two consecutive even numbers are such that the sum of their squares is 146. Find the two numbers
1. two consecutive even numbers are such that the sum of their squares is 146. Find the two numbers
Jawaban:
dua bilangan genap berurutan sehingga jumlah kuadratnya adalah 146. Tentukan kedua bilangan tersebut
Penjelasan dengan langkah-langkah:
x+(x+2)= 146
2x = 144
x = 72
bilangan genap terbesar = x+2
= 72 +2 = 74
2. The difference between two positive numbers is 7 and the square of their sum is 289. Find the two numbers.
a - b = 7
a = b + 7
a² + b² = 289
(b +7)² + b² = 289
b² + b² + 14b + 49 - 289 = 0
2b² + 14b - 240 = 0
(2b + 30)(b - 8) = 0
2b + 30 = 0
2b = -30
b = -30/2
b = -15
a = 15
b - 8 = 0
b = 8
the two numbers are 15 and 8
3. There are two numbers such that their difference is 7 and the difference of their squares is 175, sum of the numbers = ?
Jawab:
a+b = 25
Penjelasan dengan langkah-langkah:
Dik:
a-b = 7
[tex]a^{2} -b^{2} = 175[/tex]
Dit? a+b
(a+b)(a-b) = 175
(a+b)(7) = 175
a+b = 175/7
a+b = 25
4. Quiz! The quotient of two numbers is 2. Their product is 72. What are the two numbers? Sertakan langkah dan penjelasan. Thanks.
Jawab:
The two numbers are 48 and 24
Cara:
1. Persamaan 1:
x + y = 72
2. Persamaan 2:
x/y = 2 --- > maka x = 2y
3. Masukkan Persamaan 2 ke Persamaan 1:
x + y = 72
2y + y = 72
3y = 72
y = 72 : 3
y = 24
4. Masukkan nilai y ke dlm Persamaan 2 utk mendapatkan nilai x :
x = 2y
x = 2 x 24
x = 48
Jadi x = 48, dan y = 24
So the two numbers are 48 and 24
Semoga membantu...
5. Questions: 1. The sum of two numbers is 30. Their product is 81. What are the two numbers? 2. The quotient of two numbers is 3. Their product is 75. What are the two numbers? Jawabnya pakai penjelasan ya. Terima kasih.
Jawab:
Yo ndak tahu kok tanya saya?
Penjelasan dengan langkah-langkah:
Yo ndak tahu kok tanya saya?
(Xcvi)
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Just kidding
Jawab:
(1.) --> There are 2 pairs of two numbers:
{3, 27} and {27, 3}
(2.) --> There are 2 pairs of two numbers:
{-15, -5} and {15, 5}
Penjelasan dengan langkah-langkah:
(1.) a+b = 30, ab = 81 -> b = 81/a
a + (81/a) = 30 -> (a²/a) + (81a/a) = (30a/a)
a² + 81 = 30a -> a² - 30a + 81 = 0
Aa² + Ba + C = 0 ∴ A = 1, B = -30, C = 81
p and q are roots, pq = C/A, p+q = -B/A
pq = 81/1, p+q = -(-30)/1 ; therefore ,
pq = 81, p+q = 30 ∴ p = 27, q = 3
a+b = 30, ab = 81 (same) p = 3, q = 27
--> There are 2 pairs of two numbers:
{3, 27} and {27, 3}
(2.) a/b = 3 -> a = 3b
ab = 75, b(3b) = 75, 3b² = 75
b = √(75/3), b = √25, b = ±5
b = {-5, 5}
a = 3b, a = {3(-5), 3(5)}
a = {-15, 15}
--> There are 2 pairs of two numbers:
{-15, -5} and {15, 5}
(xcvi)
6. The sum of two numbers is 14 and their differences is 2. Find the numbers!
Penjelasan dengan langkah-langkah:
jumlah dua angka adalah 14 dan perbedaannya adalah 2.temukan nomornya.
number is 8 and number 6
14 bagi 2 : 88-2 : 6
the answer is 6
7. 6. The sum of two numbers is 183 and their difference is 71. Find the two numbers.
The sum of two numbers = 183
The difference of them = 71
Make into algebra :
a + b = 183
a - b = 71
We can use elimination or substitution method , we get :
2b = 112
b = 56
a = 71 + b
a = 71 + 56
a = 127
Then , the numbers are 127 and 56
8. The difference between two numbers is 10 and their sum is four times the smaller number. Find the two numbers
Jawaban:
5 dan 15
misal: bilangan terkecil □ dan bilangan terbesar ■
□+■=□□□□ maka didapat ■=□□□
■-□=□□=10 jadi □=5
karena ■=□□□ maka ■=5+5+5=15
9. The sum of two numbers are 27 and their difference are 3. The product of two numbers is…Tolong dong besok harus ke kumpul
Untuk menyelesaikan masalah ini, kita dapat menggunakan persamaan-persamaan yang sesuai dengan informasi yang diberikan. Misalnya, kita bisa menggunakan persamaan (a+b) = 27 dan (a-b) = 3, dimana a dan b adalah dua bilangan yang akan dicari.
Dari persamaan (a+b) = 27, kita bisa menyimpulkan bahwa a + b = 27. Selanjutnya, kita bisa menggunakan persamaan (a-b) = 3 untuk mencari nilai a dan b. Dari persamaan tersebut, kita bisa menyimpulkan bahwa a - b = 3.
Selanjutnya, kita bisa menggabungkan kedua persamaan tersebut untuk mencari nilai a dan b. Misalnya, kita bisa menggabungkan persamaan (a+b) = 27 dengan persamaan (a-b) = 3 dengan menggunakan rumus penjumlahan dan pengurangan. Persamaan tersebut menjadi
a + b + a - b = 27 + 3
Kemudian, kita bisa menyederhanakan persamaan tersebut dengan menambahkan kedua sisi persamaan. Persamaan tersebut menjadi
2a = 30
Selanjutnya, kita bisa menyederhanakan persamaan tersebut dengan membagi kedua sisi persamaan dengan 2. Persamaan tersebut menjadi
a = 15
Setelah itu, kita bisa menggunakan nilai a yang sudah diketahui untuk mencari nilai b. Misalnya, kita bisa menggunakan persamaan (a+b) = 27 untuk mencari nilai b. Persamaan tersebut menjadi
15 + b = 27
Kemudian, kita bisa menyederhanakan persamaan tersebut dengan mengurangi kedua sisi persamaan dengan 15. Persamaan tersebut menjadi
b = 27 - 15
Selanjutnya, kita bisa menyederhanakan persamaan tersebut dengan menjumlahkan kedua sisi persamaan. Persamaan tersebut menjadi
b = 12
Jadi, nilai a adalah 15 dan nilai b adalah 12. Kemudian, kita bisa menggunakan persamaan (ab) = x untuk mencari nilai x, yaitu produk dari dua bilangan tersebut. Persamaan tersebut menjadi
(15 * 12) = x
Kemudian, kita bisa menyederhanakan persamaan tersebut dengan mengalikan kedua sisi persamaan. Persamaan tersebut menjadi
180 = x
Jadi, nilai x adalah 180
Jawab:
Penjelasan dengan langkah-langkah:
Bilangan pertama = a
Bilangan kedua = b
a + b = 27
-a + b = 3 _
2a = 24
a = 24/2 = 12
b = 3 + a
b = 3 + 12 = 15
a x b = 12 x 15 = 180
10. the product of two rational numbers -28/21.if one of the numbers is 14/27,find the other
We suppose that the other rational number is A
14/27 x A = - 28/21
A = - 28/27 : 14/27
28 27
= - ----- x -------
21 14
28:14=2 27:3=9
= - ----------- x --------
21:3=7 14:14=1
= - 2/7 x 9/1
= -18/7
= -2 4/7
So, the other rational number is -18/7 = - 2 4/7
11. Quiz! The quotient of two numbers is 3. Their product is 75. What are the two numbers? Please explain your answer! Thanks.
× : y = 3
x = 3y
x . y = 75
y = 75 : x
x = 3y
x = 3 . 75 : x
x² = 225
x = 15
y = 75 : x
y = 75 : 15
y = 5
so the numbers are 15 and 5
12. 5. The difference between two numbers is 9 andthe product of the numbers is 162. Find thetwo numbers.
Jawaban:
HP={-18,9}
Penjelasan dengan langkah-langkah:
a-b=9...(1)
a = 9+b
a.b=162...(2)
(9+b)b=162
9b+b²=162
b²+9b-162=0
(b+18)(b-9)=0
b=-18 atau b=9
HP={-18,9}
Jawab:
Penjelasan dengan langkah-langkah:
assumed the numbers are A and B
B = A + 9
A x B = 162
A x (A + 9) = A² + 9A = 162 ⇒ A² + 9A - 162 = 0
(A + 18) (A - 9) = 0
A = {-18, 9}
B = { -9,18}
The two number are -18 and -9
or
9 and 18
13. The difference between two numbers is 10 and their sum is four times the smaller number. Find the two numbers.
Jawaban:
a = 15
b = 5
Penjelasan dengan langkah-langkah:
a - b = 10
a + b = 4b
a + b = 4b
a = 4b - b
a = 3b
a - b = 10
3b - b = 10
2b = 10
b = 5
a + 5 = 20
a = 20 - 5
a = 15
14. the sum of two numbers is 8. determine the two numbers such tat the sum of their squeres is minimized
what do you want to ask ???
15. 1. The sum of two numbers is 154 and their difference is 46. Find the two numbers. please answer
Jawaban:
x + (x + 46) = 154
x + x + 46 = 154
2x + 46 - 46 = 154 - 46
2x = 108
x = 108 ÷ 2
x = 54
x + 46
54 + 46
= 100
The two numbers are 54and 100.
16. the sum of two numbers is 138 and their difference is 88.Find the two numbers
x+y=138.....pers 1
x-y=88........pers 2
x=88+y.......pers 3
substitusi pers 3 ke pers 1
x+y=138
88+y+y=138
88+2y=138
2y=138-88
2y=50
y=50/2
y=25 substitusikan ke pers 3
x=88+y=88+25=113
itu caranya. semoga membantu
17. The sum of the squares of two consecutive positive numbers is 1105. Find the numbers.Help meee
Terdapat dua bilangan positif berurutan. Jumlah kuadrat kedua bilangan tersebut bernilai 1105. Analisis persamaan kuadrat memberikan solusi bahwa tidak ada dua bilangan positif berurutan yang memenuhi.
Penjelasan dengan langkah-langkahDiketahui:
Terdapat dua bilangan positif berurutan.
Jumlah kuadrat kedua bilangan tersebut bernilai 1105.
Ditanya: kedua bilangan tersebut
Jawab:
PemisalanMisalkan kedua bilangan tersebut termasuk bilangan bulat. Misalkan pula bilangan pertama adalah x, maka bilangan kedua yang lebih besar adalah x+1.
PersamaanJumlah kuadrat kedua bilangan tersebut bernilai 1105, dapat diubah ke dalam persamaan berikut:
x²+(x+1)² = 0
x²+x²+2x+1 = 0
2x²+2x+1 = 0
x²+x+0,5 = 0
DiskriminanKoefisien x² dan x-nya sama-sama bernilai 1, sedangkan konstantanya bernilai 0,5.
D = 1²-4·1·0,5 = 1-2 = -1 < 0
KesimpulanKarena diskriminannya bernilai negatif, maka persamaan kuadrat tersebut tidak memiliki akar real. Jadi, tidak ada dua bilangan positif berurutan yang jumlah kuadratnya bernilai 1105.
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18. The sum of three numbers in a particular arithmetic sequence is 33 and their product is 1287. Then, find the numbers that satisfied the sequence.
Jawab:
The numbers that satisfied the sequence are 9, 11, and 13.
Penjelasan dengan langkah-langkah:
If those three numbers are x, y, and z, then it is known that:
y - x = z - y ......(1) (rule of any arithmetic sequence)x + y + z = 33 .....(2)xyz = 1287 .....(3)From equation (1), we can get:
[tex]\begin{aligned}y-x &= z-y\\y-x-z+y&=0\quad\quad\text{(+z-y change\ side)}\\2y-x-z&=0\\\bf-x+2y-z&=\bf0\quad\dots\text{(4)}\end{aligned}[/tex]
The sum of equation (2) and (4) would be:
[tex]\begin{tabular}{r r r r c}x + &y + &z = &33&\ \\-x + &2y - &z = &0&\ \\\cline{1-4} &\\0 + &3y + &0 = &33&\ \\& & \bf y = & \bf 11\end{tabular}[/tex]
Substitute the value of y to equation (3):
[tex]\begin{aligned}xyz &= 1287\\11xz &= 1287\\xz &= 1287 \div 11\\\bf xz &= \bf 117\quad\dots(5)\end{aligned}[/tex]
Substitute the value of y to equation (2):
[tex]\begin{aligned}x+y+z&=33\\x+11+z&=33\\x+z&=33-11\\x+z&=22\\\bf z&=\bf 22-x\quad\dots(6)\end{aligned}[/tex]
And then, substitute the value of z from equation (6) to equation (5):
[tex]\begin{aligned}xz&=117\\x(22-x)&=117\\22x-x^2&=117\\-22x+x^2&=-117\quad\dots\text{($\times$ -1)}\\\bf x^2-22x+117&=\bf 0\quad\dots\text{(put them in order)}\\(x-9)(x-13)&=0\quad\dots\text{(factorization)}\\\bf x=9 \ \ or\ \ x&=\bf 13\end{aligned}[/tex]
At this point, we can guess that:
x = 9 and z = 13.
Let's verify this sequence:
9, 11, 13 : yes, this is an arithmetic sequencex + y + z = 9 + 11 + 13 = 33 : it is proved that the sum of these three numbers equals to 33.xyz = 9×11×13 = 99×13 = 1287 : it is also proved that their product is 1287.CONCLUSION:
The numbers that satisfied the sequence are 9, 11, and 13.
19. The sum of two numbers is 20. Their difference is 4. Find the two numbers.
Jawab:
Hari/Tanggal: Rabu, 15 Desember 2020
Jam: 08.57 WIB
____________________________________
a + b = 20
Difference is 4 (a - b)
So,
Elimination~
a + b = 20
a - b = 4
_________-
b + b = 16
2b = 16
b = 16/2
b = 8
Subtitution~
a - b = 4
a - 8 = 4
a = 4 + 8
a = 12
a = 12
b = 8
12 - 8 = 4
and
12 + 8 = 20
it's right :^
May Be Useful ^^
The two numbers are 12 and 8 .
PembahasanDiketahui :
The sum of two numbers is 20. Their difference is 4.
Ditanya :
Find the two numbers.
Jawab :
[tex]\text{Let two numbers are} \: \: x \: \: \text{and} \: \: y \: . \\ \\[/tex]
The sum of two numbers is 20.
[tex]x + y = 20 \\ \\ [/tex]
Their difference is 4.
[tex]x - y = 4 \\ \\ [/tex]
[tex]\text{Eliminate} \: \: y \: \: \text{to find} \: \: x \: .\\ \\ [/tex]
[tex]\begin{aligned}\\&x + y = 20\\ &x - y = 4\end{aligned} \\ \: \: \: \underline{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: } + \\ \begin{aligned}\\ \: 2x& = 24 \\ \: x& = 12\end{aligned} \\ [/tex]
Substitute x = 12 to equality x - y = 4
We get
[tex]x - y = 4 \\ \\ 12 - y = 4 \\ \\ y = 12 - 4 \\ \\ \boxed{y = 8} \\ \\ [/tex]
Kesimpulan :
The two numbers are 12 and 8 .
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Detail JawabanKelas : 8
Mapel : Matematika
Materi : Sistem persamaan linear dua variabel
Kode Kategorisasi : 8.2.5
Kata Kunci : SPLDV, eliminate, substitute, sum, difference
20. Is it possible to find 2005 different positive square numbers suchthat their sum is also a square number?Answer this question completely, Thank U
Jawaban:
the answer yes, it is possible.
Penjelasan dengan langkah-langkah:
it supposed to be pythagoras theorem which is a² + b² = c²
the sum of diffetent positive square numbers that is also a square number.
for example :
3² + 4⁴ = 5²
30² + 40² = 50²
or, 6² + 8² = 10²
60² + 80² = 100²
we can just find some pythagoras calculation, and add 0 to each pairs to have different answers.
in conclusion, to find such 2005 different pairs of pythagoras theorem is possible.
*cmiiw*
Jawaban:
Yes
Penjelasan dengan langkah-langkah:
Pythagoras theorem supports:
a² + b² = c²
Suppose we have a set of x integer be:
a² + b² = c²
d² + e² = f²
c² + f² = g²
and so on.
If you have 2005 numbers alike then:
2004 reduced to 1002 (Pythagoras), 1 remains
1002 reduced to 504, 1 remains
504 reduced to 252, 1 remains
252 reduced to 126, 1 remains
126 reduced to 63, 1 remains
64 (63 + remaining 1) reduced to 32
32 reduced to 16
16 reduced to 8
8 reduced to 4
4 reduced to 2
2 reduced to 1
Thus it is possible
What I mean by reduced, e.g. we have 4 numbers a, b, c, d:
a² + b² = x²
c² + d² = y²
So the 4-set numbers {a, b, c, d} here are reduced to 2-set numbers {x, y}
If we have an odd sets (e.g. 5) we can only reduce 4 of them, leaving 1 number not reduced at all.
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