Given:
• Number of male who survived = 338
,• Number if female sho survived = 316
,• Number f children who survived = 57
,• Number of male who died = 1352
,• Number of female who died = 109
,• Number of children who died = 52
,• Total number of people = 2224
Let's solve for the following:
(a). Probability of the passenger that survived:
[tex]P(\text{survived)}=\frac{nu\text{mber who survived}}{total\text{ number if people }}=\frac{711}{2224}=0.320[/tex](b). Probability of the female.
We have:
[tex]P(\text{female)}=\frac{\text{ number of females}}{total\text{ number }}=\frac{425}{2224}=0.191[/tex](c). Probability the passenger was female or a child/
[tex]P(\text{female or child)}=\frac{425}{2224}+\frac{109}{2224}=\frac{425+109}{2224}=0.240[/tex](d). Probability that the passenger is female and survived:
[tex]P(femaleandsurvived)=\frac{316}{2224}=0.142[/tex](e). Probability the passenger is female and a child:
[tex]P(\text{female and child)=}\frac{425}{2224}\times\frac{109}{2224}=0.009[/tex](f). Probability the passenger is male or died.
[tex]P(male\text{ or died) = P(male) + }P(died)-P(male\text{ and died)}[/tex]Thus, we have:
[tex]P(\text{male or died)}=\frac{1690}{2224}+\frac{1513}{2224}-\frac{1352}{2224}=0.832[/tex](g). If a female passenger is selected, what is the probability that she survived.
[tex]P(\text{survived}|\text{female)}=\frac{316}{425}=0.744[/tex](h). If a child is slelected at random, what is the probability the child died.
[tex]P(died|\text{ child)=}\frac{52}{109}=0.477[/tex](i). What is the probability the passenger is survived given that the passenger is male.
[tex]=\frac{338}{1690}=0.2[/tex]ANSWER:
• (a). 0.320
,• (b). 0.191
,• (c). 0.240
,• (d). 0.142
,• (e). 0.009
,• (f). 0.832
,• (g) 0.744
,• (h). 0.477
,• (i) 0.2
Use the graph below to answer the following questionsnegative sine graph with local maxima at about (-3,55) and local minima at (3,55)1. Estimate the intervals where the function is increasing.2. Estimate the intervals where the function is decreasing.3. Estimate the local extrema.4. Estimate the domain and range of this graph.
Answer:
1. Increasing on ( -inf, -3] and ( 3, inf)
2. decreasing on (-3, 3]
3. Local maximum: 60, Local minimum: -60
4. Domain: (-inf , inf)
Range: [-60, 60]
Explanation:
1.
A function is increasing when its slope is positive. Now, in our case we can see that the slope of f(x) is postive from - infinity to -3 and then it is negatvie from -3 to 3; it again increasing from 3 to infinity.
Therefore, we c
In a robotics competition, all robots must be at least 37 inches tall to enter the competition.Read the problem. Which description best represents the heights a robot must be?Any value less than or equal to 37Any value greater than or equal to 37Any value greater than 37Any value less than 37
Solution
Since the robots must be at least 37 inches tall to enter the competition.
Therefore, the height of any robot must be Any value greater than or equal to 37
Which of the following functions has an amplitude of 3 and a phase shift of pi over 2 question mark
Remember that f(x) = A f(Bx-C) +D
Where |A| is the Amplitude and C/B is the phase Shift
Options
A, B C all have amplitudes of |3| so we have just eliminated D with the amplitude
We need a phase shift of C/B = pi/2
A has Pi/2
B has -Pi/2
C has pi/2 /2 = pi/4
Choice A -3 cos ( 2x-pi) +4 has a magnitude of 3 and and phase shift of pi/2
I resolved this problem for a test already but it looks like the graph it’s not ok can you help me?
SOLUTION
The function given is
[tex]f(x)=2x+1[/tex]To obtain the slope, we compare the equation above with the standard form of a slope intercept form.
Hence,, slope intercept is given as
[tex]\begin{gathered} y=mx+c \\ \text{Where m=slope.c=intercept on y (0,c)} \end{gathered}[/tex]Comparing with the function given, we have
[tex]\begin{gathered} M=2,c=1 \\ \text{Hence } \\ \text{slope}=2,\text{ y-intercept=(0,1)} \end{gathered}[/tex]Therefore
The slope = 2 and the y-intercept= (0,1 )
The graph of the functionis given in the image below
If y varies directly with x and y = 48 when x = -4, write the equation that represents this direct variation relationship12 345
Answer
The equation that represents the direct variation relationship between y and x is
y = -12x
Explanation
We are told that y varies directly with x.
y = 48 when x = -4.
We are then told to write the equation that represents this direct variation relationship.
In mathematical terms, y varies directly with x is written as
y ∝ x
If we introduce a constant of variation, k, we can then write this relationship as
y = kx
To now fully write this relationship, we need to solve for k.
y = 48 when x = -4.
y = kx
48 = k × -4
48 = -4k
-4k = 48
Divide both sides by -4
(-4k/-4) = (48/-4)
k = -12
We can then put in the value of k obtained
y = kx
y = -12x
The equation given is
3y = 10x
Recall that variation is represented as
y ∝ x
And written as
y = kx
So, we can convert 3y = 10x into this form and establish the direct variation and obtain the value of k.
3y = 10x
Divide both sides by 3
(3y/3) = (10x/3)
y = (10x/3)
which is similar to y = kx
k = (10/3)
So, option A is correct.
Hope this Helps!!!
ok so the question is Write an expression to rubbers in the area of the figure the figure is a right triangle with 2X -2 and 4X plus 2 in the answer to that is 4X to the power of 2 - 2X -2 and that's part a and amp RP is what would the area be if X equals negative 2
ANSWERS
a) A = 4x² - 2x - 2
b) if x = -2, A = 18 units²
EXPLANATION
The area of a triangle is the length of the base, multiplied by its height and divided by 2:
[tex]A=\frac{b\cdot h}{2}[/tex]In this triangle, b = 4x + 2 and h = 2x - 2. The area is:
[tex]A=\frac{(4x+2)(2x-2)}{2}[/tex]We can simplify this expression. First we have to multiply the binomials in the numerator:
[tex]\begin{gathered} A=\frac{4x\cdot2x-4x\cdot2+2\cdot2x-2\cdot2}{2} \\ A=\frac{8x^2-8x+4x-4}{2} \\ A=\frac{8x^2-4x-4}{2} \end{gathered}[/tex]Now, using the distributive property for the division:
[tex]\begin{gathered} A=\frac{8x^2}{2}-\frac{4x}{2}-\frac{4}{2} \\ A=4x^2-2x-2 \end{gathered}[/tex]For part b, we just have to replace x with -2 in the expression above and solve:
[tex]\begin{gathered} A=4(-2)^2-2(-2)-2 \\ A=4\cdot4+4-2 \\ A=16+2 \\ A=18 \end{gathered}[/tex]If (2 +3i)^2 + (2 - 3i)^2 = a + bia =b=
(2 + 3i)^2 = 4 + 12i + 9(-1)
= 4 + 12i - 9
= -5 + 12i
(2 - 3i)^2 = 4 - 12i - 9(-1)
= 4 - 12i + 9
= 13 - 12i
REsult
= -5 + 12i + 13 - 12i
= 8 - 0i
Then
a = 8 and b = 0
For which pair of triangles would you use ASA to prove the congruence of the two triangles?
Solution:
Remember that the Angle-Side-Angle Postulate (ASA) states that if two angles and the included side of one triangle are congruent to two angles and the included side of another triangle, then the two triangles are congruent. According to this, the correct answer is:
C.
The one-to-one functions 9 and h are defined as follows.g={(0, 5), (2, 4), (4, 6), (5, 9), (9, 0)}h(x)X +811
Step 1: Write out the functions
g(x) = { (0.5), (2, 4), (4,6), (5,9), (9,0) }
[tex]h(x)\text{ = }\frac{x\text{ + 8}}{11}[/tex]Step 2:
For the function g(x),
The inputs variables are: 0 , 2, 4, 5, 9
The outputs variables are: 5, 4, 6, 9, 0
The inverse of an output is its input value.
Therefore,
[tex]g^{-1}(9)\text{ = 5}[/tex]Step 3: find the inverse of h(x)
To find the inverse of h(x), let y = h(x)
[tex]\begin{gathered} h(x)\text{ = }\frac{x\text{ + 8}}{11} \\ y\text{ = }\frac{x\text{ + 8}}{11} \\ \text{Cross multiply} \\ 11y\text{ = x + 8} \\ \text{Make x subject of formula} \\ 11y\text{ - 8 = x} \\ \text{Therefore, h}^{-1}(x)\text{ = 11x - 8} \\ h^{-1}(x)\text{ = 11x - 8} \end{gathered}[/tex]Step 4:
[tex]Find(h.h^{-1})(1)[/tex][tex]\begin{gathered} h(x)\text{ = }\frac{x\text{ + 8}}{11} \\ h^{-1}(x)\text{ = 11x - 8} \\ \text{Next, substitute h(x) inverse into h(x).} \\ \text{Therefore} \\ (h.h^{-1})\text{ = }\frac{11x\text{ - 8 + 8}}{11} \\ h.h^{-1}(x)\text{ = x} \\ h.h^{-1}(1)\text{ = 1} \end{gathered}[/tex]Step 5: Final answer
[tex]\begin{gathered} g^{-1}(9)\text{ = 5} \\ h^{-1}(x)\text{ = 11x - 8} \\ h\lbrack h^{-1}(x)\rbrack\text{ = 1} \end{gathered}[/tex]How many ways can we arrange five of the seven Harry Potter books on a shelf if Harry Potter and The Chamber of Secrets must be one of them?
There are 7 Harry potter books and 5 books needs to be arranged.
One of the five place is filled by book "Harry Potter and The Chamber of Secrets" and remaining 4 places must be filled by remaining 6 books.
So number of ways are,
[tex]\begin{gathered} 1\cdot^6P_4=1\cdot\frac{6!}{(6-4)!} \\ =1\cdot\frac{6\cdot5\cdot4\cdot3\cdot2\cdot1}{2\cdot1} \\ =1\cdot6\cdot5\cdot4\cdot3 \\ =360 \end{gathered}[/tex]So there are 360 ways in which 5 of 7 Harry pooter book can be arranges such that " Harry Potter and The Chamber of Secrets" must included.
The table shows the number of hours spent practicingsinging each week in three samples of 10 randomlyselected chorus members.Time spent practicing singing each week (hours)Sample 1 45873 56 579 Mean = 5.9Sample 2 68 74 5 4 8 4 5 7 Mean = 5.8Sample 3 8 4 6 5 6 4 7 5 93 Mean = 5.7Which statement is most accurate based on the data?O A. A prediction based on the data is not completely reliable, becausethe means are not the same.B. A prediction based on the data is reliable, because the means ofthe samples are close together.O C. A prediction based on the data is reliable, because each samplehas 10 data points.D. A prediction based on the data is not completely reliable, becausethe means are too close together.
The means of three samples are close together. Therefore, option B is the correct answer.
In the given table 3 sample means are given.
What is mean?In statistics, the mean refers to the average of a set of values. The mean can be computed in a number of ways, including the simple arithmetic mean (add up the numbers and divide the total by the number of observations).
Here, mean of sample 1 is 5.9, mean of sample 2 is 5.8 and mean of sample 3 is 5.7.
Thus, means of these three samples are close together.
The means of three samples are close together. Therefore, option B is the correct answer.
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2. The length of one side of the square is the square root ofits area. Use the table tofind the approximate length of one side of the square. Explain how you used thetable to find this information
we know that
the area of a square is equal to
A=b^2
where
b is the length side
Apply square root both sides of the formula we have
[tex]\sqrt{A}=b[/tex]in the figure below, RTS is an isosceles triangle with sides SR=RT, TVU is an equilateral triangle, WT is the bisected of angle STV, points S, T, and U are collinear, and c= 40 degrees.I'm completely lost and have to answer for a, b+c, f, b+f, a+d, e+g
Step 1: Concept
Triangle SRT is an isosceles triangle with equal base angles a = b
Triangle TUV is an equilateral triangle with all angles equal: g = d = h
Step 2: Apply sum of angles in a triangle theorem to find angle a and b.
[tex]\begin{gathered} a+b+c=180^o \\ c=40^o \\ \text{Let a = b = x} \\ \text{Therefore} \\ x\text{ + x + 40 = 180} \\ 2x\text{ = 180 - 40} \\ 2x\text{ = 140} \\ x\text{ = }\frac{140}{2} \\ x\text{ = 70} \\ a\text{ = 70 and b = 70} \end{gathered}[/tex]Step 3:
2) a = 70
3) b + c = 70 + 40 = 110
Step 4:
Since WT is a bisector of angle STV,
Angle f = e = x
b + f + e = 180 sum of angles on a straight line.
b = 70
70 + x + x = 180
2x = 180 - 70
2x = 110
x = 110/2
x = 55
Hence, f = 55
4) f = 55
5) f + b = 55 + 70 = 125
Step 5:
Since triangle TUV is an equilateral triangle, angle g = h = d = 60
g = 60
h = 60
d = 60
6) Angle a + d = 70 + 60 = 130
7) e + g = 55 + 60 = 115
At noon a private plane left Austin for Los Angeles, 2100 km away, flying at 500 km/h. One hour later a jet left Los Angeles for Austin at 700 km/h. At what time did they pass each other?
Need help with all of them please help me serious
we have 4,5,6
In a right triangle
c^2=a^2+b^2
where
c is the hypotenuse (greater side)
a and b are the legs
In an acte triangle
c^2 < a^2+b^2
we have
c=6
a=4
b=5
substitute
c^2=6^2=36
a^2=4^2=16
b^2=5^2=25
36 < 16+25
36 < 41
therefore
is an acute triangle
Part 2
10,24,26 and also classify the triangle
we have
c=26
a=10
b=24
so
c^2=676
a^2=100
b^2=576
in this problem
c^2=a^2+b^2
therefore
Is a right triangle
I need this practice problem answered I will provide the answer options in another pic
The inverse of a matrix can be calculated as:
[tex]\begin{gathered} \text{When} \\ A=\begin{bmatrix}{a} & {b} & {} \\ {c} & {d} & {} \\ {} & {} & \end{bmatrix} \\ \text{Then A\textasciicircum-1 is:} \\ A^{-1}=\frac{1}{ad-bc}\begin{bmatrix}{d} & -{b} & {} \\ {-c} & {a} & {} \\ {} & {} & \end{bmatrix} \end{gathered}[/tex]Then, let's start by calculating the inverse of the given matrix:
[tex]\begin{gathered} \begin{bmatrix}{4} & {1} & {} \\ {-2} & {3} & {} \\ {} & {} & \end{bmatrix}^{-1}=\frac{1}{4\cdot3-1\cdot(-2)}\begin{bmatrix}{3} & -{1} & {} \\ {-(-2)} & {4} & {} \\ {} & {} & \end{bmatrix} \\ \begin{bmatrix}{4} & {1} & {} \\ {-2} & {3} & {} \\ {} & {} & \end{bmatrix}^{-1}=\frac{1}{14}\begin{bmatrix}{3} & -{1} & {} \\ {2} & {4} & {} \\ {} & {} & \end{bmatrix} \end{gathered}[/tex]The problem says he multiplies the left side of the coefficient matrix by the inverse matrix, thus:
[tex]\begin{gathered} \begin{bmatrix}{4} & {1} & {} \\ {-2} & {3} & {} \\ {} & {} & \end{bmatrix}^{-1}\begin{bmatrix}{4} & {1} & {} \\ {-2} & {3} & {} \\ {} & {} & \end{bmatrix}\cdot\begin{bmatrix}{x} & {} & {} \\ {y} & {} & {} \\ {} & {} & {}\end{bmatrix}=\begin{bmatrix}{4} & {1} & {} \\ {-2} & {3} & {} \\ {} & {} & \end{bmatrix}^{-1}\begin{bmatrix}{2} & {} & {} \\ {-22} & {} & {} \\ {} & {} & {}\end{bmatrix} \\ \end{gathered}[/tex]*These matrices will be the options to put on the first and second boxes.
Then:
[tex]\begin{gathered} \begin{bmatrix}{x} & {} & {} \\ {y} & {} & {} \\ {} & {} & {}\end{bmatrix}=\frac{1}{14}\begin{bmatrix}{3} & -{1} & {} \\ {2} & {4} & {} \\ {} & {} & \end{bmatrix}\cdot\begin{bmatrix}{2} & {} & {} \\ {-22} & {} & {} \\ {} & {} & {}\end{bmatrix}\text{ This is for the third box} \\ \begin{bmatrix}{x} & {} & {} \\ {y} & {} & {} \\ {} & {} & {}\end{bmatrix}=\frac{1}{14}\begin{bmatrix}{3\times2+(-1)\times(-22)} & & {} \\ {2\times2+4\times(-22)} & & {} \\ {} & {} & \end{bmatrix}=\frac{1}{14}\begin{bmatrix}{28} & & {} \\ {-84} & & {} \\ {} & {} & \end{bmatrix}\text{ This is the 4th box} \\ \begin{bmatrix}{x} & {} & {} \\ {y} & {} & {} \\ {} & {} & {}\end{bmatrix}=\begin{bmatrix}{28/14} & & {} \\ {-84/14} & & {} \\ {} & {} & \end{bmatrix}=\begin{bmatrix}{2} & & {} \\ {-6} & & {} \\ {} & {} & \end{bmatrix}\text{ And finally this is the last box} \end{gathered}[/tex]What property of equity is this identify the property : if B is between O and K, BK=OK
Segment addition
The sum of the lenghts of the segments OB and Bk will give the total lenght OK
Hello, may I have help with finding the maximum or minimum of this quadratic equation? Could I also know the domain and range and the vertex of the equation?
To solve this problem, we will use the following graph as reference:
From the above graph, we get that the quadratic equation represents a vertical parabola that opens downwards with vertex:
[tex](3,5)\text{.}[/tex]The domain of the function consists of all real numbers, and the range consists of all numbers smaller or equal to 5.
Answer:
Maximum of 5, at x=3.
Vertex (3,5).
Domain:
[tex](-\infty,\infty).[/tex]Range:
[tex](-\infty,5\rbrack.[/tex]the points (-4,-2) and (8,r) lie on a line with slope 1/4 . Find the missing coordinate r.
The points (-4, -2) and (8, r) are located on a line of slope 1/4, We are asked to find the value of "r" that would make suche possible.
So we recall the definition of the slope of the segment that joins two points on the plane as:
slope = (y2 - y1) / (x2 - x1)
in our case:
1/4 = ( r - -2) / (8 - -4)
1/4 = (r + 2) / (8 + 4)
1/4 = (r + 2) / 12
multiply by 12 both sides to cancel all denominators:
12 / 4 = r + 2
operate the division on the left:
3 = r + 2
subtract 2 from both sides to isolate "r":
3 - 2 = r
Then r = 1
consider the graph of the function f(x)= 10^x what is the range of function g if g(x)= -f(x) -5 ?
SOLUTION
So, from the graph, we are looking for the range of
[tex]\begin{gathered} g(x)=-f(x)-5 \\ where\text{ } \\ f(x)=10^x \\ \end{gathered}[/tex]The graph of g(x) is shown below
[tex]g(x)=-10^x-5[/tex]The range is determined from the y-axis or the y-values. We can see that the y-values is from negative infinity and ends in -5. So the range is between
negative infinity to -5.
So we have
[tex]\begin{gathered} f(x)<-5\text{ or } \\ (-\infty,-5) \end{gathered}[/tex]So, comparing this to the options given, we can see that
The answer is option B
Hello! Need some help on part c. The rubric, question, and formulas are linked. Thanks!
Explanation:
The rate of increase yearly is
[tex]\begin{gathered} r=69\% \\ r=\frac{69}{100}=0.69 \end{gathered}[/tex]The number of lionfish in the first year is given beow as
[tex]N_0=9000[/tex]Part A:
To figure out the explicit formula of the number of fish after n years will be represented using the formula below
[tex]P(n)=N_0(1+r)^n[/tex]By substituting the formula, we will have
[tex]\begin{gathered} P(n)=N_{0}(1+r)^{n} \\ P(n)=9000(1+0.69)^n \\ P(n)=9000(1.69)^n \end{gathered}[/tex]Hence,
The final answer is
[tex]f(n)=9,000(1.69)^n[/tex]Part B:
to figure out the amoutn of lionfish after 6 years, we wwill substitute the value of n=6
[tex]\begin{gathered} P(n)=9,000(1.69)^{n} \\ f(6)=9000(1.69)^6 \\ f(6)=209,683 \end{gathered}[/tex]Hence,
The final answer is
[tex]\Rightarrow209,683[/tex]Part C:
To figure out the recursive equation of f(n), we will use the formula below
From the question the common difference is
[tex]d=-1400[/tex]Hence,
The recursive formula will be
[tex]f(n)=f_{n-1}-1400,f_0=9000[/tex]determine the -domain- and -range- of the graphanswer in interval notation
Explanation: Let's consider two things
- Domain = represented by the minimum and maximum x-values
- Range = represented by the minimum and maximum y-values
Step 1: Let's take a look at the picture below
As we can see above
max x-value = + ∞
min x-value = - ∞
max y-value = 4
min y-value = - ∞
Final answer: So the final answer is
[tex]\begin{gathered} \text{domain}\Rightarrow(-\infty,+\infty) \\ \text{range}\Rightarrow(-\infty,4) \end{gathered}[/tex].
Use properties to rewrite the given equation. Which equations have the same solution as 2.3p – 10.1 = 6.5p – 4 – 0.01p? Select two options. 2.3p – 10.1 = 6.4p – 4 2.3p – 10.1 = 6.49p – 4 230p – 1010 = 650p – 400 – p 23p – 101 = 65p – 40 – p 2.3p – 14.1 = 6.4p – 4
The required equation has the same solution as 2.3p – 10.1 = 6.5p – 4 – 0.01p is 230p – 1010 = 650p – 400 – p.
What is an equivalent expression?Equivalent expressions are even though they appear to be distinct, their expressions are the same. when the values are substituted into the expression, both expressions produce the same result and are referred to be equivalent expressions.
We have the given expression below:
⇒ 2.3p – 10.1 = 6.5p – 4 – 0.01p
Convert the decimal into a fraction to get
⇒ (23/10)p – (101/10) = (65/10)p – 4 – (1/100)p
⇒ (23p – 101)/10 = (650p – 400 – p) /100
⇒ 230p – 1010 = 650p – 400 – p
As a result, the equation that has the same answer as 230p – 1010 = 650p – 400 – p.
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Question
In a pet store, the small fishbowl holds up to 225 gallons of water. The large fishbowl holds up to 213 times as much water as the small fishbowl.
Eloise draws this model to represent the number of gallons of water the large fishbowl will hold.
How many gallons of water does the large fishbowl hold?
The number of gallons that the large fishbowl holds would be = 47,925 gallons.
What are fishbowls?The fishbowls are containers that can be used to transport liquid substance such as water and food products such as fish. This can be measured in Liters, millilitres or in gallons.
The quantity of water the small fishbowl can take = 225 gallons.
The quantity of water the large fish bowl can take = 213(225 gallons)
That is, 213 × 225= 47,925 gallons.
Therefore, the quantity of water that the large fishbowl can hold is 47,925 gallons.
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A 9-foot roll of waxed paper costs $4.95. What is the price per yard ?
Answer:
$0.55 per yard
Step-by-step explanation:
a 9 foot roll is 4.95 so you divide the cost by the amount to get the unit rate which is $0.55 per yard
Translate the sentence into an equation.Eight more than the quotient of a number and 3 is equal to 4.Use the variable w for the unknown number.
We are to translate into an equation
Eight more than the quotient of a number and 3 is equal to 4.
Let the number be w
Hence, quotient of w and 3 is
[tex]\frac{w}{3}[/tex]Therefore, eight more than the quotient of a number and 3 is equal to 4
Is given as
[tex]\frac{w}{3}+8=4[/tex]Solving for w
we have
[tex]\begin{gathered} \frac{w}{3}=4-8 \\ \frac{w}{3}=-4 \\ w=-12 \end{gathered}[/tex]Therefpore, the equation is
[tex]\frac{w}{3}+8=4[/tex]Determine the value for which the function f(u)= -9u+8/ -12u+11 in undefined
ANSWER
[tex]\frac{11}{12}[/tex]EXPLANATION
A fraction becomes undefined when its denominator is equal to 0.
Hence, the given function will be undefined when:
[tex]-12u+11=0[/tex]Solve for u:
[tex]\begin{gathered} -12u=-11 \\ u=\frac{-11}{-12} \\ u=\frac{11}{12} \end{gathered}[/tex]That is the value of u for which the function is undefined.
Translate the sentence into an inequality.Twice the difference of a number and 2 is at least −28.Use the variable x for the unknown number.
To answer this question we have to identify the elements of the inequality.
1. The difference of a number and 2 is represented by the expression: x-2.
2. Twice the difference (...) is represented by the expression: 2(x-2).
3. At least is represented by the sign greater than or equal to ≥.
4. The result is -28.
By putting these all together we obtain the inequality:
[tex]2(x-2)\ge-28[/tex]It means that the answer is 2(x-2) ≥ -28.
the sum of the measure of angle m and angle r is 90
Given:
The sum of measure of angle m and r is 90 degrees.
E is the midpoint of DF, DE = 2x + 4 and EF = 3x - 1 how do I find the value of x, DE, EF and DF
We know that
E is midpoint of DF, that means DE is equal to EF, so we can form the following equation
[tex]DE=EF[/tex]Replacing the given equations, we have
[tex]\begin{gathered} 2x+4=3x-1 \\ 4+1=3x-2x \\ x=5 \end{gathered}[/tex]Now, we replace this value in each equation to find each part of the segment.
[tex]\begin{gathered} DE=2x+4=2(5)+4=10+4=14 \\ EF=3x-1=3(5)-1=15-1=14 \end{gathered}[/tex]Therefore, each part of the segment is 14 units, and DF is 28.