Given:
M is the midpoint of AB.
M(2,0) and A(-3, 3).
To find:
The coordinates of point B.
Solution:
Midpoint formula:
[tex]Midpoint=\left(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\right)[/tex]
Let the coordinates of point B are (a,b). Then, using the midpoint formula, we get
[tex](2,0)=\left(\dfrac{-3+a}{2},\dfrac{3+b}{2}\right)[/tex]
On comparing both sides, we get
[tex]\dfrac{-3+a}{2}=2[/tex]
[tex]-3+a=2\times 2[/tex]
[tex]a=4+3[/tex]
[tex]a=7[/tex]
And,
[tex]\dfrac{3+b}{2}=0[/tex]
[tex]3+b=0[/tex]
[tex]3+b-3=0-3[/tex]
[tex]b=-3[/tex]
Therefore, the coordinates of point B are (7,-3).
When air expands adiabatically (without gaining or losing heat), its pressure P and volume V are related by the equation PV^1.4 = C, where C is a constant. Suppose that at a certain instant the volume is 400 cm^3 and the pressure is 80 kPa and is decreasing at a rate of 10 kPa/min. At what rate is the volume increasing at this instant?
Answer:
The volume increases at 35.71cm^3/min
Step-by-step explanation:
Given
[tex]PV^{1.4} = C[/tex]
[tex]V = 400cm^3[/tex]
[tex]P =80kPa[/tex]
[tex]\frac{dP}{dt} =-10kPa/min[/tex]
Required
Rate at which volume increases
[tex]PV^{1.4} = C[/tex] [tex]V = 400cm^3[/tex] [tex]P =80kPa[/tex]
Differentiate: [tex]PV^{1.4} = C[/tex]
[tex]P*\frac{dV^{1.4}}{dt} +V^{1.4}*\frac{dP}{dt} = \frac{d}{dt}C[/tex]
By differentiating C, we have:
[tex]P*\frac{dV^{1.4}}{dt} +V^{1.4}*\frac{dP}{dt} = 0[/tex]
Rewrite as:
[tex]P*(1.4)*V^{0.4}* \frac{dV}{dt} + V^{1.4}*\frac{dP}{dt} = 0[/tex]
Solve for [tex]\frac{dV}{dt}[/tex]
[tex]P*(1.4)*V^{0.4}* \frac{dV}{dt} =- V^{1.4}*\frac{dP}{dt}[/tex]
[tex]\frac{dV}{dt} =- \frac{V^{1.4}*\frac{dP}{dt} }{P*(1.4)*V^{0.4}}[/tex]
Substitute values
[tex]\frac{dV}{dt} =- \frac{400^{1.4}*-10 }{80*(1.4)*400^{0.4}}[/tex]
[tex]\frac{dV}{dt} =\frac{400*10 }{80*1.4}[/tex]
[tex]\frac{dV}{dt} =\frac{4000 }{112}[/tex]
[tex]\frac{dV}{dt} =35.71cm^3/min[/tex]
Five less than Greg's age is less than 23
Answer:
Greg is 28.
Step-by-step explanation:
Greg's age is y.
y - 5 = 23
y - 5 + 5 = 23 + 5
y = 28
What is the decimal multiplier to increase by 4.3%?
Answer:
1.043
Step-by-step explanation:
Decimal multiplier to increase by x% is computed as:-
100% + x%
The decimal multiplier to increase by 4.3% would be
100% +4.3% = 104.3%
To remove% we divide the number by 100, so the number will become
Hence, The decimal multiplier to increase by 4.3% = 1.043
Please Mark as Brainliest
Hope this Helps
I need help solving for x
Answer:
x = 25
Step-by-step explanation:
Given a line parallel to a side of the triangle and intersecting the other 2 sides then it divides those sides proportionally, that is
[tex]\frac{40}{24}[/tex] = [tex]\frac{x}{15}[/tex] ( cross- multiply )
24x = 600 ( divide both sides by 24 )
x = 25
Two circles are concentric if they have the same center.
On a coordinate plane, a circle has center (4, 6) and has a radius of 2 units.
Which equation represents a circle that is concentric with the circle shown but has a radius that is twice as large?
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Answer:
(b) (x – 4)^2 + (y – 6)^2 = 16
Step-by-step explanation:
The equation for a circle with center (h, k) and radius r is ...
(x -h)^2 +(y -k)^2 = r^2
You want the equation for a circle with center (4, 6) and radius 4. That equation is ...
(x -4)^2 +(y -6)^2 = 16
Answer:
B; (x – 4)2 + (y – 6)2 = 16
Step-by-step explanation:
correct on edge
Choose the pair of sets which is equivalent.
Answer and I will give you brainiliest
ABC is an isosceles triangle in which AC =BC.
D and E are points on BC and AC such that CE=CD.
prove triangle ACD and BCE are congruent
Given:
ABC is an isosceles triangle in which AC =BC.
D and E are points on BC and AC such that CE=CD.
To prove:
Triangle ACD and BCE are congruent.
Solution:
In triangle ACD and BCE,
[tex]AC=BC[/tex] (Given)
[tex]AC\cong BC[/tex]
[tex]\angle C\cong m\angle C[/tex] (Common angle)
[tex]CD=CE[/tex] (Given)
[tex]CD\cong CE[/tex]
In triangles ACD and BCE two corresponding sides and one included angle are congruent. So, the triangles are congruent by SAS congruence postulate.
[tex]\Delta ACD\cong \Delta BCE[/tex] (SAS congruence postulate)
Hence proved.
Answer:
Given ABC is an isosceles triangle with AB=AC .D and E are the point on BC such that BE=CD
Given AB=AC∴∠ABD=∠ACE (opposite angle of sides of a triangle ) ....(1)
Given BE=CDThen BE−DE=CD−DE
ORBC=CE......................................(2)
In ΔABD and ΔACE
∠ABD=∠ACE ( From 1)
BC=CE (from 2)
AB=AC ( GIven)
∴ΔABD≅ΔACE
So AD=AE [henceproved]
Which choice shows (40+10)+30 correctly rewritten using the associative property and then correctly simplified?
40 + (10 + 30) = 40 + 40 = 80
40 + 30 + 10 = 70 + 10 = 80
40 + (10 +30) = 50 + 30 =80
(10 +40) + 30 = 50 + 30 =80
Which of the following ordered pairs is a solution to the equation 4x+6y=12? Select all that apply. Select all that apply: (−3,4) (1,3) (−6,6) (−13,10) (0,2)
WILL GIVE BRAINLIST! PUT THESE NUMBERS ON THE PLOT
Answer:
"fair" srry im only in 8th so im d.u.m
Step-by-step explanation:
Please help!!! Urgent ….
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Answer:
ΔWZT ~ ΔWXY
Step-by-step explanation:
Angle XWY and angle ZWT are vertical angles, so congruent.
The sides on either side of those angles are proportional:
WZ/WX = WT/WY
11/22 = 10/20 = 1/2
so, we can claim similarity by the SAS Theorem.
ΔWZT ~ ΔWXY
What should my answer be?
The owner of a football team claims that the average attendance at games is over 74,900, and he is therefore justified in moving the team to a city with a larger stadium. An independent investigator will conduct a hypothesis test to determine whether his claim is accurate. This test is ________. Group of answer choices Two-sided Lower-tailed Upper-tailed
Answer:
Option c (Upper tailed) is the correct choice.
Step-by-step explanation:
Given that:
The average attendance is:
= 74,900
We will have to test:
⇒ [tex]H_0:\mu \leq \mu_0[/tex]
or,
[tex]H_0: \mu \leq 74,900[/tex]
Verses,
⇒ [tex]H_1: \mu> \mu_0[/tex]
or,
[tex]H_1: \mu >74,900[/tex]
The other given alternatives aren't connected to the given scenario. So the above is the correct one.
Which of the following represents the factorization of the trinomial below?
x2 - 17x+ 30
A. (x-3)(x-10)
B. (x-2)(x+15)
C. (x-2)(x-15)
D. (x-3)(x +10)
What is the probability of tossing a penny and landing on heads 3 times in a row
Answer:
Suppose you have a fair coin: this means it has a 50% chance of landing heads up and a 50% chance of landing tails up. Suppose you flip it three times and these flips are independent. What is the probability that it lands heads up, then tails up, then heads up? So the answer is 1/8, or 12.5%.
why is the mean of the set of data is always greater than it's median?
Answer:
One of the basic tenets of statistics that every student learns in about the second week of intro stats is that in a skewed distribution, the mean is closer to the tail in a skewed distribution. So in a right skewed distribution (the tail points right on the number line), the mean is higher than the median.
Step-by-step explanation:
The distribution is said to be right-skewed. In such a distribution, usually (but not always) the mean is greater than the median, or equivalently, the mean is greater than the mode; in which case the skewness is greater than zero.
Plz help me solve this algebra problem
Answer:
Exact Form:
x = 2 √ 2 − 1 3 , − 2 √ 2 − 1 3
Decimal Form:
x = 2.49509379 , − 3.16176045
Step-by-step explanation:
Simplify the decimals as needed
Write the following expression in standard place-value form.
Answer:
3408
Step-by-step explanation:
what pairs of degree measures NOT belong to the
group?
Answer:
B. 56° and 38°
Step-by-step explanation:
The sum of each of the two angles sums up to 90° ;
A.) 13° and 77°
Sum of A ; (13 + 77)° = 90°
B.) 56° + 38°
Sum = 56° + 38° = 94°
C) 37° and 53°
Sum = 37° + 53° = 90°
D.) 42° and 48°
Sum = 42° + 48° = 90°
Hence, the pair of degree measure which does not belong is B. 56° and 38°
(4-i)-(3-i) in standard form
Answer:
ifeywifheb
Step-by-step explanation:
hkefewib
On a coordinate plane, a trapezoid has points H prime (negative 3, negative 2), J prime (negative 2, negative 3), K prime (negative 3, negative 4), G prime (negative 5, negative 2).
Trapezoid GHJK was rotated 180° about the origin to determine the location of G'H'J'K', as shown on the graph. What are the coordinates of pre-image point H?
(2, 3)
(–2, 3)
(3, 2)
(3, –2)On a coordinate plane, a trapezoid has points H prime (negative 3, negative 2), J prime (negative 2, negative 3), K prime (negative 3, negative 4), G prime (negative 5, negative 2).
Trapezoid GHJK was rotated 180° about the origin to determine the location of G'H'J'K', as shown on the graph. What are the coordinates of pre-image point H?
(2, 3)
(–2, 3)
(3, 2)
(3, –2)
Answer:
the coordinates of pre - image point H is (3,2)
Answer:
c time for learning
Step-by-step explanation:
I need help anyone please
Answer:
Base: 7
Height:10
BH divided by two so
7 times 10 = 70 divided by two 35
Base: 7 yd
Height: 10 yd
Area: 35 square yd
([tex]A=\frac{h_{b}b }{2}[/tex] = [tex]\frac{10*7}{2}[/tex] = 35)
hope this helps....
Please answer the question in order please.
Answer:
Step-by-step explanation:
how many matches are needed to form figure number 9 and 17? explain
Answer:
Please find the complete question in the attached file.
Step-by-step explanation:
For point 1:
At this point we make up triangles for the first one makes up by 3 matches figure 2 make up by 5 matches figure 3 makes up by 7 matches.
For point 2:
[tex]4 \ \ 5\ \ 6\ \ 7\ \ 8[/tex] you can try to
[tex]9 \ \ 11\ \ 13 \ \ 15 \ \ 17[/tex] draw to find values
For point 3:
number of matches[tex]= 1+2 \times \ figure\ number\\\\[/tex]
For point 4:
[tex]1+2 \times 9=19 \ \ (use \ law \ of\ (iii)) \\\\1+2 \times 17=35[/tex]
Identify the relationship between the graphs of these two equations. y = 5/6x + 5 y = 5/6x - 1
parallel
perpendicular
neither
Answer:
parallel
Step-by-step explanation:
the slope of each line is 5/6 which means the lines are parallel
In ΔTUV, t = 7 inches, u = 9.4 inches and ∠V=18°. Find the area of ΔTUV, to the nearest 10th of a square inch.
Answer:
10 square inches
Step-by-step explanation:
that is the procedure above
Answer:
Area=10.167=10.2
Step-by-step explanation:
Area=1/2ab sin c
area= 1/2(7)(9.4)sin18
area = 10.167=10.2
this is for delta math
Each sample of water has a 30% chance of containing a particular organic pollutant. Assume that the samples are independent with regard to the presence of the pollutant. Find the probability that, in the next 10 samples, two or less contain the pollutant. (Hint: using the tables in Appendix A will be faster than doing it by hand!)
a) 0.678
b) 0.028
c) 0.041
d) 0.383
Answer:
d) 0.383
Step-by-step explanation:
For each sample, there are only two possible outcomes. Either they contain the pollutant, or they do not. The probability of a sample containing the pollutant is independent of any other sample. Thus, the binomial probability distribution is used to solve this question.
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
And p is the probability of X happening.
30% chance of containing a particular organic pollutant.
This means that [tex]p = 0.3[/tex]
Next 10 samples
This means that [tex]n = 10[/tex]
Probability that two or less contain the pollutant.
This is:
[tex]P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)[/tex]. So
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 0) = C_{10,0}.(0.3)^{0}.(0.7)^{10} = 0.028[/tex]
[tex]P(X = 1) = C_{10,1}.(0.3)^{1}.(0.7)^{9} = 0.121[/tex]
[tex]P(X = 2) = C_{10,2}.(0.3)^{2}.(0.7)^{8} = 0.233[/tex]
[tex]P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.028 + 0.121 + 0.233 = 0.382[/tex]
A little rounding difference, but the correct answer is given by option d.
Find the measure of each angle listed below. Enter only numbers without space.
(1) ∠PTQ
(2) ∠QTR
(3) ∠PTS
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Answer:
∠PTQ = 45°∠QTR = 15°∠PTS = 120°Step-by-step explanation:
Parallelogram PTSR is divided into two equilateral triangles by diagonal RT. All of the acute angles in that quadrilateral are 60°, and the obtuse angles are 120° (2×60° and also the supplement of 60°).
Triangle PTQ is an isosceles right triangle, so its acute angles are 45°.
∠PTQ = 45°
∠QTR = ∠PTR -∠PTQ = 60° -45°
∠QTR = 15°
∠PTS = 120° . . . . . . obtuse angle in PTSR; sum of ∠PTR and ∠RTS
A restaurant with 25 tables has 4 tables by the window. Parties are seated randomly at the tables as they come in.
What is the probability that the first 3 parties of the night are all seated at window seats?
Answer:
50/50
Step-by-step explanation:
It's random
What is the 8th term of a(n)=6•3^(n-1)
Answer:
13122
Step-by-step explanation:
a(n) = 6 * 3^(n -1)
a(8) = 6* 3^7
3^7 = 2187
a(8) =6 * 2187
a(8) = 13122
