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Elza [17]
3 years ago
6

Is a number is blank then the number is less than it’s absolute value

Mathematics
1 answer:
svlad2 [7]3 years ago
6 0

Answer:

If a number is <u>negative</u> then the number is less than its absolute value.

Step-by-step explanation:

If a number is <u>negative</u> then the number is less than its absolute value.

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Solve: XXV - IV Show your answer in standard form.​
storchak [24]

Step-by-step explanation:

the answer is xxl in roman number or 21

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2 years ago
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 find all possible value of the given variable 
mamaluj [8]
1.\\ \\ h^2+5h=0 \\ \\h(x+5)=0\\ \\x=0 \ \ \ or \ \ \ x+5 =0\ \ |-5\\ \\x+5-5=0-5\\ \\x=0 \ \ \ or \ \ \ x=-5


2.\\ \\ z^2-z=0\\ \\z(x-1)=0\\ \\z=0 \ \ \ or \ \ \ z-1 =0 \ \ | +1\\ \\z-1+1 =0 +1 \\ \\x=0 \ \ \ or \ \ \ z=1


3.\\ \\m^2+13m+40=0 \\ \\a=1 ,\ b=13, \ c=40 \\ \\\Delta =b^2-4ac =13^2-4\cdot 1\cdot 40=169 - 1600=-1431 \\ \\and \ we \ know \ when \ \Delta \ is \ negative, \ theres \ no \solution


4.\\ \\z^2-3z=0 \\ \\ (z-3)=0\\ \\z=0 \ \ \ or \ \ \ z-3 =0\ \ |+3\\ \\ z-3+3=0+3\\ \\z=0 \ \ \ or \ \ \ z=3


5.\\ \\q^2+7q=0 \\ \\q(q+7)=0\\ \\q=0 \ \ \ or \ \ \ q+7 =0\ \ |-7\\ \\q+7-7=0-7\\ \\q=0 \ \ \ or \ \ \ q=-7


6.\\ \\k^2+2k=0\\ \\k(k+2)=0\\ \\k=0 \ \ \ or \ \ \ k+2 =0\ \ |-2\\ \\k+2-2=0-2\\ \\k=0 \ \ \ or \ \ \ k=-2


7. \\ \\ x^2-3x-70=0 \\ \\a=1,\ b=-3, \ c=-70 \\ \\\Delta =b^2-4ac = (-3)^2-4\cdot 1\cdot (-70)= 9+280=289\\ \\ x_{1}=\frac{-b-\sqrt{\Delta} }{2a}=\frac{3-\sqrt{289}}{2 }=\frac{ 3-17}{2}=\frac{-14}{2}=-7

x_{2}=\frac{-b+\sqrt{\Delta} }{2a}=\frac{3+\sqrt{289}}{2 }=\frac{ 3+17}{2}=\frac{20}{2}=10\\ \\(x+7)(x-10)=0


8.\\ \\q^2+7q-60=0 \\ \\a=1,\ b=7, \ q=-60 \\ \\\Delta =b^2-4ac = 7^2-4\cdot 1\cdot (-60)=49+240=289 \\ \\ x_{1}=\frac{-b-\sqrt{\Delta} }{2a}=\frac{-7-\sqrt{289}}{2 }=\frac{ -7-17}{2}=\frac{-24}{2}=-12

x_{2}=\frac{-b+\sqrt{\Delta} }{2a}=\frac{-7+\sqrt{289}}{2 }=\frac{ -7+17}{2}=\frac{ 10}{2}= 5\\ \\(x+12)(x-5)=0


9.\\ \\z^2+9z-36=0 \\ \\a=1,\ b=9, \ q=-36 \\ \\\Delta =b^2-4ac = 9^2-4\cdot 1\cdot (-36)= 81+144=225\\ \\ x_{1}=\frac{-b-\sqrt{\Delta} }{2a}=\frac{-9-\sqrt{225}}{2 }=\frac{ -9-15}{2}=\frac{-24}{2}=-12

x_{2}=\frac{-b+\sqrt{\Delta} }{2a}=\frac{-9+\sqrt{225}}{2 }=\frac{ -9+15}{2}=\frac{6}{2}=3\\ \\(x+11)(x-3)=0


10.\\ \\d^2-13d+22=0 \\ \\a=1,\ b=-13, \ q=22 \\ \\\Delta =b^2-4ac = (-13)^2-4\cdot 1\cdot 22= 169-88=81\\ \\ d_{1}=\frac{-b-\sqrt{\Delta} }{2a}=\frac{13-\sqrt{81}}{2 }=\frac{ 13-9}{2}=\frac{4}{2}=2

d_{2}=\frac{-b+\sqrt{\Delta} }{2a}=\frac{13+\sqrt{81}}{2 }=\frac{ 13+9}{2}=\frac{22}{2}=11\\ \\(d-2)(d-11)=0


7 0
3 years ago
4.5x – 2y = 15<br> 3x - y =10<br><br> (answer as a combined pair)
prisoha [69]

Answer:

x = 10/3 , y = 0

Step-by-step explanation:

Solve the following system:

{4.5 x - 2 y = 15

3 x - y = 10

In the first equation, look to solve for x:

{4.5 x - 2 y = 15

3 x - y = 10

4.5 x - 2 y = (9 x)/2 - 2 y:

(9 x)/2 - 2 y = 15

Add 2 y to both sides:

{(9 x)/2 = 2 y + 15

3 x - y = 10

Multiply both sides by 2/9:

{x = (4 y)/9 + 10/3

3 x - y = 10

Substitute x = (4 y)/9 + 10/3 into the second equation:

{x = (4 y)/9 + 10/3

3 ((4 y)/9 + 10/3) - y = 10

3 ((4 y)/9 + 10/3) - y = ((4 y)/3 + 10) - y = y/3 + 10:

{x = (4 y)/9 + 10/3

y/3 + 10 = 10

In the second equation, look to solve for y:

{x = (4 y)/9 + 10/3

y/3 + 10 = 10

Subtract 10 from both sides:

{x = (4 y)/9 + 10/3

y/3 = 0

Multiply both sides by 3:

{x = (4 y)/9 + 10/3

y = 0

Substitute y = 0 into the first equation:

Answer:  {x = 10/3 , y = 0

7 0
2 years ago
I thought of a number. My number doubled is 20 bigger than 200. Find my number.
Illusion [34]

Answer:

110

Step-by-step explanation:

200+20=220

220/2=110

5 0
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KATRIN_1 [288]

Answer:

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Step-by-step explanation:

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