Answer:
Dilation
Step-by-step explanation:
it’s a dilation because the dilation changes how big the image is or what it looks like after the original image.
Multiply:
(x+y)by (x+y)
a+b by a^2-b^2
(a+5) by (a^2-2a-3)
(a^2-ab+b^3) by (a+b)
Answer:
Multiply:
[tex](x+y)by (x+y)[/tex]
[tex] : \implies(x + y)(x + y)[/tex]
[tex] : \implies \: x(x + y) + y(x + y)[/tex]
[tex] : \implies {x}^{2} + xy + xy + {y}^{2} [/tex]
[tex] : \implies{x}^{2} + 2xy + {y}^{2} [/tex]
Multiply:
[tex]a+b \: by \: a^2-b^2[/tex]
[tex]: \implies( {a}^{2} + {b}^{2} ) \times (a + b)[/tex]
[tex]: \implies \: {a}^{2} (a + b) - {b}^{2} (a + b)[/tex]
[tex]: \implies \: {a}^{3} + {a}^{2} b - {ab}^{2} - {b}^{3} [/tex]
Multiply:
[tex](a+5) by (a^2-2a-3)[/tex]
[tex]: \implies{(a + 5) \times ( {a}^{2} - 2a - 3) }[/tex]
[tex]: \implies \: a({a}^{2} - 2a - 3) + 5( {a}^{2} - 2a - 3)[/tex]
[tex]: \implies(a \times {a}^{2} - a \times 2a - a \times 3) + (5 \times {a}^{2} - 5 \times 2a - 5 \times 3)[/tex]
[tex]: \implies{a}^{3} - {2a}^{2} - 3a + 5 {a}^{2} - 10a - 15 [/tex]
[tex]: \implies{ {a}^{3} + {3a}^{2} - 13a - 15}[/tex]
Multiply:
[tex](a^2-ab+b^3) by (a+b)[/tex]
[tex]: \implies{(a + b) \times ( {a}^{2} - ab + {b}^{3} )}[/tex]
[tex]: \implies \: a( {a}^{2} - ab + {b}^{3}) + b( {a}^{2} - ab + {b}^{3} ) [/tex]
[tex]: \implies {a}^{3} - {a}^{2} b + a {b}^{3} + {a^2b} - {ab}^{2} + {b}^{4} [/tex]
[tex]: \implies{ {a}^{3}+ab^3 - ab^2+ {b}^{4} }[/tex]
Step-by-step explanation:
[tex] \blue{ \frak{Seolle_{aph.rodite}}}[/tex]
Draw and set up the integrals for the area enclosed by the y–axis, the curve y = (x + 1)1/2 and y = 2. Compute one of them.
Region II only please
If the definitions of type I and type II regions is the same as in the link provided, then as a type I region the integration domain is the set
[tex]R_{\rm I} = \left\{(x,y) \mid 0 \le x \le 3 \text{ and } \sqrt{x+1} \le y \le 2\right\}[/tex]
and as a type II region,
[tex]R_{\rm II} = \left\{(x,y) \mid 0 \le x \le y^2-1 \text{ and } 1 \le y \le 2\right\}[/tex]
where we solve y = √(x + 1) for x to get x as a function of y.
A. The area of the type I region is
[tex]\displaystyle \iint_{R_{\rm I}} dA = \int_0^3 \int_{\sqrt{x+1}}^2 dy \, dx = \int_0^3 (2 - \sqrt{x+1}) \, dx = \boxed{\frac43}[/tex]
B. The area of the type II region is of course also
[tex]\displaystyle \iint_{R_{\rm II}} dA = \int_1^2 \int_0^{y^2-1} dx \, dy = \int_1^2 (y^2-1) \, dy = \boxed{\frac43}[/tex]
I've attached a plot of the type II region to give an idea of how it was determined. The black arrows indicate the domain of x as it varies from the line x = 0 (y-axis) to the curve y = √(x + 1).
im lost can someone help?
If you help me you get a lot of points
Answer:
Step-by-step explanation:
#a
pattern 0 will include 4 reds in square
Because it's independent of pattern no
#b
Figure 1 has 4+4=8
Figure 2=4+8+2=14
Figure 3=4+12+3=19
The pattern n rule is
n²+3n+4So for 13th n
13²+3(13)+4169+39+4212squares#c
attached
y=x²+3x+4#d
Already given in c
please help 35 points!!
Answer:
67m²
Step-by-step explanation:
12m + 6m + 4m + 36 m 8m = 67m²
WILL MARK BRAINLIEST 50 POINTS Find the area of the regular pentagon if the apothem is 7 ft and a side is 10 ft. Round to the nearest whole number.
175 ft2
350 ft2
35 ft2
70 ft2
Answer:
175 ft^2
Step-by-step explanation:
split the pentagon into 5 triangles with base length 10ft and height 7ft. each triangle then has an area of 10 * 7 * 1/2 = 35 ft^2
then the pentagon has an area 35*5 = 175 ft^2
This list shows the lengths in feet of the 25 longest bridges in the United States.
1010
1470
1600
2000
2800
1053
1495
1632
2150
3500
1200
1596
1750
2150
3800
1207
1600
1800
2300
4200
1380
1600
1850
2310
4260
Jamie made a frequency table of the bridge data.
U.S. Bridges
Length (ft)
Tally
Frequency
(first interval)
|||| ||
7
(last interval)
||
2
Based on the tallies and frequencies for the first and last intervals, how many intervals of what length did Jamie use?
a.
4 intervals of 1000 feet
c.
6 intervals of 500 feet
b.
5 intervals of 1000 feet
d.
7 intervals of 500 feet
Please select the best answer from the choices provided
A
B
C
D
The length of the intervals of the given frequency table is
500 ft.
The given list shows the lengths in feet of the 25 longest bridges in the United States.
we have to determine the median and mode of the bridge data.
The median is the middle number of the data set arranged in ascending order.
1010, 1053, 1200, 1207, 1380, 1470, 1495, 1596, 1600, 1600, 1600, 1632, 1750, 1800, 1850, 2000, 2150, 2150, 2300, 2310, 2800, 3500, 3800, 4200, 4260.
The median is 1750.
Mode is the number that appears most often in the data set.
1010, 1053, 1200, 1207, 1380, 1470, 1495, 1596, 1600, 1600, 1600, 1632, 1750, 1800, 1850, 2000, 2150, 2150, 2300, 2310, 2800, 3500, 3800, 4200, 4260
The mode is 1600.
Therefore the correct option is Median: 1750, Mode: 1600
So the length of the intervals of the given frequency table is
500 ft.
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A group of kids just finished trick-or-treating. The number of pieces of candy collected by each of the 5 kids is listed below.
31,33,36,41,34
Find the standard deviation of the data set. Round your answer to the nearest hundredth.
what are the soltuions to the quadratic equation below? 12x squared + 4x -5=0
Answer: 0.5 or - 0.834
Step-by-step explanation: Here is the explanation!
At a certain college, 49% of the students are female, and 21% of the students major in civil engineering. Furthermore, 8% of the students both are female and major in civil engineering.
The probability that a student is a female or major in civil engineering is 62%
Complete questionAt a certain college, 49% of the students are female, and 21% of the students major in civil engineering. Furthermore, 8% of the students both are female and major in civil engineering. What is the probability that a randomly selected female student majors in civil engineering?
How to determine the probability?Let A represent Female and B represents civil engineering.
The above representation means that the given parameters are:
P(A) = 49%P(B) = 21%P(A and B) = 8%The required probability is calculated as:
P(A or B) = P(A) + P(B) - P(A and B)
This gives
P(A or B) = 49% + 21% - 8%
Evaluate
P(A or B) = 62%
Hence, the probability that a student is a female or major in civil engineering is 62%
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17.
select the correct answer
what is the equation of the problem shown with its focus on this graph?
Options are in photo!
Answer:
B
Step-by-step explanation:
( PLEASE HELP WITH THIS QUESTION)
You are studying a single-celled organism under a microscope. Is it possible for this organism to be classified as fungi?
Answer: Yes it is possible
Example: Yeast is a single-celled fungus.
There are probably other types of fungus that are single-celled. However, some other fungi are multi-celled. You will likely need more information about the organism under the microscrope before you can classify it properly.
can someone help me with this worksheet please!!!!
(1) The missing term in the sequence, a₁₂ = 0.8.
(2) The missing term in the sequence, a₈ = 102.5.
(3) The missing term in the sequence, a₈ = 111.
(4) The missing term in the sequence, a₁₂ = -19.
(5) The missing term in the sequence, a₁₂ = 94.
(6) The missing term in the sequence, a₆ = 40.
(7) The missing term in the sequence, a₃₆ = -52.
(8) The missing term in the sequence, a₂₁ = -58.
Missing term of the sequenceThe missing term in the sequence is determined as follows;
Tₙ = a + (n - 1)d
1.0 a₄ = 18.4 and a₅ = 16.2, a₁₂ = ?T₄ = a + 3d
18.4 = a + 3d ---(1)
T₅ = a + 4d
16.2 = a + 4d ---(2)
subtract (1) from (2)
-2.2 = d
18.4 = a + 3(-2.2)
a = 25
a₁₂ = a + 11d
a₁₂ = 25 + 11(-2.2)
a₁₂ = 0.8
2.0 a₂ = 57.5 and a₅ = 80, a₈ = ?a₂ = a + d
57.5 = a + d -- (1)
a₅ = a + 4d
80 = a + 4d --- (2)
solve (1) and (2)
d = 7.5
a = 50
a₈ = a + 7d
a₈ = 50 + 7(7.5)
a₈ = 102.5
3.0 a₁₀ = 141 and a₁₃ = 186, a₈ = ?a₁₀ = a + 9d
141 = a + 9d --- (1)
a₁₃ = a + 12d
186 = a + 12d --- (2)
Subtract (1) from (2)
d = 15
a = 6
a₈ = a + 7d
a₈ = 6 + 7(15)
a₈ = 111
4.0 a₂₂ = -49 and a₂₅ = -58, a₁₂ = ?a₂₂ = a + 21d
-49 = a + 21d ---- (1)
a₂₅ = a + 24d
-58 = a + 24d --- (2)
subtract (1) from (2)
d = -3
a = 14
a₁₂ = a + 11d
a₁₂ = 14 + 11(-3)
a₁₂ = -19
5.0 a₄ = -2 and a₈ = 46, a₁₂ = ?a₄ = a + 3d
-2 = a + 3d --- (1)
a₈ = a + 7d
46 = a + 7d ---- (2)
Subtract (1) from (2)
d = 12
a = -38
a₁₂ = a + 11d
a₁₂ = -38 + 11(12)
a₁₂ = 94
6.0 a₉ = 64 and a₁₂ = 88, a₆ = ?a₉ = a + 8d
64 = a + 8d --- (1)
a₁₂ = a + 11d
88 = a + 11d --- (2)
Subtract (1) from (2)
d = 8
a = 0
a₆ = a + 5d
a₆ = 0 + 5(8)
a₆ = 40
7.0 a₂₀ = -4 and a₂₃ = -13, a₃₆ = ?a₂₀ = a + 19d
-4 = a + 19d ---- (1)
a₂₃ = a + 22d
-13 = a + 22d --- (2)
Subtract (1) from (2)
d = -3
a = 53
a₃₆ = a + 35d
a₃₆ = 53 + 35(-3)
a₃₆ = -52
8.0 a₂₈ = 5 and a₃₃ = 50, a₂₁ = ?a₂₈ = a + 27d
5 = a + 27d ---- (1)
a₃₃ = a + 32d
50 = a + 32d --- (2)
Subtract (1) from (2)
d = 9
a = -238
a₂₁ = a + 20d
a₂₁ = -238 + 20(9)
a₂₁ = -58
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Adult tickets to a basketball game cost $5. Student tickets cost $1. A total of $3,128 was collected on the sale of 1,336 tickets. How many of each type of ticket were sold?
Answer:
Adults = 448
Students = 888
Step-by-step explanation:
Write equations with info given
A = Adult tickets
S = Student tickets
5A+1S=3,128
A+S=1336
Subtract equations from each other
4A=1792
Solve for A
A=448
Plug A into second equitation
448+S=1336
Solve for S
S=888
find the exact value of sin 15 degrees
Answer:
Hi,
Step-by-step explanation:
sin(a-b)=sin(a) cos(b)+ cos(a) sin(b)
a=45° and b=30°
[tex]sin(45^o-30^o)=sin(45^o)*cos(30^o)+cos(45^o)*sin(30^o)\\\\=\dfrac{\sqrt{2}}{2}* \dfrac{\sqrt{3}}{2}+\dfrac{\sqrt{2}}{2}*\dfrac{1}{2}\\\\\\\boxed{sin(15^o)=\dfrac{\sqrt{2}}{4}*(1+\sqrt{3} )}\\[/tex]
I need a little help here
need help with this graphing question please
Step-by-step explanation:
12 . The x-intercept is where a line crosses the x-axis, and the y-intercept is the point where the line crosses the y-axis. Thinking about intercepts helps us graph linear equations..
Find the probability that a randomly
selected point within the circle falls
in the red shaded area.
r = 4 cm
[?]%
Round to the nearest tenth of a percent.
Enter
Answer:
[tex]31.8\%[/tex]
Step-by-step explanation:
The area of the circle is [tex]A=\pi r^2=\pi(4)^2=16\pi[/tex]
The area of the triangle is [tex]A=\frac{bh}{2}=\frac{8*4}{2}=\frac{32}{2}=16[/tex]
Hence, the probability of a randomly selected point within the circle falls in the red shaded area is [tex]\frac{16}{16\pi}=\frac{1}{\pi}\approx0.318\approx31.8\%[/tex]
A bacteria population has been doubling each day for the past 5 days. It is currently
100000. What was the population 5 days ago?
find the solution set. 4x^2+x=3
Answer:
[tex]x=\frac{3}{4},\:x=-1[/tex]
Keys:
For this problem, you need the quadratic formula(listed below).
[tex]x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex][tex]1^a=1[/tex][tex]\sqrt[n]{a}^n=a[/tex]When you see ± in a quadratic equation, you must know there is going to be at least 2 solutions.
Step-by-step explanation:
solving for x₁ and x₂
[tex]4x^2+x=3\\4x^2+x-3=3-3\\4x^2+x-3=0\\x_{1,\:2}=\frac{-1\pm \sqrt{1^2-4\cdot 4\left(-3\right)}}{2\cdot 4}\\[/tex]
[tex]1^2=1\\=\sqrt{1-4\cdot \:4\left(-3\right)}\\=\sqrt{1+4\cdot \:4\cdot \:3}\\=\sqrt{1+48}\\=\sqrt{49}\\=\sqrt{7^2}\\\sqrt{7^2}=7\\=7[/tex]
[tex]x_{1,\:2}=\frac{-1\pm \:7}{2\cdot \:4}\\x_1=\frac{-1+7}{2\cdot \:4},\:x_2=\frac{-1-7}{2\cdot \:4}\\[/tex]
solve for x₁
[tex]\frac{-1+7}{2\cdot \:4}[/tex]
[tex]=\frac{6}{2\cdot \:4}[/tex]
[tex]=\frac{6}{8}[/tex]
[tex]= \frac{6\div2}{8\div2}[/tex]
[tex]=\frac{3}{4}[/tex]
solve for x₂
[tex]\frac{-1-7}{2\cdot \:4}[/tex]
[tex]=\frac{-8}{2\cdot \:4}[/tex]
[tex]=\frac{-8}{8}[/tex]
[tex]=-\frac{8}{8}[/tex]
[tex]=-1[/tex]
Hope this helps!
nth term formula? maths quickly
[tex]\text{Nth term of an arithmetic series} = a +(n-1)d \\\\\text{Nth term of an geometric series}= ar^{n-1}\\\\\text{where,}\\\\\text{a = first term.}\\\\\text{d = common difference.}\\\\\text{r = common ratio.}[/tex]
A distribution has the five-number summary shown below. What is the
interquartile range (IQ) of this distribution?
Answer:
tiookvgvc. jbjvth kivtcth jjvf h. bkbgv
Answer:
The IQR of the given distribution is
Step-by-step explanation:
The given distribution has the five-number
28, 34, 43, 59, 62
Divide these numbers in two equal parts.
(28, 34), 43,( 59, 62)
Now divide each parenthesis in two equal parts.
(28), (34), 43,( 59), (62)
It means first quartile is the average of 28 and 34. Third quartile is the average of 59 and 62.
The interquartile range (IQR) of this distribution is
Therefore the IQR of the given distribution is 29.5.
Deion has 62 m of fencing to build a four-sided fence around a rectangular plot of
land. The area of the land is 228 square meters. Solve for the dimensions (length and
width) of the field?
The dimensions of the the plot of land is a length of 19 meters and width of 12 meters.
What is an equation?An equation is an expression that shows the relationship between two or more numbers and variables.
Let x represent the length of the field and y represent the width of the field, hence:
Deion has 62 m of fencing:
2(x + y) = 62
x + y = 31
y = 31 - x (1)
The area of the land is 228 square meters. hence:
xy = 228
x(31 - x) = 228
x² - 31x + 228 = 0
x = 12 or x = 19
when x = 12; y = 31 - 12 = 19
when x = 19; y = 31 - 19 = 12
The dimensions of the the plot of land is a length of 19 meters and width of 12 meters.
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Please help me with this!
Answer:
Yes, x = 0 is a solution to the given equation
Step-by-step explanation:
[tex]1^3+1(1-1)=1-x^2[/tex] (Given)[tex]L.H.S.=1^3+1(1-1)[/tex] [tex]= 1 +1(0)[/tex] [tex]= 1+0 [/tex] [tex]=1[/tex] [tex]R.H.S. =1-x^2[/tex] [tex]=1-(0)^2[/tex] (Plug x = 0)[tex]=1-0[/tex] [tex]=1[/tex] [tex]\implies L.H.S. = R.H.S.[/tex]Thus, x = 0 is a solution to the equation [tex]1^3+1(1-1)=1-x^2[/tex]Hi Student!
The goal of this question is to determine x = 0 is a solution of the expression that was provided. The first step that we must take is input 0 into all of the x's that we have in the expression. Then we just simplify both sides and determine if the end expression is true and if it is then x = 0 is a solution.
Plug in the values
[tex]1^3 + 1(1 - 1) = 1 - x^2[/tex][tex]1^3 + 1(1 - 1) = 1 - (0)^2[/tex]Simplify both sides
[tex]1^3 + 1(0) = 1 - 0[/tex][tex]1 + 0 = 1[/tex][tex]1 = 1[/tex]Looking at the final expression, we can see that 1 is indeed equal to 1 and since the expression is true, we can say that x = 0 is a solution of the expression that was provided in the problem statement.
Without calculating the cubes find 1 cube+2cube+2(4) cube+(-5)cube+(-6)cube .
Calculate the area of the alarm clock.
Given the diameter of the surface of the clock, the area of the surface of the alarm clock is 3846.5cm².
What is the area of the alarm clock?Note that: Area of a circle is expressed as;
A = πr²
Where r is radius and π is constant pi ( π = 3.14 )
Given that;
Diameter d = 70cm Radius r = d/2 = 70cm/2 = 35cmArea = ?A = πr²
A = 3.14 × ( 35cm )²
A = 3.14 × 1225cm²
A = 3846.5cm²
Therefore, given the diameter of the surface of the clock, the area of the surface of the alarm clock is 3846.5cm².
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Which is an x-intercept of the continuous function in the table ? (0, - 6); (3, 0); (- 6, 0) O (0, 3)
An x-intercept of the continuous function in the table is (-1, 0)
Intercept of a lineThe x-intercept of a line is the point where the line crossed the x-axis or the point where the value of y is zero.
From the table, the x-intercept are all the point where the value of f(x) is zero. Hence the Which is an x-intercept of the continuous function in the table is (-1, 0)
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Given Segment AC with point B contained on the segment, as shown below.
Write a complete two-column proof for following information:
Given: Segment AB = x + 16, Segment BC = 4x + 11
Segment AC = 77
Prove: AB = 26
It is true that the line segment AB equals 26
How to prove that line segment AB = 26?The given parameters are:
AB = x +16
BC = 4x + 11
AC = 77
The two-column proof is as follows:
AC = AB + BC Line segment formula
77 = x + 16 + 4x + 11 Substitution property of equation
77 = 5x + 27 Addition property of equation
5x = 50 Subtraction property of equation
x = 10 Division property of equation
AB = 10 +16 Substitution property of equation
AB = 26
Hence, the line segment AB has been proved to equal 26
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Select ALL the correct answers. Consider the following graph of function f. Which transformations will change function f into function g given below. a vertical shift down 3 units a vertical shift down 5 units a vertical shift up 5 units a horizontal shift left 7 units a horizontal shift right 7 units a horizontal shift left 4 units
If the population is highly skewed, the sample size needed for the central limit theorem to apply usually has to be ______ that when the population is not highly skewed a. different from b. the same as c. larger than d. smaller than
Answer:
2
Step-by-step explanation:
the same as...
(2) is the answer
If the population is highly skewed, the sample size needed for the central limit theorem to apply usually has to be the same as that when the population is not highly skewed.
What is the central limit theorem?The central limit theorem states in probability theory that, in many instances, when independent random variables are added together, their correctly normalized sum tends toward a normal distribution, even if the original variables are not normally distributed.
If the population is highly skewed, the sample size needed for the central limit theorem to apply usually has to be the same as that when the population is not highly skewed.
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