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ad-work [718]
3 years ago
6

Question 5(Multiple Choice Worth 5 points)

Mathematics
1 answer:
andriy [413]3 years ago
7 0

Answer:

A is the correct answer have a nice day

Step-by-step explanation:

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the sum of three consecutive integers must not be more than 12 what are the integer.grade 11 linear inequalities.
ICE Princess25 [194]
[tex]x+(x+1)+(x+2)\leq 12\\
3x+3\leq 12\\
3x\leq 9\\
x\leq 3\\
\boxed{\text{3,4,5\ or\ lesser}}
4 0
3 years ago
An integer minus 5 times its reciprocal is<br> 76<br> 9<br> What is the integer?
GaryK [48]

Answer: The only one of these that's an integer is 9.

Explanation: Let X be the integer. Then

X - 3/X = 26/3. Multiplying by 3X on both sides:

3X^2 - 9 = 26X

3X^2 - 26X - 9 = 0

X = (26 +/- SQRT(26^2-(4*(-27))))/6 = (26 +/- 28) / 6,

so the two solutions for X are 1/3 and 54/6 = 9.
4 0
3 years ago
Can you help me on my homework
marshall27 [118]
The answer is #1. The mean of Team A is less than the mean of Team B, and the standard deviation of Team A is greater than the standard deviation of Team B.
6 0
3 years ago
Read 2 more answers
Step-by-step how-to evaluation
Orlov [11]
Least common multiple=lcm
lcm(9,7)=63

126(8/9  +  (x-4)/7 )=
126(7*8+9(x-4)/63=
126(56+9x-36)/63=
126(20+9x)/63=                      (126/63=2)
2(20+9x)=
40+18x

Answer: 126(8/9  +  (x-4)/7 )=  40+18x
3 0
3 years ago
Read 2 more answers
Xsquared+1/20x-1/20 so I need that to be solved any way possible
VARVARA [1.3K]
Hallo ^_^

Quadratic Equation :
ax² + bx + c = 0

From the task :
x² + 1/20 x = 1/20

Switch sides :
x² + 1/20x - 1/20 = 0

Multiple each sides by 20
20(x² + 1/20x - 1/20) = 0(20)
⇒ 20x² + x - 1 = 0

From that :
a = 20, b = 1 and c = - 1

Solve x using quadratic formula :

x_1x_2 = \frac{-b\pm\sqrt{b^2-4ac}}{2a}

Substitution using the formula :

x_1x_2 = \frac{-1 \pm \sqrt{1^2-4(20)(-1)}}{2(20)} \\\\&#10;x_1x_2 = \frac{-1 \pm \sqrt{81}}{40} \\\\&#10;x_1x_2 = \frac{-1 \pm 9}{40} \\\\&#10;&#10;x_1 =  \frac{- 1 + 9}{40} \\\\&#10;x_1 =  \frac{8}{40} \\\\&#10;x_1 =  \frac{1}{5} \\\\&#10;&#10;x_2 =  \frac{- 1 - 9}{40} \\\\&#10;x_2 =  \frac{-10}{40} \\\\&#10;x_2 = -  \frac{1}{4} \\\\&#10;&#10;\therefore \boxed{x = \frac{1}{5} \ or \ x =-  \frac{1}{4}}

Sorry for my bad english :)
5 0
3 years ago
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