse floor is a rectangle with dimensions of 6 by 8.5 feet. To determine the amount of carpet needed, multiply the length and width to calculate the area, which is 51 square feet.
Karl and his dad need to know the amount of carpeting required to cover the floor of a playhouse, which is a practical mathematics problem involving area calculation. To find out how much carpet they need, they have to calculate the area of the rectangular floor, which is the product of its length and width.
The floor measures 6 feet in length and 8.5 feet in width. Multiplying these two dimensions gives us the area:
Area = Length times Width
Area = 6 ft times 8.5 ft
Area = 51 square feet
Therefore, Karl and his dad would need to purchase 51 square feet of carpeting to cover the playhouse floor.
Every year, 50 million flea collars are thrown away. How many flea collars are thrown away per day, rounded to the nearest thousand
Answers
Flea collars thrown away per day = 50 million/Average number of days in a year
Average number of days in a year = 365 days
Therefore,
Flea thrown away per day = 50,000,000/365 = 136,986.3
Rounding the answer to nearest thousand,
Number of flea thrown away per day = 137,000
Use the unit circle to find the value of each trigonometric function at the angle indicated
Answers
-1
-inf
1
0
0
repectively
Answer:
Step-by-step explanation:
We have to find the values of the given trigonometric ratios at the angle indicated. Thus,
(A) The given trigonometric function is:
[tex]cos270^{\circ}[/tex]
Since, the function lies in Quadrant III, therefore the value of the function will be negative.
Also, [tex]cos270^{\circ}=-(0)=0[/tex]
(B) The given trigonometric function is:
[tex]sin270^{\circ}[/tex]
Since, the function lies in Quadrant III, therefore the value of the function will be negative.
Also, [tex]sin270^{\circ}=-1[/tex]
(C) The given trigonometric function is:
[tex]tan270^{\circ}[/tex]
Since, the function lies in Quadrant III, therefore the value of the function will be negative.
Also, [tex]tan270^{\circ}=undefined[/tex]
(D) The given function is:
[tex]cos0^{\circ}[/tex]
Since, the function lies in Quadrant I, therefore the value of the function will be positive.
Also, [tex]cos0^{\circ}=1[/tex]
(E) The given function is:
[tex]sin0^{\circ}[/tex]
Since, the function lies in Quadrant I, therefore the value of the function will be positive.
Also, [tex]sin0^{\circ}=0[/tex]
(F) The given function is:
[tex]tan0^{\circ}[/tex]
Since, the function lies in Quadrant I, therefore the value of the function will be positive.
Also, [tex]tan0^{\circ}=0[/tex]
Can someone help me solve 2√x−4=10 I'm confused.
Answers
let's analyze two cases
case A) 2(√x)−4=10
2√x=10+4----> 2√x=14----> √x=14/2----> square---> x=7²----> x=49
case B)
2√(x-4)=10
√(x-4)=10/2-----> square---> x-4=5²----> x=25+4-----> x=29