To answer this questions we need to remember the standard score formula given by:
[tex]z=\frac{x-\mu}{\sigma}[/tex]where x is the value we are looking for, mu si the mean and sigma is the standard deviation.
a.
We need the probability:
[tex]P(IQ>95)[/tex]using the standard score this is equivalent to:
[tex]\begin{gathered} P(IQ>95)=P(z>\frac{95-100}{15}) \\ =P(z>-0.3333) \end{gathered}[/tex]Using a normal distribution table we have:
[tex]P(z>-0.3333)=0.6306[/tex]Therefore the probability to select a person with more than 95 IQ points is 63.1%.
b.
Following the same reasoning as before we have:
[tex]\begin{gathered} P(IQ<125)=P(z<\frac{125-100}{15}) \\ =P(z<1.6667) \\ =0.9522 \end{gathered}[/tex]Therefore the probability to select a person with less than 125 IQ points is 95.2%
c.
To find how many people of this sample have more less than 110 points we need to find that probability:
[tex]\begin{gathered} P(IQ<110)=P(z<\frac{100-110}{15}) \\ =P(z<-0.666) \\ =0.7475 \end{gathered}[/tex]Multiplying this value with the sample size we have
[tex](800)(0.7475)=598[/tex]Therefore 598 people will have an IQ less than 110.
d.
By the same reasoning as before we have:
[tex]\begin{gathered} P(IQ>140)=P(z>\frac{140-100}{15}) \\ =P(z>2.6667) \\ =0.0038 \end{gathered}[/tex]Multiplying this value with the sample size we have
[tex](800)(0.0038)=3[/tex]Therefore 3 people will have an IQ greater than 140.
4) A cannonball is shot out of a cannon at a 459angle with an approximatecannon from which the ball was fired sits on the edge of a cliff, and its he20 meters. The equations given below represent the cannonball's heighand its horizontal distance (x) from the face of the cliff, (E)seconds afterHow many seconds after the ball was fired does its verticat height abovehorizontal distance from the cliff?
Let t be the amunt of seconds that have passed when the height of the cannonball above the ground is the same as its horizontal fistance from the cliff.
Since the height of the cannonball above the ground is represented using the variable y and the horizontal distance from the cliff is represented using the variable x, then, the condition that the height equals the horizontal distance can be expressed as:
[tex]y=x[/tex]Replace the expressions for y and x in terms of t into the equation:
[tex]-5t^2+2t+20=2t[/tex]We obtained a quadratic equation on the variable t.
Notice that the term 2t appears in both members of the equation. Then, it can be cancelled out:
[tex]-5t^2+20=0[/tex]Solve for t²:
[tex]\begin{gathered} \Rightarrow-5t^2=-20_{} \\ \Rightarrow t^2=\frac{-20}{-5} \\ \Rightarrow t^2=4 \end{gathered}[/tex]Take the square root to solve for t:
[tex]\begin{gathered} \Rightarrow t=\pm\sqrt[]{4} \\ =\pm2 \end{gathered}[/tex]Since t must be greater or equal to 0, then the negative solution should be discarded.
Therefore, the vertical height of the cannonball equals its horizontal distance from the cliff 2 seconds after the ball is fired.
The correct choice is option B) 2
Find the sum: (502 + 8d + )+(502 + 3d + 4)
The sum:
(502 + 8d ) + ( 502 + 3d + 4 )
Clearing the brackets, we get,
502 + 8d + 502 + 3d + 4
Collecting the like terms, we get,
8d + 3d + 502 + 502 + 4
11d + 1008
The correct answer: 11d + 1008
Which formula is used to determine the standard normal random variable (Z)?
The standard normal random variable Z can be calculated using the formula:
[tex]Z=\frac{x-\mu}{\sigma}[/tex]Where x is the input, μ is the mean and σ is the standard deviation.
Therefore the correct option is the first one.
two cylinders have the same volume the first cylinder has a diameter of 10 cm and a height of 30 cm The second cylinder has a diameter of 8 cm what is the height of the second cylinders the nearest tenth of a centimer
Solution
Given that two cylinders have the same volume but different dimension
For Cylinder 1
Diameter is 10 cm Height is 30 cm
For Cylinder 1
Diameter is 8 cm Height is h cm
The volume of a cylinder is given as;
[tex]V=\pi r^2h[/tex]Since the two cylinders have the same volume,
Since radius = Diameter/2
[tex]\begin{gathered} \pi\times(\frac{10}{2})^2\times30=\pi\times(\frac{8}{2})^2\times h \\ \Rightarrow5^2\times30=4^2\times h \\ \Rightarrow h=\frac{25\times30}{16}\approx46.9 \end{gathered}[/tex]Hence, the second cylinder the nearest tenth of a centimeter is 46.9 cm
Part A: which of the following can be used to find the measure of angles
We shall begin by calculating the value of x, as that would help us to find the angle measure of each angle.
[tex]\begin{gathered} \angle C+\angle D+\angle E=180 \\ x-5+2x-3+x=180 \\ 4x-8=180 \\ 4x=180+8 \\ 4x=188 \\ x=\frac{188}{4} \\ x=47 \\ \angle C=x-5 \\ \angle C=47-5 \\ \angle C=42 \\ \angle D=2x-3 \\ \angle D=2(47)-3 \\ \angle D=94-3 \\ \angle D=91 \\ \angle E=x \\ \angle E=47 \end{gathered}[/tex]The triangle is a scalene triangle (all angles are different in measure)
(1) Part A; we can find the angles using the Triangle angle-sum theorem
(2) Part B; measure of each angle as shown as;
The angles are;
C = 42
D = 91
E = 47
Divide 1/4 ÷ 2/3 and express the answer in simplest terms.
Given the expression
1/4 ÷ 2/3
This is expressed as 1/4 * 3/2
multiply the numerator and denominator together to have;
1/4 * 3/2
= (1*3)/(4*2)
= 3/8
Hence the expression in its simplest form is 3/8
When purchasing bulk orders of batteries, a toy manufacturer uses this acceptance sampling plan: Randomly select and test 42 batteries and determine whether each is within specifications. The entire shipment is accepted if at most 2 batteries do not meet specifications. A shipment contains 7000 batteries, and 2% of them do not meet specifications. What is the probability that this whole shipment will be accepted? Will almost all such shipments be accepted, or will many be rejected? The probability that this whole shipment will be accepted is (Round to four decimal places as needed.) The company will accept % of the shipments and will reject % of the shipments, so (Round to two decimal places as needed.)
Solution
Here, we would calculate the probability that one is defective, two are defective, three are defective and the probability that none are defective and sum them together.
The probability that any individual battery is not defective = 1 - 2% = 0.98
We need 42 of them,
Therefore, P(that none are defective) = (0.98)^42 = 0.4281
The probability that a specific battery will be the only defective battery is (0.02)*(0.98)^41
Since we have 42 of them, ((0.02)*(0.98)^41)*42 = 0.367
Note that
[tex]42C2=\frac{42!}{(42-2)!2!}=861[/tex]=> ((0.02)^2*(0.98)^40)*861 = 0.1535
Therefore, 0.4281 + 0.367 + 0.1535 = 0.9486
The probability shows that about 94.86% of all shipments will be accepted
what are the roots of the quadratic equation below?[tex]3 {x}^{2} + 9x - 2 = 0[/tex]
Given:
A quadratic equation is:
[tex]3x^2+9x-2=0[/tex]Find-:
The roots of the quadratic equation
Explanation-:
Use quadratic formula:
[tex]ax^2+bx+c=0[/tex]Roots of the equation,
[tex]x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]The roots of the given equation are:
[tex]3x^2+9x-2=0[/tex][tex]\begin{gathered} x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a} \\ \\ x_{1,2}=\frac{-9\pm\sqrt{9^2-4(3)(-2)}}{2(3)} \\ \\ x_{1,2}=\frac{-9\pm\sqrt{81-(-24)}}{6} \\ \\ x_{1,2}=\frac{-9\pm\sqrt{81+24}}{6} \\ \\ x_{1,2}=\frac{-9\pm\sqrt{105}}{6} \end{gathered}[/tex]The roots of a quadratic equation are:
[tex]\begin{gathered} x_{1,2}=\frac{-9\pm\sqrt{105}}{6} \\ \\ x_1=\frac{-9+\sqrt{105}}{6},x_2=\frac{-9-\sqrt{105}}{6} \end{gathered}[/tex][tex]\begin{gathered} x_1=\frac{-9+\sqrt{105}}{6},x_2=\frac{-9-\sqrt{105}}{6} \\ \\ x_1=0.2078,x_2=-3.2078 \end{gathered}[/tex]The roots of a quadratic equation are 0.2078 and -3.2078.
A standard deck has 52 cards. Half the cards are black and half are red. There are four suits: clubs, spades, hearts, and diamonds. Each suit has 13 cards. There are 4 of each number card, and 4 of each face card (jack, queen, king).Use the image below to help you answer (Part A and Part B)about a deck of cards.PART A: You randomly choose one card from a standard deck. What is the probability that you’ll get a queen? Simplify the fraction.1/161/131/41/2PART B: You randomly choose one card from a standard deck. What is the probability that you will choose a heart? 10%15%25%33%
Pr(queen) = 1/13
Pr(heart) = 25%
Explanation:Part A:
Total cards in a standard deck = 52
from the diagram, there are 4 queens:
1 club queen, 1 spade queen, 1 heart queeen, 1 diamond queen
Probability of getting a queen = number of queens/total number of cards
[tex]\begin{gathered} Pr(queen)\text{ = }\frac{4}{52} \\ Pr(\text{queen) = 1/13} \end{gathered}[/tex]Part B:
from the diagram, there are 13 hearts:
There are 4 suits, each of them has 13 cards. One of the suits is a heart. This means there are 13 cards with hearts
Probability of getting a heart = number of hearts/total number of cards
[tex]\begin{gathered} Pr(\text{heart) = }\frac{13}{52} \\ Pr(\text{heart) = 1/4} \\ In\text{ percentage, 1/4 = }\frac{1}{4\text{ }}\times\text{ 100\% = 25\%} \\ \\ Pr(\text{heart) = 25\%} \end{gathered}[/tex]hi can someone help me solve this thank you!!
For the given inequality, the smallest value of x is a whole number which is exactly divisible by both 2 and 5.
As given in the question,
Given inequality :
[tex]4x/5 - 1x/2 > 18[/tex]
[tex][2(4x) + 5(-x)]/5.2 > 18[/tex]
[tex](2.4x - 5x) / 5.2 > 18[/tex]
[tex](8x-5x)/10 > 18[/tex]
[tex]3x/10 > 18[/tex]
[tex]3x > 180 \\3x/3 > 180/3\\x > 60\\[/tex]
x ∈ (60, ∞)
∵ Since 60 is whole number and divisible by 2 and 5 both.
therefore answer is option e.
Therefore, for the given inequality, the smallest value of x is a whole number which is exactly divisible by both 2 and 5.
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Drag the tiles to the correct boxes to complete the pairs. Not all tiles will be used.Match each quadratic equation with its solution set.2x^2–8x+5=02x^2-10x-3=02x^2-8x-3=02x^2-9x-1=02x^2-9x+6=0
Solution
For this case we have the following equations:
[tex]2x^2-8x+5=0[/tex]The solutions are:
[tex]x=\frac{4\pm\sqrt[]{6}}{2}[/tex][tex]2x^2-10x-3=0[/tex]Solutions are:
[tex]x=\frac{5\pm\sqrt[]{31}}{2}[/tex][tex]2x^2-8x-3=0[/tex]Solutions are:
[tex]x=\frac{4\pm\sqrt[]{22}}{2}[/tex][tex]2x^2-9x-1=0[/tex]Solutions are:
[tex]x=\frac{9\pm\sqrt[]{89}}{4}[/tex][tex]2x^2-9x+6=0[/tex]Solutions are:
[tex]x=\frac{9\pm\sqrt[]{33}}{4}[/tex]Then final solutions are:
[tex]\frac{9\pm\sqrt[]{33}}{4}\Rightarrow2x^2-9x+6=0[/tex][tex]\frac{4\pm\sqrt[]{6}}{2}\Rightarrow2x^2-8x+5=0[/tex][tex]\frac{9\pm\sqrt[]{89}}{4}\Rightarrow2x^2-9x-1=0[/tex][tex]\frac{4\pm\sqrt[]{22}}{2}\Rightarrow2x^2-8x-3=0[/tex]Graph triangle ABC with vertices A(0,5) B(4,3) and C(2,-1) and it’s image after a reflection in the line y=2
By reflecting the given points about y=2, the x-coordinates remain the same, then we have
[tex]\begin{gathered} (0,5)\longrightarrow(0,-1) \\ (4,3)\longrightarrow(4,1) \\ (2,-1))\longrightarrow(2,5) \end{gathered}[/tex]then, the new triangle is:
where the red line represents the line y=2. We found the new points by searching the points which are at the same distance of the original poinst to the line:
Then, the preimage and the image after the reflection are:
vertices abc are a(-4,5), b(-2,4), c(-3,2) if abc is reflected across the line y= -2 to produce the image abc; find the coordinates of vertex A
So, the coordinates of the new vertex A must be
[tex](-4,5-14)=(-4,-9)[/tex]So, the coordinates of vertex A is (-4,-9)
graph f(x)=(x-4)^3-2
Explanation:
Before we plot the graph, you need to understand the transformation process of the parent function x³
x - 4 inside the parenthesis talks about the horizontal shift of the graph. -4 means that the graph will move to the right by 4.
The constant value outside the parenthesis talks about the graph movement. The constant -2 shows that the graph moves downward by 2.
Find the graph of the function below:
3.8 times 24 long multipilcalion
Craig like to collect vinyl records. Last year he ahead 10 records in his collection. Now he has 12 records. What is the percent increase?
Last year, Craig had 10 records.
Now, he has 12 records.
What is the percent increase?
The percent increase is given by
[tex]\%\: increase=\frac{\text{new value-old value}}{\text{old value}}\times100[/tex]In this case,
Old value = 10 records
New value = 12 records
[tex]\begin{gathered} \%\: increase=\frac{\text{new value-old value}}{\text{old value}}\times100 \\ \%\: increase=\frac{12-10}{10}\times100 \\ \%\: increase=\frac{2}{10}\times100 \\ \%\: increase=20 \end{gathered}[/tex]Therefore, there is a 20% increase in his record collection.
Each of the four graphs below represents a function.Which function has an inverse that is also a function?
For the given question, we will choose the function that has an inverse that will be a function
So, the function must be a one-to-one function
We will use the horizontal line test to check if the function is a one-to-one function or not
the horizontal line will intersect with the graph of the function at only one point
So, the answer will be option 4 as shown in the following figure:
In the figure below, c || d. Classify each of the following angle pairs, and tell whether they arecongruent or supplementary.6. 21 and 23Supp.7. 26 and 23Supp.8. 21 and 283/47/8Supp.5/6t9. 27 and 24Supp.10. 22 and 21& Supp.
Two angles are congruent if they are equal and supplementary if there sum is 180.
Given data:
c and d are parallel.
Now
[tex]\angle1,\angle3[/tex]are corresponding angles, so they are equal.
So,
[tex]\angle1,\angle3[/tex]form a congruent pair.
Now since
[tex]\angle1=\angle6[/tex]since they are vertically oppsoye angles.
And,
[tex]\angle1=\angle3(\text{corresponding angles)}[/tex]So,
[tex]\angle6=\angle3[/tex]So,
[tex]\angle6,\angle3[/tex]form a congruent pair.
Now,
[tex]\begin{gathered} \angle3=\angle8(vertically\text{ opposite angles)} \\ \angle1=\angle3(corresponding\text{ angles)} \\ \Rightarrow\angle1=\angle8 \end{gathered}[/tex]So,
[tex]\angle1,\angle8[/tex]form a congruent pair.
[tex]\angle7=\angle4(vertically\text{ opposite angles)}[/tex]So,
[tex]\angle7,\angle4[/tex]form a congruent pair.
Now,
[tex]\angle1+\angle2=180(linear\text{ pair)}[/tex]So,
[tex]\angle1,\angle2[/tex]form a supplementary pair.
Find the length of the legs.(4,4)[](-2,-2)[?]Enter the number thatbelongs in the green boxEnter
We have a right triangle and we have to find the length of the legs.
We first analyse the hypotenuse.
It is a line that is defined by two points: (4,4) and (-2,-2).
The line passes through the center of coordinates, but we will check it:
The slope can be calculated as:
[tex]m=\frac{y_2-y_1}{x_2-x_1}=\frac{4-(-2)}{4-(-2)}=1[/tex]We can now write the slope-point equation as:
[tex]\begin{gathered} y-y_0=m(x-x_0) \\ y-4=1(x-4) \\ y-4=x-4 \\ y=x \end{gathered}[/tex]Then, we know that the equation of the line that defines the hypotenuse is y=x.
If the legs are parallel to the axis, we can find the angle that is defined by the hypotenuse and the horizontal leg as:
[tex]\begin{gathered} \tan (\theta)=m=1 \\ \theta=\arctan (1)=45\degree \end{gathered}[/tex]As we have an angle of 45 degree for one leg, the other has to have an equal angle, as 180-90-45=45 (Note: we are substracting from the sum of the 3 angles, 180 degrees, the already known angles, 90 and 45 degrees, so we are left with 45 degrees that correspond to the third angle).
If the two angles for the legs are equal, their lengths are equal too.
So we can start by calculating the length of the hypotenuse: it is the distance between (4,4) and (-2,-2).
[tex]\begin{gathered} D=\sqrt[]{(x_2-x_1)^2+(y_2-y_1)^2}=\sqrt[]{(4-(-2))^2+(4-(-2))^2} \\ D=\sqrt[]{6^2+6^2}=\sqrt[]{2\cdot6^2}=6\sqrt[]{2} \end{gathered}[/tex]Then, we can write the Pythagorean theorem as (a and b are the legs, and c is the hypotenuse):
[tex]\begin{gathered} a^2+b^2=c^2 \\ a=b\longrightarrow2a^2=c^2=(6\sqrt[]{2})^2=36\cdot2=72 \\ a^2=\frac{72}{2} \\ a^2=36 \\ a=\sqrt[]{36} \\ a=b=6 \end{gathered}[/tex]Answer: the length of the legs is 6 units.
you have two fractions 2/5 and 3/10 and you want to rewrite them so that they have the same denominator what numbers could you use for the denominator is it A 20 or B 10 or C5 or D 15
By definition, a fraction has the following form:
[tex]\frac{a}{b}[/tex]When "a" is the numerator and "b" is the denominator.
In this case you have the following fractions:
[tex]\begin{gathered} \frac{2}{5} \\ \\ \frac{3}{10} \end{gathered}[/tex]Notice that the denominator of the first fraction is 5 and the denominator of the second fraction is 10.
The steps to find the a Common denominator are shown below:
1. Descompose 5 and 10 into their Prime factors:
[tex]\begin{gathered} 5=5 \\ 10=2\cdot5 \end{gathered}[/tex]2. In this case, you need to choose 5, because it is the common one. It will be the Common denominator of the fractions.
3. Divide the original denominator of the first fraction by the Common denominator 5. Multiply the the numerator by the result. Then:
[tex]\frac{2}{5}=\frac{2\cdot1}{5}=\frac{2}{5}[/tex]4. Apply the procedure explained in step 3 to the second fraction:
[tex]\frac{3}{10}=\frac{3\cdot2}{5}=\frac{6}{5}[/tex]5. You can identify that you can also get a common denominator multiplying the denominators and the numerators of both fractions by
The answer is: Option C.
O DATA ANALYSIS AND STATISTICSFinding the mode and range of a data setEach day, Susan records the number of news articles she reads. Here are her results for the last eight days.3, 6, 6, 9, 1, 6, 9,6Find the range and the mode for the data.
We are given the following data set:
[tex]3,6,6,9,1,6,9,6[/tex]We are asked to determine the range. To do that we will arrange the data from smallest to greatest values, like this:
[tex]1,3,6,6,6,6,9,9[/tex]Now, the range is the difference between the smallest and greatest value, therefore:
[tex]\text{Rage}=9-1=8[/tex]Therefore, the range is 8
Now, the mode is the value that repeats the most. In this case, we have that the value "6" is repeated 4 times, therefore, the mode is:
[tex]\mod e=6[/tex]a 2 ft by 2 ft square is divided into smaller squares and portions are shaded. What is the are of the portion and shades portion.?
Answer:
[tex]1.5ft^2[/tex]Explanation:
Here, we want to get the area of the shaded portion
To get this, we need the entire area
The entire area would be the product of the sides of the big square:
[tex]2\text{ }\times2=4ft^2[/tex]Now, let us count the number of shaded small squares.
6 out of 16 squares are shaded
The area of the shaded porion is thus:
[tex]\frac{6}{16}\times4=1.5ft^2[/tex]Let R be the event that a randomly chosen athlete runs. Let W be the event that a randomly chosen athlete lifts weights.Identify the answer which expresses the following with correct notation: The probability that a randomly chosen athlete liftsweights, given that the athlete runs.
I need help with homework If angle CVD is 4x-72 and angle BVA is 2x+18, then the value of x is......Also find, angle CVD, angle DVA, angle AVB , angle BVC... I got the picture with the questions
Given,
[tex]\begin{gathered} \angle CVD\text{ = 4x-72} \\ \angle AVB=2x+18 \end{gathered}[/tex][tex]\angle CVD=\angle AVB\text{ (vertically opposite angles.)}[/tex]That is,
[tex]\begin{gathered} 4x-72=2x+18 \\ 2x=90 \\ x=45 \end{gathered}[/tex]Therefore,
[tex]\angle CVD=180-72=108[/tex][tex]\begin{gathered} \angle DVA=180-\angle CVD\text{ (linear pair)} \\ =180-108 \\ =72 \end{gathered}[/tex][tex]\begin{gathered} \angle AVB=2x+18 \\ =90+18 \\ =108 \end{gathered}[/tex][tex]\begin{gathered} \angle BVC=\angle DVA\text{ (vertically opposite angles)} \\ \angle BVC=72 \end{gathered}[/tex]The difference between an integer and its square root is 12. What is the integer?
ANSWER:
16
STEP-BY-STEP EXPLANATION:
From the statement we can establish the following equation (where x is the integer):
[tex]x\:-\:\sqrt{x}\:=\:12[/tex]We solve for x:
This means that the integer that satisfies the statement is 16
Write an equation in slope-intercept form for the line that is perpendicular to y = 3x + 7 and passes through the point (-6, 9).
y = -x/3 +11 is the line perpendicular to y = 3x +7 and passes through the point (-6,9)
What is a slope-intercept form?It gives the graph of a straight line and it is represented in the form
y= mx +c. It is one of the form used to calculate the equation of a straight line. We have to calculate the slope of the line from the equation. The slope calculated can be used in the slope-intercept form. It is the most popular form of a straight line.
We need to find the perpendicular slope to the line y = 3x +7.
The slope of a line perpendicular to m is -1/m
Here, from the equation y=3x+7, m=3
So,-1/m = -1/3
The slope-intercept form is,
y-y1=m(x-x1)
y - 9 = -1/3 * (x+6)
Now, simplify the above equation
y-9= -x/3 +6/3
By adding 9 on both sides, we get
y= -x/3 +11
y = -x/3 +11 is the line perpendicular to y = 3x +7 and passes through the point (-6,9)
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If a triangle ABC is at: A = ( 2, 9 ) B = ( 5, 1 ) C = ( - 6, - 8 ) and if it is translated right 2 and down 7, find the new point B'.
Solution
Step 1
Triangle ABC is at: A = ( 2, 9 ) B = ( 5, 1 ) C = ( - 6, - 8 )
Step 2
If it is translated right 2 and down 7
B = (5, 1)
B' = ( 5+2, 1-7)
B' = ( 7, -6)
Final answer
B' = ( 7, -6)
Task: Find the value of x and y that proves these triangles congruent. Instructions In one part you will find the value of x that proves the triangles congruent. In the second part you will find the value ofy that proves the triangles congruent. (G.6) (2 point) Complete each of the 2 activities for this Task. Activity 1 of 2 Find the value of x.(G.6)(1 point) 24 HI 31 7x-4 to 4y+8
Activity 1:
We are given two triangles. The two side lengths of one triangle are known but of the other are not. Our task is to choose the value of x and y that will make the triangles congruent.
Now, the side lengths that are congruent are with 31 in the rightmost triangle and 7x -4 in the left-most triangle; therefore, equating them gives
[tex]7x-4=31[/tex]Similarly, side length 24 must equal 4y+8; therefore,
[tex]4y+8=24[/tex]Now we have to choose the values of x and y that will make both equations true.
Let us solve for x in the first equation by first adding 4 to both sides. Doing this gives
[tex]7x=35[/tex]Finally, dividing both sides by 7 gives
[tex]x=5.[/tex]Activity 2:
Now, for the value of y.
To solve for y, we first subtract 8 from both sides to get
[tex]4y=16[/tex]Finally, dividing both sides by 4 gives
[tex]y=4.[/tex]Hence, to conclude x = 5 and y = 4.
what are the first five terms of the recursive sequence aₙ = 3aₙ₋₁ + 3 where a₁ = 9
The expression for the recursive sequence is :
[tex]a_n=3a_{n-1}+3[/tex]where a1 = 9
First term:
Since first term is already given:
[tex]a_1=9[/tex]Second Term :
Substitute n =2 in the recursive expression and simlify
[tex]\begin{gathered} a_n=3a_{n-1}+3 \\ a_2=3(a_{2-1})+3 \\ a_2=3(a_1)+3 \\ a_2=3(9)+3 \\ a_2=27+3 \\ a_2=30 \end{gathered}[/tex]Second Term : 30
Third Term:
Substitute n = 3 in the given recursive expression:
[tex]\begin{gathered} a_n=3a_{n-1}+3 \\ a_3=3(a_{3-1})+3 \\ a_3=3(a_2)+3 \\ a_3=3(30)+3 \\ a_3=90+3 \\ a_3=93 \end{gathered}[/tex]Third Term = 93
Fourth Term:
Substitute n = 4 in the given recursive expression:
[tex]\begin{gathered} a_n=3a_{n-1}+3 \\ a_2=3(a_{2-1})+3 \\ a_2=3(a_1)+3 \\ a_2=3(9)+3 \\ a_2=27+3 \\ a_2=30 \end{gathered}[/tex]
Find the solution of the system of equations.5 +2g = 185x-Y=36
Ok we have the following system of equations:
[tex]\begin{gathered} 5x+2y=18 \\ 5x-y=36 \end{gathered}[/tex]So the first thing to do is take one of the equations above and clear either x or y. I'm going to pick the second equation and clear y:
[tex]\begin{gathered} 5x-y=36 \\ 5x=36+y \\ 5x-36=y \\ y=5x-36 \end{gathered}[/tex]Now we substitute this result in the first equation:
[tex]\begin{gathered} 5x+2y=5x+2\cdot(5x-36)=18 \\ 5x+10x-72=18 \\ 15x=18+72=90 \\ x=\frac{90}{15}=6 \end{gathered}[/tex]Now that we know x we take the result of clearing y from the second equation and find its value:
[tex]\begin{gathered} y=5x-36 \\ y=5\cdot6-36=30-36 \\ y=-6 \end{gathered}[/tex]So in the end x=6 and y=-6.