the sum of the reciprocals of two consecutive even intergers is 9/40 this can be represented bby the equation shown 1/x+1/x+2=9/40 use the rational equation to determine the integers
SHOW YOUR WORK PLEASE!!!!!!!
Answers
Answer:
In explanation
Step-by-step explanation:
To solve the rational equation 1/x + 1/(x + 2) = 9/40, we can start by finding a common denominator and then simplifying the equation. The common denominator for the two fractions is (x)(x + 2). Multiplying each term by this common denominator, we get:
[(x + 2) + x] / (x)(x + 2) = 9/40
Simplifying the numerator, we have:
(2x + 2) / (x)(x + 2) = 9/40
Now, we can cross-multiply to eliminate the denominators:
40(2x + 2) = 9(x)(x + 2)
Expanding both sides, we have:
80x + 80 = 9x^2 + 18x
Rearranging the equation and setting it equal to zero:
9x^2 - 62x - 80 = 0
Now, we can factor the quadratic equation:
(9x + 10)(x - 8) = 0
Setting each factor equal to zero, we have:
9x + 10 = 0 or x - 8 = 0
Solving for x in each case:
9x = -10 or x = 8
Dividing both sides of the first equation by 9, we find:
x = -10/9 or x = 8
Since we are looking for even integers, we can disregard the negative solution. Therefore, the value of x is 8.
Hence, the two consecutive even integers can be represented by x and x + 2, which gives us 8 and 10. So, the integers are 8 and 10.
Related Questions
Please help! -screenshot-
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A chemist wants to dilute a 30% phosphoric acid solution to a 15% solution. He needs 10 liters of the 15% solution. How many liters of the 30% solution and water must the chemist use?
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The chemist needs to use 5 liters of the 30% solution and 5 liters of water.
Answer:
30% solution x liters
30%(water) 10-x liters
15% 10 liters
Step-by-step explanation:
15 points to whoever gets this right <3
Answers
Answer:
77.5
Step-by-step explanation:
Given:
Scores: 65, 70, 75, 80, 85, 90, 95
Number of Students: 8, 5, 3, 6, 6, 4, 2
Step 1: Calculate the sum
Sum = (65 * 8) + (70 * 5) + (75 * 3) + (80 * 6) + (85 * 6) + (90 * 4) + (95 * 2)
= 520 + 350 + 225 + 480 + 510 + 360 + 190
= 2635
Step 2: Calculate the mean
Mean = Sum / Total Number of Students
= 2635 / (8 + 5 + 3 + 6 + 6 + 4 + 2)
= 2635 / 34
≈ 77.5
Answer:77.5
Step-by-step explanation:
HELP ASAP!! A student is painting a brick for his teacher to use as a doorstop in the classroom. He is only painting the front of the brick. The vertices of the face are (−4, 2), (−4, −11), (4, 2), and (4, −11). What is the area, in square inches, of the painted face of the brick?
A. 42 in2
B. 52 in2
C. 104 in2
D. 128 in2
Answers
Answer:
104in2
Step-by-step explanation:
The area, in square inches, of the painted face of the brick is Option C 104 [tex]in^{2}[/tex].
Using the Quadrant system of the Number line
B (-4,2) lies in the 2nd quadrant.
C (-4,-11) lies in the 3rd quadrant
A (4,2) lies in the 1st quadrant
D (4,-11) lies in the 4th quadrant
Length of Side AB= Length of Side CD= Distance between x coordinate of A & B OR Between C & D i.e. 8 in
Length of Side BC= Length of side AD= Distance between y coordinate of A & D OR Between B & C i.e. 13 in
Area of the Square ABCD= (length of Side AB or CD) X (Length of Side BC or AD)
= (8 x 13) [tex]in^{2}[/tex]
= 104 [tex]in^{2}[/tex]
Thus, Area of square is 104 [tex]in^{2}[/tex].
To learn more about area: https://brainly.com/question/25292087
Sanderson is having trouble with his assignment. His shown work is as follows:− 8/9 div 9/8 = − 89 × 98 = − 72/72 = −1 v\However, his answer does not match the answer that his teacher gives him.
Complete the description of Sanderson's mistake. Find the correct answer.
He forgot to multiply by the
(select)
, and instead just multiplied by the
(select)
that was in the denominator. The answer should look like this:
− 8/9 div 9/8 = - 8/9 × 8/9 =
need this for final exams
Answers
Answer:
He forgot to flip the second fraction.
Step-by-step explanation:
He has:
−8/9 ÷ 9/8 = −8/9 × 9/8 = −72/72 = −1
He made a mistake. When you change a division to a multiplication, you must use the reciprocal of the second fraction. You have to flipt the second fraction. He did not flip it.
This is correct:
−8/9 ÷ 9/8 = −8/9 × 8/9 = −64/81
Surface Area Gift Wrapping Drag and Drop Activity What is the answer for 2,5,6, & seven? I need it quick please !!
Answers
In this Drag and Drop Activity Answer 2nd is 162 [tex]in^{2}[/tex], Answer 5th is 384 [tex]in^{2}[/tex], Answer 6th is 1392 [tex]in^{2}[/tex] , Answer 7th is 1 roll.
2) In the 2nd figure, we have 5 faces as two equal Right angle Triangles (on two opposite faces) and three rectangles(on three sides of triangles)
Dimensions of both Right Angle Triangle:
Perpendicular = 9in
Base = 12 in ;
Dimensions of Rectangles:
Rectangle 1 : L1 = 15 in
B1 = 3 in
Rectangle 2: L2 = 9 in
B2 = 3 in
Rectangle 3: L3 = 12 in
B3 = 3 in
Area of 2nd figure = Sum of the area of its 5 faces
= Area of two Right angle Triangles + Area of Rectangle 1 + Area of rectangle 2 + Area of Rectangle 3
= [tex]\frac{1}{2}[/tex] * perpendicular * Base + L1*B1 + L2*B2 + L3*B3 ...................... (Formulas for particular Figures)
= [tex]\frac{1}{2}[/tex]* 9*12 + 15*3 + 9*3 + 12*3 [tex]in^{2}[/tex]
= 54+ 45+ 27 + 36 [tex]in^{2}[/tex]
=162 [tex]in^{2}[/tex]
Area of 2nd figure = 162 [tex]in^{2}[/tex]
5) In the 5th figure, we have a cube with all sides equals to 8in. Cube has 6 faces.
The area of the figure is = [tex]6(side)^{2}[/tex]
= [tex]6(8)^{2}[/tex][tex]in^{2}[/tex]
=384 [tex]in^{2}[/tex]
Area of 5th figure = 384 [tex]in^{2}[/tex]
6) Total Amount of wrapping paper needed is the sum of the Area of all Figures
= 364 + 162 + 290 + 192+ 384 [tex]in^{2}[/tex]
= 1392 [tex]in^{2}[/tex]
The total amount of wrapping paper needed is = 1392 [tex]in^{2}[/tex]
7) One Roll of wrapping paper with measurements 30 in and 60 in has an Area
= 30*60 [tex]in^{2}[/tex]
= 1800 [tex]in^{2}[/tex]
As we need only 1392 [tex]in^{2}[/tex] of wrapping paper which is less than one roll of wrapping paper so only 1 roll is required
hence, In this Drag and Drop Activity Answer 2nd is 162 [tex]in^{2}[/tex], Answer 5th is 384 [tex]in^{2}[/tex], Answer 6th is 1392 [tex]in^{2}[/tex] , Answer 7th is 1 roll.
To learn more about Area of Rectangle
https://brainly.com/question/2607596
Keira said that if 6 tickets are sold, there should be 2 tickets left. Is Keira correct?
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According to Keira, if 6 tickets are sold, there should be 2 tickets left. However, we don't have any information about the initial number of tickets available.
If the total number of tickets initially was 8, and 6 tickets were sold, then there would indeed be 2 tickets left, supporting Keira's statement.
However, if the total number of tickets initially was less than 8, or if the statement doesn't specify the initial number of tickets, we cannot determine if Keira's statement is correct.
Therefore, without additional information, we cannot determine the accuracy of Keira's statement.
Please help, screenshot!
Answers
Answer:
4/7
Step-by-step explanation:
class total = 21
chance of picking one of 12 boys out of 21.
12/21 which can be simpilfied to 4/7
9 girls + 12 boys = 21
Boys/total = 12/21
12/21 divided by 3 = 4/7
Does the series converge or diverge? if it converges, what is the sum? SHOW YOUR WORK!!!!!!
equation is in picture.
Answers
=============================================
Reason:
This is a geometric series with
a = -4 = first termr = -1/3 = common ratioThe template is [tex]a(r)^{n-1}[/tex]
If -1 < r < 1, then the infinite geometric series converges to a finite number. This is because we add on smaller and smaller pieces, which prevents the sum going off to infinity.
In the case of r = -1/3, it fits the interval -1 < r < 1. In other words -1 < -1/3 < 1 is true.
We'll plug those values into the formula below to wrap things up.
[tex]S = \frac{a}{1-r}\\\\S = \frac{-4}{1-(-1/3)}\\\\S = \frac{-4}{1+1/3}\\\\S = \frac{-4}{3/3+1/3}\\\\[/tex]
[tex]S = \frac{-4}{4/3}\\\\S = -4 \div \frac{4}{3}\\\\S = \frac{-4}{1} \times \frac{3}{4}\\\\S = \frac{-4*3}{1*4}\\\\S = -3\\\\[/tex]
Therefore,
[tex]\displaystyle \sum_{n=1}^{\infty} -4\left(-\frac{1}{3}\right)^{n-1} = -3[/tex]
The final answer is -3.
You can verify the answer by generating partial sums with a spreadsheet. The partial sums should steadily get closer to -3.
Here's a few partial sums.
[tex]\begin{array}{|c|c|c|} \cline{1-3}\text{n} & \text{a}_{\text{n}} & \text{S}_{\text{n}}\\\cline{1-3}1 & -4 & -4\\\cline{1-3}2 & 1.333333 & -2.666667\\\cline{1-3}3 & -0.444444 & -3.111111\\\cline{1-3}4 & 0.148148 & -2.962963\\\cline{1-3}5 & -0.049383 & -3.012346\\\cline{1-3}6 & 0.016461 & -2.995885\\\cline{1-3}7 & -0.005487 & -3.001372\\\cline{1-3}8 & 0.001829 & -2.999543\\\cline{1-3}9 & -0.00061 & -3.000153\\\cline{1-3}10 & 0.000203 & -2.99995\\\cline{1-3}\end{array}[/tex]
The interesting thing is that the partial sums [tex]S_n[/tex] bounce around -3 while also getting closer to it.