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Find the solution/s of the following mathematical sentences. Describe these mathematical sentences.
1.b+3<10
2. 4c-5>19
3. 13-7m>-1
4. r^2+6r-7=0
5. 2m^2=50
6. n^2-7n= - 10


Categorize the mathematical sentences below using the given table.
☞︎︎ǫᴜᴀᴅʀᴀᴛɪᴄ ᴇǫᴜᴀᴛɪᴏɴs

☞︎︎︎ǫᴜᴀᴅʀᴀᴛɪᴄ ɪɴᴇǫᴜᴀʟɪᴛɪᴇs

☞︎︎︎ɴᴏᴛ ᴀ ǫᴜᴀᴅʀᴀᴛɪᴄ ᴇǫᴜᴀᴛɪᴏɴ ɴᴏʀ ᴀ ǫᴜᴀᴅʀᴀᴛɪᴄ ɪɴᴇᴀǫᴜᴀʟɪᴛʏ


FOR: Ate @CzarinaTrix
​

Find The Solutions Of The Following Mathematical Sentences Describe These Mathematical Sentences1b3lt102 4c5gt193 137mgt14 R26r705 2m2506 N27n 10Categorize The class=

Sagot :

AGboiiidn

Answer:

  1. b < 7
  2. c > 6
  3. m < 2
  4. r = -7 and 1
  5. m = 5 and -5
  6. n = 5 and 2

☞︎︎ǫᴜᴀᴅʀᴀᴛɪᴄ ᴇǫᴜᴀᴛɪᴏɴs

  • 5+6k = 7k²
  • x²-4x+8 = 0

☞︎︎︎ǫᴜᴀᴅʀᴀᴛɪᴄ ɪɴᴇǫᴜᴀʟɪᴛɪᴇs

  • 3x²+x+10 > 0
  • (2n²+5)(3n-1) > 0
  • 2s²-3s+5 < 0

☞︎︎︎ɴᴏᴛ ᴀ ǫᴜᴀᴅʀᴀᴛɪᴄ ᴇǫᴜᴀᴛɪᴏɴ ɴᴏʀ ᴀ ǫᴜᴀᴅʀᴀᴛɪᴄ ɪɴᴇᴀǫᴜᴀʟɪᴛʏ

  • m+5 <_ -12
  • 7d² <_ 14
  • (2r-3)(r+1) >_ 2

Step-by-step explanation:

1. b+3 < 10

b < 10 - 3

b < 7

2. 4c-5 > 19

4c > 19 + 5

4c > 24

c > 24/4

c > 6

3. 13-7m > -1

-7m > -1-13

-7m > -14

m < -14/-7

m < 2

4. r²+6r-7 = 0

(r+7)(r-1) = 0

r+7 = 0 , r-1 = 0

r = -7 , r = 1

5. 2m² = 50

m² = 50/2

m² = 25

m = √25

m = 5 and -5

6. n²-7n = -10

n²-7n+10 = 0

(n-5)(n-2) = 0

n-5 = 0 , n-2 = 0

n = 5 , n = 2

Hanimeyaidn

[tex]\red{⊱┈────────────────────┈⊰}[/tex]

DIRECTIONS :

[tex] \\ [/tex]

  • Find the solution/s of the following mathematical sentences. Describe these mathematical sentences.

[tex]\red{⊱┈────────────────────┈⊰}[/tex]

SOLUTION :

1) b+3 < 10

[tex] \\ [/tex]

  • Subtract 3 from both sides to get b < 10 - 3 then subtract 3 from 10 to have 7,so therefore the answer is [tex]\tt\red{\underline{ \: b < 7}}[/tex]

[tex]\red{⊱┈────────────────────┈⊰}[/tex]

2) 4c - 5 > 19

[tex] \\ [/tex]

  • First, let's add 5 to both sides,4c > 19 + 5, then add 19 + 5 to get 24,4c > 24.divide both sides by 4,since 4 is positive, the inequality direction remains the same , [tex]c>\frac{24}{4}[/tex] , then divide 24 by 4 to get 6,[tex] \tt\red{\underline{ \: c>6 }}[/tex]

[tex]\red{⊱┈────────────────────┈⊰}[/tex]

3) 13 - 7m > -1

[tex] \\ [/tex]

  • Subtract 13 from both sides, -7m > -1 -13, subtract 13 from -1 to get -14, -7m > -14,then divide both sides by -7, since -7 is negative, the inequality direction changed, [tex]m<\frac{-14}{-7},[/tex] divide -14 by -7 to get 2, [tex]\tt\red{\underline{ \: m<2 }}[/tex]

[tex]\red{⊱┈────────────────────┈⊰}[/tex]

4) r² + 6r - 7 = 0

[tex] \\ [/tex]

FORMULA :

  • r² + ( a + b) r + ab = ( r + a ) ( r + b )

SOLUTION :

[tex] \\ [/tex]

now let's set up a system to be solved :

  • [tex]\tt{a+b=6, ab=-7 }[/tex]
  • [tex]\tt{a=-1,b=7 }[/tex]
  • [tex]\tt{\left(r-1\right)\left(r+7\right) }[/tex]
  • [tex] \tt\red{{ \:r=1 \: or \:r=-7 }}[/tex]

[tex] \\ [/tex]

Therefore,the answer is [tex] \tt\red{\underline{ \:r=1,r=-7 }}[/tex]

[tex]\red{⊱┈────────────────────┈⊰}[/tex]

5) 2m² = 50

[tex] \\ [/tex]

  • Divide both sides by 2,[tex]m^{2}=\frac{50}{2} [/tex], then divide 50 by 2 to get 25,m² = 25, take the square root of both sides of the equation to get the answer, [tex]\tt\red{\underline{ \: m=5 \:or \:m=-5 }}[/tex]

[tex]\red{⊱┈────────────────────┈⊰}[/tex]

6) n² - 7n = - 10

[tex] \\ [/tex]

FORMULA :

  • n² + ( a + b) n + ab = ( n + a ) ( n + b )

SOLUTION :

[tex] \\ [/tex]

now let's set up a system to be solved :

  • [tex]\tt{ a+b=-7 \:,\: ab=10 }[/tex]
  • [tex]\tt{ -1,-10\:, \:-2,-5 }[/tex]
  • [tex]\tt{ -1-10=-11 \:,\:-2-5=-7 }[/tex]
  • [tex]\tt{ a=-5\:,\:b=-2 }[/tex]
  • [tex]\tt\red{ \left(n-5\right)\:,\:\left(n-2\right) }[/tex]

[tex] \\ [/tex]

Therefore, the answer is [tex]\tt\red{\underline{ \: n=5 \:,\:n=2 }}[/tex]

[tex]\red{⊱┈────────────────────┈⊰}[/tex]

[tex]\tiny\begin{array}{|c|c|c|}\hline \\ \sf \: Quadratic \: Equations & \sf \: Quadratic \: Inequalities & \sf \: Not \: a \: Quadratic \: Equation \: nor \: a \: Quadratic \: Inequalities \\ \hline \\ \red{5 + 6k = 7 {k}^{2} } \: & \red{3 {x}^{2} + x + 10 > 0 } & \red{m + 5 <_ { - 12} } \\ \hline \\ \red{ {x}^{2} - 4x + 8 = 0} \: & \red{( \: 2 {n}^{2} + 5 \: )( \: 3n - 1 \: ) } & \red{7 {d}^{2} < _ { 14} } \\ \hline \\ \: & \red{2 {s}^{2} - 3s + 5 < 0 } & \red{(2r - 3)(r + 1) < _ { 2} } \\ \hline \end{array}[/tex]

[tex]\red{⊱┈────────────────────┈⊰}[/tex]

#CarryOnLearning

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