In this problem you can reflect the small triangle and you will see that the angle D is equal to the angle x, and the angle E is equal to the angle B so we can sum tyhe internal angles of the big triangle to find x so:
[tex]x+81+30=180[/tex]And we solve for x so:
[tex]\begin{gathered} x=180-81-30 \\ x=69 \end{gathered}[/tex]the angles x is equal to 69º
Brennan puts 600.00 into an account to use for school expenses the account earns 11%interest compounded annually how much will be in the account after 6 years
Here,
P = 600
t = 6
n = 1 (annually)
r = 11% = 0.11
Applying the fromula to calculate compound interest we have,
[tex]\begin{gathered} A=P(1+\frac{r}{n})^{nt} \\ \text{ =600(1+0.11)}^6 \\ \text{ =1122.248} \end{gathered}[/tex]The answer is 1122.248.
suppose you are buying CDs and DVDs from AMAZON for gifts CDs cost $4 each and DvDs cost 8$ each you want to spend less than 40$ on all of the gifts you need at least 4 gifts altogether graph a system of linear inequalities to model the scenario and give two solutions combinations of CDs and DVDs they could sell to meet their goal
x + y ≤ 4
4x +8y ≤ 40
(1,2) and (2,2) are two possible solutions.
Check the graph below, please.
1) Gathering the data
CD = $4
DVD = $8
budget: $40
2) Notice the word at least. We can write two inequalities, one relating the price of each item and the budget. And the other one relating the number of CDs and DVDs to be bought.
x=CD, y= DVD
x + y ≤ 4 4 CDs and DVDs, altogether
4x +8y ≤ 40 The price of each item, as coefficient and the budget of $40
2.2) Let's plot those inequalities
Let's pick two solutions, for the common region shaded by both graphs.
Like (2, 2) and (1, 2)
3) If we plug into those inequalities we can verify them. So the answers are:
x + y ≤ 4
4x +8y ≤ 40
(1,2) and (2,2) are two possible solutions.
To factor 9x^2 - 4, you can first rewrite the expression as: A. (3x-2)^2B. (x)^2 - (2)^2C. (3x)^2 - (2)^2D. None of the above
We have the expression 9x^2-4.
We know that both terms are squares, so we can express this as:
[tex]9x^2-4=(3x)^2-2^2[/tex]This is as much as we can transform this expression.
The answer is C.
19. If p(x) = 3x^2 - 4 and r(x) = 2x^2 - 5x+1 find -5r*(2a)
We have two polynomials:
[tex]\begin{gathered} p(x)=3x^2-4 \\ r(x)=2x^2-5x+1 \end{gathered}[/tex]We have to find -5*r*(2a). This can be written as:
[tex]-5\cdot r\cdot(2a)=(-5\cdot2a)\cdot r=-10a\cdot r[/tex][tex]\begin{gathered} -10a\cdot r(x)=-10a\cdot(2x^2-5x+1) \\ -10a\cdot2x^2-10a(-5x)-10a\cdot1 \\ -20ax^2+50ax-10a \end{gathered}[/tex]Answer: -5r(2a) = -20ax^2+50ax-10a
y=f(x) is the particular solution to the differential equation dy/dx=(x(y-1))/4, with the initial condition of f(1)=3. write an equation for the line tangent to the graph of f at the point (1,3) and use it to approximate f(1,4)
We are given the following differential equation:
[tex]\frac{dy}{dx}=\frac{x(y-1)}{4}[/tex]Since this equation gives the value of the slope of the tangent line at any point (x,y). To determine the equation of such line we need to use the general form of a line equation:
[tex]y=mx+b[/tex]Since in a tangent line the slope is equivalent to the derivative we may replace that into eh line equation like this:
[tex]y=\frac{dy}{dx}x+b[/tex]Now we determine the value of dy/dx at the point (1,3):
[tex]\frac{dy}{dx}=\frac{(1)(3-1)}{4}=\frac{2}{4}=\frac{1}{2}[/tex]Replacing into the equation of the line:
[tex]y=\frac{1}{2}x+b[/tex]Now we replace the point (1,3) to get the value of "b":
[tex]3=\frac{1}{2}(1)+b[/tex]Solving for "b":
[tex]\begin{gathered} 3=\frac{1}{2}+b \\ 3-\frac{1}{2}=b \\ \frac{5}{2}=b \end{gathered}[/tex]Replacing into the line equation:
[tex]y=\frac{1}{2}x+\frac{5}{2}[/tex]And thus we get the equation of the tangent line.
To approximate the value of f(1.4) we replace the value x = 1.4 in the equation of the tangent line:
[tex]y=\frac{1}{2}(1.4)+\frac{5}{2}[/tex]Solving the operation:
[tex]y=3.2[/tex]Therefore, the approximate value of f(1.4) is 3.2
Mikel creates the table below to help her determine 40 percent of 70
We want to determine the 40 percent of 70, so we have to multiply 70 by 40%:
[tex]\begin{gathered} 40\text{ percent=}\frac{40}{100} \\ 70\cdot40\text{ percent=70}\cdot\frac{40}{100}=\frac{2800}{100}=28 \end{gathered}[/tex]
Ellie earned a score of 590 on Exam A that had a mean of 650 and a standarddeviation of 25. She is about to take Exam B that has a mean of 63 and a standarddeviation of 20. How well must Ellie score on Exam B in order to do equivalently wellas she did on Exam A? Assume that scores on each exam are normally distributed.
Answer
15
Ellie needs to score 15 on Exam B in order to do equivalently as well as she did on Exam A.
Explanation
Since both distributions are normally distributed, we will use the standardized scores of Ellie on both exams to answer this question.
The standardized score for any value is the value minus the mean then divided by the standard deviation.
z = (x - μ)/σ
where
z = standardized score or z-score
x = score in the distribution or in the exam
μ = Mean
σ = Standard deviation
So, we will find the standardized score on Exam A and use that standardized score to find the equivalent score on exam B.
For exam A,
z = ?
x = 590
μ = 650
σ = 25
z = (x - μ)/σ
z = (590 - 650)/25
z = (-60/25) = -2.4
For exam B, to find the equivalent score with a standardized score of -2.4
z = -2.4
x = ?
μ = 63
σ = 20
z = (x - μ)/σ
-2.4 = (x - 63)/20
(x - 63)/20 = -2.4
Cross multiply
x - 63 = (20) (-2.4)
x - 63 = -48
x = 63 - 48
x = 15
Hope this Helps!!!
[tex]x + y = - 2 \\ 3x - y = - 2[/tex]draw each line and estimate the solution.
We can draw each line by assuming that x = 0 and y = 0 and solve each case.
In the first equation, we have the following:
[tex]\begin{gathered} x+y=-2 \\ x=0\Rightarrow y=-2 \\ y=0\Rightarrow x=-2 \end{gathered}[/tex]notice that we have a pair of coordinate points (0,-2) and (-2,0). These two points will be useful when we draw the line.
Next, for the second equation we have:
[tex]\begin{gathered} 3x-y=-2 \\ x=0\Rightarrow-y=-2\Rightarrow y=2 \\ y=0\Rightarrow3x=-2\Rightarrow x=-\frac{2}{3} \end{gathered}[/tex]in this case we have the points (0,2) and (-2/3,0). Now, if we draw both lines on the coordinate plane we get the following:
notice that both lines intersect on the point (-1,-1). Thus, the solution of the system of equations is the point (-1,-1)
I need help with a homework
Consider the triangle PAM and triangle PBM.
[tex]\begin{gathered} \angle PMA=\angle PMB\text{ (Each angle is right angle)} \\ AM=BM\text{ (M is perpendicular bisector of AB)} \\ PM\cong PM\text{ (Common side)} \\ \Delta\text{PMA}\cong\Delta\text{PMB (By SAS similarity)} \\ PA\cong PB\text{ (Corresponding part of Congurent triangle)} \end{gathered}[/tex]Hence it is proved that,
[tex]PA\cong PB[/tex]Solve equations: 1. 3x-4=232. 9-4x=173.6(x-7)=364. 2(x-5)-8= 34
1.
3x-4=23
First, add 4 to both sides of the equation:
3x-4+4 =23+4
3x =27
Divide both sides of the equation by 3.
3x/3 = 27/3
x= 9
A circle has a radius of 5.5A. A sector of the circle has a central angle of 1.7 radians. Find the area of the sector. Do not round any intermediate computations. Round your answer to the nearest tenth
Answer:
The circle has the following parameters:
[tex]\begin{gathered} \text{Radius = }5.5ft \\ \text{Angle = 1.7 Radians} \end{gathered}[/tex]We have to figure out the area of the sector of this circle that has the given angle and radius.!
[tex]\begin{gathered} A(\text{sector) = }\frac{1.7r}{2\pi r}\times2\pi(5.5)^2ft^2 \\ =(1.7\times5.5)ft^2 \\ =9.35ft^2 \end{gathered}[/tex]This is the area of the sector that we were interested in.!
given two sides of a triangle, find a range of possible lengths for the third side.9yd, 32yd
We have to use the Triangle Inequality Theorem, which states that any of the 2 sides of a triangle must be a greater sum than the third side.
So, to find the correct range of lengths, we have to use the difference of the two sides and their addition to calculate the interval.
[tex]\begin{gathered} 32-9Therefore, the range of possible lengths is 23I need help with my math prep
Answer:
I can help!
Step-by-step explanation:
The table below gives the grams of fat and calories in certain food items. Use this data to complete the following 3 question parts.Fat (x)31391934432529Calories580680410590660520570b. Describe the correlation seen in the scatter plot.is it positve or negative or no correlation?
b. We can see throught the scatter plot that as the grams of fat increaseas, so does the calories in certain food, therefore there is a directly proportion relationship between them.
It is a positive relationship because while one increases, the other one too.
which value of your makes the equation 8.6 + y = 15 true?y=23.6y=1.7y=129y=6.4
y=6.4
1) Let's find out then, solving that equation.
8.6 + y = 15 Subtract 8.6 from both sides
y= 15 -8.6
y=6.4
8.6 + 6.4 = 15
15 = 15 True
2) Hence the value of y that makes it true is 6.4
My questions are: #1) Determine the account balance at 4 years if $20,000 was invested in an account that compounds daily at 4.5% per year.#2) Determine the account balance at 5 years if $20,000 was invested in an account that compounds continuously at 4.5% per year.#3) A bacterial culture grows from 10 bacteria at 1.5% per minute starting at 7:00 a.m. find bacteria count after 12 hours if continues growth is assumed. (round down to the nearest whole bacterium)
We have the following formula:
[tex]P(t)=10\cdot(1.015)^t[/tex]where t is the amount of minutes we have waited. So in this case we have 12hours, therefore we have waited 12*60=720 minutes
so we have that after 720 minutes the population of bacteria is
[tex]10\cdot(1.015)^{720}=452428.98\approx452429[/tex]so the answer is 452429
19. We saw 10 ladybugs. We saw 5 butterflies. How many more ladybugs than butterflies did we see?
Given that,
Total ladybugs = 10
Total butterflies = 5
More ladybugs than butterflies are = ladybugs - butterflies
=> 10 - 5
=> 5
Therefore, there will be 5 more ladybugs than butterflies.
Rich is attending a 4-year college. As a freshman, he was approved for a 10-year, federal unsubsidized student loan in the amount of $7,900 at 4.29%. He knows he has the
option of beginning repayment of the loan in 4.5 years. He also knows that during this non-payment time, Interest will accrue at 4.29%.
If Rich decides to make no payments during the 4.5 years, the Interest will be capitalized at the end of that period.
a. What will the new principal be when he begins making loan payments?
b. How much interest will he pay over the life of the loan?
The $7,900 10 year, federal unsubsidized student loan has a new principal and interest paid as follows;
a. The new principal of the loan after 4.5 years is approximately $9,543.75
b. The interest on the loan is approximately $3,016.27
What are unsubsidized student loans?Unsubsidized loans are loans that are not based on financial need of undergraduate and graduate students.
The future value of the loan is found using the formula;
[tex]FV = PV\cdot \left(1+\dfrac{r}{100} \right)^n[/tex]
Where;
FV = The future value of the loan
PV = The present value of the loan = $7,900
r = The interest rate of the loan = 4.29%
n = The number of years = 4.5 years
Which gives;
[tex]FV = 7900\times \left(1+\dfrac{4.29}{100} \right)^{4.5}\approx 9543.75[/tex]
The new principal of the loan when he begins to make loan payments is $9,543.75
b. The payment (amortization) formula is presented as follows;
[tex]A = P\cdot \dfrac{r\cdot (1+r)^n}{(1+r)^n-1}[/tex]
Which gives;
[tex]A = 9543.75\times \dfrac{0.0429\cdot (1+0.0429)^{5.5}}{(1+0.0429)^{5.5}}{-1} \approx 1984.78[/tex]
The amount paid annually ≈ $1,984.78
The amount paid in 5.5 years ≈ 1984.78 × 5.5 = 10,916.27
The interest paid = $10,916.27 - $7,900 = $3,016.27
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R(-2,3) S(4,4) T(2,-2) state the coordinates of R'S'T' after a dilation of 2
We are given the following coordinates.
R(-2,3)
S(4,4)
T(2,-2)
We are asked to state the coordinates of R'S'T' after dilation of 2
A dilation of 2 means that we have to multiply the original coordinates (RST) by 2 to get the new coordinates (R'S'T')
Since the scale factor is 2 (greater than 1) the new image will result in enlargement.
Please note that with dilation the figure remains the same only the size of the image changes.
The new coordinates (R'S'T') are
R'(-2×2, 3×2) = (-4. 6)
S'(4×2, 4×2) = (8, 8)
T'(2×2, -2×2) = (4, -4)
Therefore, the
consider the relationship between f(x)=2^x and g(x)=log2 x.g is a reflection of f over the line y=x.True or False
the function
[tex]\log _2x[/tex]is the inverse function of
[tex]2^x^{}[/tex]On the graph, the inverse of a function is the reflection of the original function over the line y = x. Then, the statement is true
Angel's bank gave her a 4 year add on interest loan for $7,640 to pay for new equipment for her antiques restoration business. The annual interest rate is 5.27% How much interest will she pay on the loan? $_______How much would her monthly payments be? $________(round to nearest cent)
From the information given,
Principal = 7640
Number of years = 4
Recall, 1 year = 12 months
4 years = 4 * 12 = 48 months
rate = 5.27/100 = 0.0527
The amount of the principal to be paid each month = 7640/48 = 159.17
Amount of interest owed each month = (7640 * 0.0527)/12 = 33.55
Amount required to be paid by the borrower each month = 159.17 + 33.55 = 192.72
Total interest to be paid on the loan = 7640 * 0.0527 * 4 = 1610.51
The first answer is $1610.51
The second answer is $33.55
A job placement agency advertised that last year its clients, on average, had a starting salary of $39,500. Assuming that average refers to the mean, which of the following claims must be true based on this information?Note: More than one statement could be true. If none of the statements is true, mark the appropriate box.Last year some of their clients had a starting salary of at least $39,500 .Two years ago some of their clients had a starting salary of at least $39,500 .Last year, the number of their clients who had a starting salary of more than $39,500 was equal to the number of their clients who had a starting salary of less than $39,500.Last year at least one of their clients had a starting salary of more than $42,000.Last year at least one of their clients had a starting salary of exactly $39,500.None of the above statements are true.
In the question, it is given that the average salary is $39,500.
In consideration of the first statement
(a) , last year, some of their clients had a starting of atleast $39500 ...this is true
(b) they have mentioned the case of last two years, this is also incorrect.
( c )if the client has lesser than $39,500 salary, and the average salary is $39,500 , then average will be less than $39,500, then statement a is not true..
(d) Last year at least one of their clients had a starting salary of more than $42,000., this is more than the average , but could be true, but it is false as $42000 will be more than the average of $39,500
(e) Last year at least one of their clients had a starting salary of exactly $39,500. ... this is not true, as exactly would not allow the $39,500 to be any less.
• So correct options would be A
English Do the head bean to see how many Ms Elkot has gallons of gas in her, and the car uses 1 of a gallon of gas on the drive to work How can Ms Emo Egure out how many trips to work she can make? Check all that apply use the expression 6/8 / 1/4 to find the answer 3 orange parts fit on the blue parts 2 blue parts fit on the orange part Ms Elliot can make 2 trips to school Ms Ellot can make 3 trips to school
Since she needs 1/4 of gallons and she has 6/8 gallons, then she can use the expression
[tex]\frac{6}{8}\text{ \%}\frac{1}{4}[/tex]to find out haw many trips she can make
Function gis represented by the equation.915) = –18(3) *+ 2Which statement correctly compares the two functions on the interval [-1, 2]?
step 1
Find out the average rate of change function f over the interval [-1,2]
[tex]\frac{f(b)-f(a)}{b-a}[/tex]we have
a=-1
b=2
f(a)=f(-1)=-22
f(b)=f(2)=-1
substitute
[tex]\frac{-1-(-22)}{2-(-1)}=\frac{21}{3}=7[/tex]step 2
Find out the average rate of change function g(x) over the interval [-1,2]
we have
a=-1
b=2
g(a)=g(-1)=-18(1/3)^-1+2=-52
g(b)=g(2)=-18(1/3)^2+2=0
substitute
[tex]\frac{0-(-52)}{2-(-1)}=\frac{52}{3}=17.3[/tex]therefore
17>7
the answer is option AFind the area of quadrilateral math with vertices M(7, 6), A(3, - 2), T(- 7, 1) and H(- 1, 9)
Lets draw a picture of our quadrilateral:
In order to find the area, we can divide our parallelogram in 2 triangles:
The area of triangle AHT is given by
[tex]\text{Area }\Delta AHT=\frac{1}{2}(x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2))[/tex]where
[tex]\begin{gathered} (x_1,y_1)=(3,-2)=A \\ (x_2,y_2)=(-1,9)=H \\ (x_3,y_3)=(-7,1)=T \end{gathered}[/tex]By substituting these points into the given formula, we get
[tex]\text{Area }\Delta AHT=\frac{1}{2}(3_{}(9_{}-(-7))-1((-7)-(-2))-7((-2)-9))[/tex]which gives
[tex]\begin{gathered} \text{Area }\Delta AHT=\frac{1}{2}(3_{}(16)-1(-5)-7(-11)) \\ \text{Area }\Delta AHT=\frac{1}{2}(48+5+77) \\ \text{Area }\Delta AHT=\frac{130}{2} \\ \text{Area }\Delta AHT=65 \end{gathered}[/tex]Similarly, for the area of triangle AHM, we can choose
[tex]\begin{gathered} (x_1,y_1)=(3,-2)=A \\ (x_2,y_2)=(-1,9)=H \\ (x_3,y_3)=(7,6)=M \end{gathered}[/tex]By substuting in our area formula, we get
[tex]\text{Area }\Delta AHM=\frac{1}{2}(3_{}(9_{}-6)-1(6-(-2))+7((-2)-9))[/tex]which gives
[tex]\begin{gathered} \text{Area }\Delta AHM=\frac{1}{2}(3_{}(3)-1(8)+7(-11) \\ \text{Area }\Delta AHM=\frac{1}{2}(9-8-77) \\ \text{Area }\Delta AHM=\frac{76}{2} \\ \text{Area }\Delta AHM=38 \end{gathered}[/tex]Then, the total area is given by
[tex]\begin{gathered} A=\text{Area }\Delta AHT+\text{Area }\Delta\text{AHM} \\ A=65+38 \\ A=103 \end{gathered}[/tex]then, the answer is 103 units squared.
If the side adjacent to the 55° angle is five units, what equation should be used to solve for the hypotenuse
For a given Right angled triangle, equation for hypotenuse is,
AB = 5/ Sin55°
Right angled triangle :
A right triangle or right-angled triangle, is a triangle in which one angle is a right angle, i.e., in which two sides are perpendicular. The relation between the sides and other angles of the right triangle is the basis for trigonometry. Right angled triangle is also known as an orthogonal triangle, or rectangled triangle.
In a right angle triangle ΔABC
m∠A = 55°
BC = 5 units
SinA = BC / AB
sin55° = 5/AB
or,
AB = 5/ Sin55°
as, Sin55° = 0.81915204
AB = 5/0.81915204
AB = 6.1038
Thus,
For a given Right angled triangle, equation for hypotenuse is,
AB = 5/ Sin55°
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What is the probability it lands between birds B and C?
B. 1/9
Explanation
The probability of an event is the number of favorable outcomes divided by the total number of outcomes.
[tex]P(A)=\frac{favorable\text{ outcomes}}{\text{total outcomes}}[/tex]Step 1
Let A represents the event that the birds lands between b and c
a)so, in this case the favorable outcome is that the birds lands between b and c, the length bewtween b and c is
[tex]BC=2\text{ in}[/tex]and , the total outcome is the total lengt, so total outcome = AD
[tex]\begin{gathered} AD=10\text{ in+ 2 in +6 in} \\ AD=18\text{ in} \end{gathered}[/tex]b) now,replace in the formula
[tex]\begin{gathered} P(A)=\frac{favorable\text{ outcomes}}{\text{total outcomes}} \\ P(A)=\frac{2i\text{n }}{18\text{ in}}=\frac{1}{9} \\ P(A)=\frac{1}{9} \end{gathered}[/tex]therefore, the answer is
B. 1/9
I hope this helps you
1-3. Adaptive Practice Powered by KnewtonDue 1Goal ♡ ♡ ♡A square rug has an area of 121 ft2. Write the side length as a square root. Then decide if the sidelength is a rational number.The rug has side length7 ft.Is the side length a rational number?YesNoView progressSubmit and continue
Area of a square: Side length ^2
121 = s^2
Solve for s ( side length)
√121 ft= s
Rational numbers can be expressed as a fraction of 2 integers.
√121=11 =11/1
The side length is √121 ft and is a rational number.
The quotient of 93 and x
The quotient of ;
[tex]undefined[/tex]Question 3. Y=(1/5)^xSketch the graph of each of the exponential functions and label three points on each graph.
Given exponential function:
[tex]y\text{ = (}\frac{1}{5})^x[/tex]Let us obtain three points including the y-intercept so that we can plot the function y = f(x)
When x =0:
[tex]\begin{gathered} y\text{ = (}\frac{1}{5})^0 \\ =\text{ 1} \end{gathered}[/tex]when x =1:
[tex]\begin{gathered} y\text{ = (}\frac{1}{5})^1 \\ =\text{ }\frac{1}{5} \end{gathered}[/tex]when x =2:
[tex]\begin{gathered} y\text{ = (}\frac{1}{5})^2 \\ =\text{ }\frac{1}{25} \end{gathered}[/tex]We have the points : (0, 1), (1, 1/5), and (2, 1/25)
Using these points, let us provide a sketch of the plot of y =f(x). We have the plot as shown below: