437,160 is read as four hundred thirty-seven thousand one hundred sixty and 43,716 is read as fourty-three thousand seven hundred sixteen.
The first number has hundreds of thousand and the second one only has tens of thousand. It means that the first number is greater than the second one.
Another way to know this is because of the number of figures before the decimal point. The first number has 6 figures before the decimal point and the second one only has 5.
That way, we know that 437,160 is greater than 43,716.
Clark and Lindsay Banks have agreed to purchase a home for $225,000. They made a down payment of 15%. They have obtained a mortgage loan at a 6.5% annual interest rate for 25 years. What is the mortgage total if they finance the closing costs?
SOLUTION
We will be using the annual compound interest formula to solve this question.
[tex]\begin{gathered} A=P(1+\frac{R}{100})^{mn} \\ \text{where m=1, n=25years, R=6.5,} \end{gathered}[/tex]After a down payment of 0.15 x $225,000 = $33750
The principal value will be $225,000 - $33750 = $191250
Put all these values into the compound interest formula above,
we will have:
[tex]\begin{gathered} A=191250(1+\frac{6.5}{100})^{1\times25} \\ A=191250(1+0.065)^{25} \end{gathered}[/tex][tex]\begin{gathered} A=191250(1.065)^{25} \\ \text{ = 191250}\times4.8277 \\ \text{ =923,297.63} \end{gathered}[/tex]The mortgage total if they finance the closing costs will be:
$923,297.63
Can you please help me out with a question
Notice that the arc GH has the same measure as the angle GFH, which also has the same measure as the angle EFI since they are vertical angles.
On the other hand, EFI and IFS are adjacent angles, then:
[tex]m\angle EFI+m\angle IFS=m\angle EFS[/tex]Observe that the measure of the angle EFS is 90°. Since the measure of IFS is 20°, substitute those values into the equation to find the measure of EFI:
[tex]\begin{gathered} m\angle EFI+20=90 \\ \Rightarrow m\angle EFI=90-20 \\ \Rightarrow m\angle EFI=70 \end{gathered}[/tex]Thereore, the measure of GH is 70°.
Find the area to the right of x=71 under a normal distribution curve with the mean=53 and standard deviation=9
Answer:
[tex]Area=0.0228\text{ or 2.28\%}[/tex]Explanation:
We were given the following information:
This is a normal distribution curve
Mean = 53
Standard deviation = 9
We are to find the area right of x = 71
This is calculated as shown below:
[tex]\begin{gathered} z=\frac{x-\mu}{\sigma} \\ x=71 \\ \mu=53 \\ \sigma=9 \\ \text{Substitute these into the formula, we have:} \\ z=\frac{71-53}{9} \\ z=\frac{18}{9} \\ z=2 \end{gathered}[/tex]We will proceed to plot this on a graph as sown below:
The area to the right of x = 71 (highlighted in red above) is given by using a Standard z-score table:
[tex]\begin{gathered} =1-0.9772 \\ =0.0228 \\ =2.28\text{\%} \end{gathered}[/tex]Therefore, the area that lies to the right of x = 71 is 0.0228 or 2.28%
You play a game where you toss a die. If the die lands on a 6, you win $6. It costs $2 toplay. Construct a probability distribution for your earnings. Find your expected earnings.
SOLUTION
Now from the question, if the die lands on 6, I win $6. So probability of landing on 6 is
[tex]\frac{1}{6}\text{ since a die has 6 faces }[/tex]Since I will pay $2 to play, we subtract this from $6 that we will win.
And probability of losing becomes
[tex]\frac{5}{6}\text{ }[/tex]The table becomes
From the table the expected earnings is calculated as
[tex]\begin{gathered} E=\sum_^xP(x) \\ =4(\frac{1}{6})-2(\frac{5}{6}) \\ =\frac{4}{6}-\frac{10}{6} \\ =-\frac{6}{6} \\ =-1 \end{gathered}[/tex]Hence expected earnings is -$1
5.) y = -5/4 x + 10 (Use Slope Int. Method Make apparent your Int. Point AND the point from your slope = RISE/RUN)
Which are correct representations of the inequality –3(2x – 5) < 5(2 – x)? Select two options. x < 5 –6x – 5 < 10 – x –6x + 15 < 10 – 5x A number line from negative 3 to 3 in increments of 1. An open circle is at 5 and a bold line starts at 5 and is pointing to the right. A number line from negative 3 to 3 in increments of 1. An open circle is at negative 5 and a bold line starts at negative 5 and is pointing to the left.
The correct representations of the inequality -3(2x-5)<5(2-x) are -6x + 15 < 10 - 5x and "an open circle is at 5 and a bold line that starts at 5 and is pointing to the right" , the correct option is (c) and (d) .
In the question ;
it is given that
the inequality -3(2x-5)<5(2-x)
on solving this inequality further , we get
-3(2x-5)<5(2-x)
-6x+15<10-5x
which is option (c) .
Further solving
Subtracting 15 from both the sides of the inequality , we get
-6x + 15 -15 < 10 -5x -15
-6x < -5 -5x
-6x +5x < -5
-x < -5
multiplying both sides by (-1) ,
we get
x > 5 .
x> 5 on number line means an open circle is at 5 and a bold line starts at 5 and is pointing to the right .
Therefore , the correct representations of the inequality -3(2x-5)<5(2-x) are -6x + 15 < 10 - 5x and "an open circle is at 5 and a bold line that starts at 5 and is pointing to the right" , the correct option is (c) and (d) .
The given question is incomplete , the complete question is
Which are correct representations of the inequality -3(2x - 5) < 5(2 - x)? Select two options.
(a) x < 5
(b) –6x – 5 < 10 – x
(c) –6x + 15 < 10 – 5x
(d) A number line from negative 3 to 7 in increments of 1 , An open circle is at 5 and a bold line that starts at 5 and is pointing to the right.
(e) A number line from negative 7 to 3 in increments of 1, An open circle is at negative 5 and a bold line that starts at negative 5 and is pointing to the left.
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Omar has $84 and maryam has $12. how much money must Omar give to maryam so that maryam will have three times as much as omar? let x be the amount of dollars Omar will give maryam. which equation best represents the situation described above? A.) 84 - x = 3(12) + xB.) 3(84 - x) = 12 - xC.) 3(84 - x ) = 12 + xD.) 3x = 84 - (12 + x)
Given data:
The given money Omar has $84.
The given money maryam has $12.
The expression for the money Omar give to maryam so that maryam will have three times as much as omar.
[tex]3(84-x)=12+x[/tex]Thus, the final expression is 3(84-x)=12+x.
the coordinates of two points on a line are (-4,8) and (2,2). Find the slope of the line.
the coordinates of two points on a line are (-4,8) and (2,2). Find the slope of the line.
Applying the formula to calculate the slope
we have
m=(2-8)/(2+4)
m=-6/6
m=-1
slope is -1Kuta Software - Infinite Precalculus Angles and Angle Measure Find the measure of each angle.
Explanation:
We are to draw the angle that is equivalent to 5pi/4
First we need to convert the radian value to degree
Since pi rad = 180degrees
5pi/4 = x
Cross multiply
pi * x = 5pi/4 * 180
x = 5/4 * 180
x = 5 * 45
x = 225 degrees
This can also be written as 225 = 180 + 45
225degrees = 180 + pi/4
Note that 180degrees is an angle on a straight line. Find the digaram attached
The remaining angle which is pi/4 is the reason for the angle extensionon for the angle extension
5pi/4 = x
Cross multiply
pi * x = 5pi/4 * 180
x =
Quadrilateral PQRS is plotted in the coordinate plane. The quadrilateral is dilated by a scale factor of 3/4. What are the new ordered pairs for P'Q'R'S'?
Explanation:
The first thing is to state the coordinates of Quadrilateral PQRS
P (5, 5), Q (3, 5), R (3, 1), S (5, 1)
Then we find the distance between two points using the distance formula
[tex]dis\tan ce\text{ = }\sqrt[]{(y_2-y_1)^2+(x_2-x_1)^2}[/tex][tex]\begin{gathered} P(5,5),Q(3,5)\text{ = (x1, y1) and (x2, y2)} \\ \text{distance PQ = }\sqrt[]{(5-5)^2+(3-5)^2}\text{ = }\sqrt[]{0+(-2)^2}\text{ =}\sqrt[]{4} \\ \text{distance PQ = }2 \end{gathered}[/tex][tex]\begin{gathered} Q(3,5),R(3,1)\text{= (x1, y1) and (x2, y2)} \\ \text{distance QR = }\sqrt[]{(1-5)^2+(3-3)^2}\text{ = }\sqrt[]{(-4)^2+0}\text{ = }\sqrt[]{16} \\ \text{distance QR = 4} \end{gathered}[/tex]It is a quadrilateral, meaning the two lengths are equal. Like wise the two widths are equal.
length PQ = length SR = 2
Length QR = length PS = 4
Scale factor = 3/4
Scale factor = corresponding side of new image/ corresponding side of original image
PQRS = original image, P'Q'R'S' = new image
3/4 = P'Q'/PQ
3/4 = P'Q'/2
P'Q' = 2(3/4) = 6/4 = 3/2
Since P'Q' = S'R'
S'R' = 3/2
3/4 = Q'R'/QR
3/4 = Q'R'/4
Q'R' = 3/4 (4) = 12/4 = 3
Since Q'R' = P'S
P (5, 5), Q (3, 5), R (3, 1), S (5, 1)
PQRS to P'Q'R'S' = 3/4(
P' = 3/4 (5, 5) = (15/4, 15/4)
Q' = 3/4 (3, 5) = (9/4, 15/4)
R' = 3/4 (3, 1) = (9/4, 3/4)
S' = 3/4 (5, 1)
please help me with this. four potential solutions.450, 780, 647, 354
So first of all let's take:
[tex]x_1=x\text{ and }x_2=y[/tex]Then we get:
[tex]\begin{gathered} \text{Min}z=1.5x+2y \\ x+y\ge300 \\ 2x+y\ge400 \\ 2x+5y\leq750 \\ x,y\ge0 \end{gathered}[/tex]The next step would be operate with the inequalities and the equation so we end up having only the term y at the left side of each:
[tex]\begin{gathered} \text{Min}z=1.5x+2y \\ 1.5x+2y=\text{Min}z \\ 2y=\text{Min}z-1.5x \\ y=\frac{\text{Min}z}{2}-0.75x \end{gathered}[/tex][tex]\begin{gathered} x+y\ge300 \\ y\ge300-x \end{gathered}[/tex][tex]\begin{gathered} 2x+y\ge400 \\ y\ge400-2x \end{gathered}[/tex][tex]\begin{gathered} 2x+5y\leq750 \\ y\leq150-\frac{2}{5}x \end{gathered}[/tex]So now we have the following inequalities and equality:
[tex]\begin{gathered} y=\frac{\text{Min}z}{2}-0.75x \\ y\ge300-x \\ y\ge400-2x \\ y\leq150-\frac{2}{5}x \end{gathered}[/tex]If we take the three inequalities and replace their symbols by "=' we'll have three equations of a line:
[tex]\begin{gathered} y=300-x \\ y=400-2x \\ y=150-\frac{2}{5}x \end{gathered}[/tex]The following step is graphing these three lines and delimitating a zone in the grid that meets the inequalities:
Where the blue area is under the graph of y=150-(2/5)x which means that it meets:
[tex]y\leq150-\frac{2}{5}x[/tex]And it is also above the x-axis, y=400-2x and y=300-x which means that it also meets:
[tex]\begin{gathered} x\ge0 \\ y\ge0 \\ y\ge400-2x \\ y\ge300-x \end{gathered}[/tex]All of this means that the values of x and y that give us the correct minimum of z are given by the coordinates of a point inside the blue area. The next thing to do is take the four possible values for Min(z) and use them to graph four lines using this equation:
[tex]y=\frac{\text{Min}z}{2}-0.75x[/tex]Then we have four equations of a line:
[tex]\begin{gathered} y=\frac{450}{2}-0.75x \\ y=\frac{780}{2}-0.75x \\ y=\frac{647}{2}-0.75x \\ y=\frac{354}{2}-0.75x \end{gathered}[/tex]The line that has more points inside the blue area is the one made with the closest value to Min(z). Then we have the following graph:
As you can see there are two lines that have points inside the blue area. These are:
[tex]\begin{gathered} y=-\frac{3}{4}x+\frac{450}{2} \\ y=-\frac{3}{4}x+\frac{354}{2} \end{gathered}[/tex]That where made using:
[tex]\begin{gathered} \text{Min }z=450 \\ \text{Min }z=354 \end{gathered}[/tex]Taking a closer look you can see that the part of the orange line inside the blue area is larger than that of the red line. Then the value used to make the orange line would be a better aproximation for the Min z. The orange line is -(3/4)x+450/2 which means that the answer to this problem is the first option, 450.
Find the set A n Φ.U = {1, 2, 3, 4, 5, 6, 7, 8, 9)A = 2, 3, 8, 9)Selectthe correct choice below and, if necessary, fill in the
Answer
Option B is the correct answer.
A n Φ = {}
A n Φ is the empty set.
Explanation
We are told to find the intersection between set A and the empty set Φ.
The intersection of two sets refers to the elements that belong to the two sets, that is, the elements that they both have in common.
Set A = (2, 3, 8, 9)
Set Φ = {}
What the elements of set A and set Φ (an empty set) have in common is nothing.
Hence, the intersection of set A and set Φ is an empty set.
A n Φ = {}
Hope this Helps!!!
Problem Solving: Fraction Division For exercises 1 and 2, write three problem situations for each division 56÷1/3 and 6/1/2÷1/2/3
56÷1/3
We have to model a problem where the solution is 56÷1/3.
So, we take something that is 56 and we have to divide it by 1/3rd.
So, we can say:
George had 56 large cakes.
Giving 1/3rd of each cake to each person is enough.
If George used all of the cake, how many person could he feed?
A trail mix brand guarantees a peanut to raisin ratio of 5:2. If a bag of that trail mix contains 30 peanuts, how many raisins are in the bag?
Answer:
12
Explanation:
In the bag, the guaranteed ratio of peanut to raisin = 5:2
Number of peanuts = 30
Let the number of raisins =x
We therefore have that:
[tex]\begin{gathered} 5\colon2=30\colon x \\ \frac{5}{2}=\frac{30}{x} \\ 5x=30\times2 \\ x=\frac{30\times2}{5} \\ x=12 \end{gathered}[/tex]The number of raisins in the bag is 12.
What number is 75% of 96?
The number 96 is equivalent to the 100%. So we can state the following rule of three:
[tex]\begin{gathered} 96\text{ ------ 100 \%} \\ x\text{ -------- 75 \%} \end{gathered}[/tex]By cross-multiplying these numbers, we have
[tex]\text{ (100\%)}\times x=(96)\times\text{ (75 \%)}[/tex]So, x is given by
[tex]\begin{gathered} x=\frac{(96)\times\text{ (75 \%)}}{\text{ 100\%}} \\ x=72 \end{gathered}[/tex]Therefore, the answer is 72
A local road has a grade of 5%. The grade of a road is its slope expressed as a percent. What is the slope? What is the rise? What is the run?
a) Since the grade is given by the slope, and the grade has a 5%.
We can rewrite it as a fraction, like this:
[tex]\frac{5}{100}=\frac{1}{20}[/tex]Note that we have simplified this to 1/20 by dividing the numerator and the denominator (bottom number) by 5
So, the slope is:
[tex]\frac{1}{20}[/tex]b) The "rise" is the difference between two coordinates on the y-axis and the "run" is the subtraction between two coordinates on the x-axis. Let's remember the slope formula and the Cartesian plane:
[tex]m=\frac{y_2-y_1}{x_2-x_1}=\frac{1}{20}[/tex]So the "rise" for this grade is 1 foot and the run is 20 feet.
3) Hence, the answers are:
[tex]\begin{gathered} a)\text{ }\frac{1}{20} \\ b)\text{ }Rise\colon\text{ }1\text{ Run: 20} \end{gathered}[/tex]When 6 is subtracted from the 5 times of a number the sum becomes 9 find the number
Let that unknown number be x
⇒Mathematically this is written as
[tex]5(x)-6=9\\5x-6=9\\5x=9+6\\5x=15\\\frac{5x}{5} =\frac{15}{5} \\x=3[/tex]
This just means that the unknown number is 3
GOODLUCK!!
Answer:
nine plus six
= 15 ÷ five
answer Three
Please show me how to solve this step by step im really confused
Given
[tex]-16t^2+v_0t+h_0[/tex]initial velocity = 60 feet per second
initial height = 95 feet
Find
Maximum height attained by the ball
Explanation
we have given
[tex]\begin{gathered} h(t)=-16t^2+60t+95 \\ h^{\prime}(t)=-32t+60 \end{gathered}[/tex]put h'(t) = 0
[tex]\begin{gathered} -32t+60=0 \\ -32t=-60 \\ t=\frac{60}{32}=1.875sec \end{gathered}[/tex]to find the maximum height find the value of h(1.875)
[tex]\begin{gathered} h(1.875)=-16(1.875)^2+60(1.875)+95 \\ h(1.875)=-56.25+112.5+95 \\ h(1.875)=-56.25+207.5 \\ h(1.875)=151.25 \end{gathered}[/tex]Final Answer
Therefore , the maximum height attained by the ball is 151.25 feet
I need help with the entire problem. The question is about a sketchy hotel.
Let d and s be the cost of a double and single- occupancy room, respectively. Since a double-occupancy room cost $20 more than a single room, we can write
[tex]d=s+20\ldots(A)[/tex]On the other hand, we know that 15 double-rooms and 26 single-rooms give $3088, then, we can write
[tex]15d+26s=3088\ldots(B)[/tex]Solving by substitution method.
In order to solve the above system, we can substitute equation (A) into equation (B) and get
[tex]15(s+20)+26s=3088[/tex]By distributing the number 15 into the parentheses, we have
[tex]15s+300+26s=3088[/tex]By collecting similar terms, it yields,
[tex]41s+300=3088[/tex]Now, by substracting 300 to both sides, we obtain
[tex]41s=2788[/tex]then, s is given by
[tex]s=\frac{2788}{41}=68[/tex]In order to find d, we can substitute the above result into equation (A) and get
[tex]\begin{gathered} d=68+20 \\ d=88 \end{gathered}[/tex]Therefore, the answer is:
[tex]\begin{gathered} \text{ double occupancy room costs: \$88} \\ \text{ single occupancy room costs: \$68} \end{gathered}[/tex]Read the following scenario and write two equations we could use to solve to find for the number of cars and trucks washed. Use the variables C for cars washed and T for trucks washed. (Hint: both equations should have T and C). SCENARIO: Western's eSports Team raised money for charity by organizing a car wash. They washed a total of 80 vehicles and raised a total of $486. They charged $5 to wash a car and $7 to wash a truck.
Let:
C = Number of cars washed
T = Number of trucks washed
They washed a total of 80 vehicles, so:
[tex]C+T=80[/tex]They raised a total of $486. They charged $5 to wash a car and $7 to wash a truck. so:
[tex]5C+7T=486[/tex]Let:
[tex]\begin{gathered} C+T=80_{\text{ }}(1) \\ 5C+7T=486_{\text{ }}(2) \end{gathered}[/tex]From (1) solve for T:
[tex]T=80-C_{\text{ }}(3)[/tex]Replace (3) into (2):
[tex]\begin{gathered} 5C+7(80-C)=486 \\ 5C+560-7C=486 \\ -2C=486-560 \\ -2C=-74 \\ C=\frac{-74}{-2} \\ C=37 \end{gathered}[/tex]Replace the value of C into (3):
[tex]\begin{gathered} T=80-37 \\ T=43 \end{gathered}[/tex]They washed 37 cars and 43 trucks
lily ordered a set of green and brown pin.she received 35 pins, and 80% of them were green.How many green pins did lily receive?
In total there are 35 pins so that correspound to the 100%, so we can use a rule of 3 to solve it so:
[tex]\begin{gathered} 35\to100 \\ x\to80 \end{gathered}[/tex]So the equation is:
[tex]x=\frac{35\cdot80}{100}=28[/tex]So there are 28 green pins
I got 4089 for the answer but it was incorrect
Let A be the event "person under 18" and B be the event "employed part-time". So, we need to find the following probability
[tex]P(A\text{ or B) =P(A}\cup B)[/tex]which is given by
[tex]P(A\text{ or B) =P(A}\cup B)=P(A)+P(B)-P(A\cap B)[/tex]Since the total number od people in the table is equal to n=4089, we have that
[tex]P(A)=\frac{28+174+395}{4089}=\frac{597}{4089}[/tex]and
[tex]P(B)=\frac{174+194+71+179+173}{4089}=\frac{791}{{4089}}[/tex]and
[tex]P(A\cap B)=\frac{174}{4089}[/tex]we have that
[tex]P(A\text{ or B) =}\frac{597}{4089}+\frac{791}{{4089}}-\frac{174}{4089}[/tex]which gives
[tex]P(A\text{ or B) =}\frac{597+791-174}{4089}=\frac{1214}{4089}=0.29689[/tex]Therefore, the answer the searched probability is: 0.296
Bill Jensen deposits $8500 with Bank of America in an investment paying 5% compounded semiannually. Find the interest in 6 years
Amount deposited = $8500
Rate = 5%
time for interest = 6years
Compounded semiannually
The formula for semiannually is
[tex]A=P(1+\frac{r}{100n})^{nt}[/tex]From the given information
P = $8500
r = 5
t = 6
Since the investment was compounded semiannually then
n = 2
Substitute the values into the formula
This gives
[tex]A=8500(1+\frac{5}{100\times2})^{6\times2}[/tex]Solve for A
[tex]\begin{gathered} A=8500(1+0.025)^{12} \\ A=8500(1.025)^{12} \\ A=11431.56 \end{gathered}[/tex]To find the interest
Recall
[tex]I=A-P[/tex]Where I, is the interest
Hence
[tex]\begin{gathered} I=\text{\$}11431.56-\text{\$}8500 \\ I=\text{\$}2931.56 \end{gathered}[/tex]What’s the correct answer asap for brainlist
Answer:
Step-by-step explanation:its a 69420 dum as
1 point Esther thinks she understands how to find the midpoint of a segment on a graph. "I always look for the middle of the line segment. But what should I do if the coordinates are not easy to graph?" she asks. Find the midpoint of KL if (2.125) and L(98, 15). *
In this case, we can write out the parameters
[tex]\begin{gathered} x_1=2,_{}y_1=125, \\ x_2=98,y_2=15 \end{gathered}[/tex]Thus, substitute the coordinates in the mid-point formula and simplify
[tex]\begin{gathered} x_m=\frac{98+2}{2}=\frac{100}{2}=50 \\ y_m=\frac{125+15}{2}=\frac{140}{2}=70 \end{gathered}[/tex]Hence, the coordinate of the mid-point is (50, 70)
The table represents a linear function.What is the slope of the function?y08-2.04х-4-2-112-10-14-22-26O 2O 5
Answer
Option B is correct.
The slope of this function = -4
Explanation
For a linear function, the slope of the line can be obtained when the coordinates of two points on the line or the values of the linear function (y) at different values of x are known. If the two points are described as (x₁, y₁) and (x₂, y₂), the slope is given as
[tex]Slope=m=\frac{Change\text{ in y}}{Change\text{ in x}}=\frac{y_2-y_1}{x_2-x_1}[/tex]Using the two extreme points, (x₁, y₁) and (x₂, y₂) are (-4, -2) and (2, -26).
x₁ = -4
y₁ = -2
x₂ = 2
y₂ = -26
[tex]\text{Slope = }\frac{-26-(-2)}{2-(-4)}=\frac{-26+2}{2+4}=\frac{-24}{6}=-4[/tex]Hope this Helps!!!
Multiply.(2x + 4)(2x - 4)A. 4x2 + 16x- 16B. 4x2 - 16C. 4x2 - 16x - 16D. 4x2 + 16
We have to multiply the expression (2x + 4)(2x - 4):
[tex]\begin{gathered} \left(2x+4\right)\left(2x-4\right) \\ 2x\cdot2x+2x\cdot(-4)+4\cdot2x+4\cdot(-4) \\ 4x^2-8x+8x-16 \\ 4x^2+(8-8)x-16 \\ 4x^2-16 \end{gathered}[/tex]The answer is:
B. 4x^2 - 16
The sum of two numbers is ten. One number is
twenty less than four times the other. Find the
numbers.
Note: List numbers with a comma separating
them, e.g. 5, 12.
By solving the equations, we can conclude that the two numbers are 4 and 6.
What are equations?An equation is a mathematical statement that contains the symbol "equal to" between two expressions with identical values. As in 3x + 5 = 15, for example. There are many different types of equations, including linear, quadratic, cubic, and others. The three primary forms of linear equations are point-slope, standard, and slope-intercept.So, the two numbers are:
Let the 2nd number be 'x'.Then, the 1st number will be '4x - 20'.The equation will be:
4x - 20 + x = 10Now, solve this equation for 'x' as follows:
4x - 20 + x = 105x = 10 + 205x = 30x = 6Now, 4x - 20:
4(6) - 2024 - 204Therefore, by solving the equations, we can conclude that the two numbers are 4 and 6.
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-14.4 + x = -8.2what does x equal?I NEED ANSWERS ASAPi will give brainliest
the given expression is,
-14.4 + x = -8.2
x = 14.4 - 8.2
x = 6.2
thus, the answer is x = 6.2
Is y-x+wz=5 linear? And not, why and if so, can you put it in slope intercept form?
A linear equation is an equation in which the highest power of the variable is always 1. It is also known as a one-degree equation. The standard form of this kind of equation is given by:
[tex]Ax+By=C[/tex]For the equation:
[tex]y-x+wz=5[/tex]We can conclude is not a linear equation since there is a product between two variables.