The initial point of vector v if it has a magnitude of 10 and the terminal point (-3,3) is option B (-9,-5).
How to find the vector component?If we have given a vector v of initial point A and terminal point B
v = ai + bj
then the components form as;
AB = xi + yj
Here, xi and yj are the components of the vector.
Given;
The magnitude of 10 and the terminal point (-3,3).
v = ai + bj
v = [tex]\sqrt{a^2 + b^2}[/tex]
10 = [tex]\sqrt{a^2 + b^2}[/tex]
AB = (-3 - a)i + (3 - b)j
If we put (-9,-5)then
AB = (-3 - a)i + (3 - b)j
AB = (-3 + 9)i + (3 + 5)j
AB = 6i + 8j
Put the value in the vector magnitude
10 = [tex]\sqrt{a^2 + b^2}[/tex]
= [tex]\sqrt{6^2 + 8^2}\\\\\sqrt{36+ 64}\\\\\sqrt{100}[/tex]
10 = 10
Thus, The correct option is B (-9,-5).
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The solution for which of the following is neither a interger nor a fraction 4x+7=x+2 ,5x-8=x+4, 3x+2=5x+2, none of the above
Answer:
none of the above
Step-by-step explanation:
4x + 7 = x + 2
4x - x + 7 = x - x + 2
3x + 7 = 2
3x + 7 - 7 = 2 - 7
3x = -5
3x / 3 = -5 / 3
x = -5 / 3 = -1.667
5x - 8 = x + 4
5x - x - 8 = x - x + 4
4x - 8 + 8 = 4 + 8
4x = 12
4x / 4 = 12 / 4
x = 3
3x + 2 = 5x + 2
3x - 5x + 2 = 5x - 5x + 2
-2x + 2 = 2
-2x + 2 - 2 = 2 - 2
-2x = 0
x = 0
Integers, As a whole number that can be written without a remainder,
x = 3
x = 0
(Mixed) Fraction
x = -5 / 3 = -1.667
congruent figures someone please help
( 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.
x − y = 12
x + 2y = 21
Answer: x=15, y=3
Step-by-step explanation:
Subtracting the two equations, we get -3y=-9, meaning y=3.
Substituting this into the first equation, we get that x-3=12, and thus, x=15.
please help 35 points!!
Answer:
67m²
Step-by-step explanation:
12m + 6m + 4m + 36 m 8m = 67m²
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).
suppose y varies inversely as x and y=12 when x=6 find y if x=8
Answer:
y = 9
Step-by-step explanation:
given that y varies inversely as x then the equation relating them is
y = [tex]\frac{k}{x}[/tex] ← k is the constant of variation
to find k use the condition y = 12 when x = 6
12 = [tex]\frac{k}{6}[/tex] ← multiply both sides by 6 to clear the fraction
72 = k
y = [tex]\frac{72}{x}[/tex] ← equation of variation
when x = 8 , then
y = [tex]\frac{72}{8}[/tex] = 9
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
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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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 bacteria population has been doubling each day for the past 5 days. It is currently
100000. What was the population 5 days ago?
3 quick questions 28 points
For the first image, I would recommend using Desmos; it is an online graphing calculator. You can find it on online and it is free; I use it all the time for my school work
I would input each function and sees which one matches the image on the test.
As for the second image, I believe the answer is C; x is greater than or equal to -1.
For the third and last image, I believe the answer is A; x is less than -1.
I hope this helps .
I need a little help here
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!
You plan to build a house that is 1 ½ times as long as it is wide. You want the land around the house to be 20 feet wider than the width of the house, and twice as long as the length of the house, as shown at the right.
The total area with land based on the information given is 3x² + 60x.
How to find the area?Let the width = x
Let the length = (1.5 × x) = 1.5x
Area = 1.5x × x = 1.5x²
The area along with land:
Width = x + 20
Length = 3 × x = 3x
The total area with land:
= Length × Width
= 3x(x + 20)
= 3x² + 60x.
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im lost can someone help?
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!
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]
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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Without calculating the cubes find 1 cube+2cube+2(4) cube+(-5)cube+(-6)cube .
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]
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]
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
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.
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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e) 35-12 + 6 x 36 /4
[tex]35 - 12 + \frac{6 - 36}{4 } \\ 35 - 12 + \frac{216}{4} [/tex]
2. Divide the number[tex] 35 - 12 + \frac{216}{4} \\ 35 - 12 + 54[/tex]
3. Add the numbers[tex]35 - 12 + 54 \\ 77[/tex]
Solution : 77find the sum. 1-2+3-4+5-6+7-8...99-100
Answer:
-50
Step-by-step explanation:
1 - 2 + 3 - 4 + 5 - 6 + 7 - 8...99 - 100
just means,
1 - 2 = -1
3 - 4 = -1
5 - 6 = - 1
7 - 8 = - 1
99 - 100 = -1
basically, -1 + -1 + -1 + -1 + ....+ -1 or -1 - 1 - 1 - 1 - ... - 1
[tex](-1)^{n}[/tex]
Treating Grandi's series as a divergent geometric series we may use the same algebraic methods that evaluate convergent geometric series to obtain a third value:
S = 1 − 1 − 1 + ..., so
1 − S = 1 − (1 − 1 − 1 ...) = 1 − 1 − 1 ... = S
1 - S = S
1 = 2*S
resulting in S = 1/2. The same conclusion results from calculating −S, subtracting the result from S, and solving 2S = 1
1/2 of the last term is 100 x 1/2 = 50
This because you are using 2 numbers per -1
example: 1 - 2 = -1 or 3 - 4 = -1 resulting in 50 of the -1 being subtracted
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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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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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.