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goblinko [34]
3 years ago
12

Help me please ....​

Mathematics
1 answer:
trasher [3.6K]3 years ago
5 0

i think that this is wrong and your asswer 56b x56i +78 t and that this maybe wrong but atleast i tried to help!

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Using no digits other than 5 and 8, make a 6-digit number, that has the properties below. If it is impossible, write so. Of cour
Anna11 [10]
Since it's a multiple of 24, it has to be a multiple of the factors of 24.

Factors of 24:
2,3,4,6,8,12

You can use some of this knowledge to help create the number.
Since the # needs to be a multiple off 2, the last digit needs to be an 8

All numbers that are multiples of 3 have the property that all of their digits added together have to be a number that is evenly divisible by 3.

so your number will look like:
_ _ _ _ _ 8

so start trying combinations for the other 5 digits that give you a number that is a multiple of 3: 3,6,9,12,15, ect. If you can't find one, then it's impossible
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Consider these numbers, ordered from least to greatest. Negative 1.8, negative 1 and one-fourth, negative 0.2, blank, one-half,
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Step-by-step explanation:

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3 years ago
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Please help me someone with this question
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4 0
4 years ago
Find k so that the distance from (–1, 1) to (2, k) is 5 units. k= k= *there are two solutions for 2*
dalvyx [7]

Answer:

k = -3

k =5

Step-by-step explanation:

d = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\d = 5\\(-1,1) =(x_1,y_1)\\(2,k)=(x_2,y_2)\\

5=\sqrt{\left(2-\left(-1\right)\right)^2+\left(k-1\right)^2}\\\\\mathrm{Square\:both\:sides}:\quad 25=k^2-2k+10\\25=k^2-2k+10\\\\\mathrm{Solve\:}\:25=k^2-2k+10:\\k^2-2k+10=25\\\\\mathrm{Subtract\:}25\mathrm{\:from\:both\:sides}\\k^2-2k+10-25=25-25\\k^2-2k-15=0\\\\\mathrm{Solve\:by\:factoring}\\\\\mathrm{Factor\:}k^2-2k-15:\quad \left(k+3\right)\left(k-5\right)\\\mathrm{Solve\:}\:k+3=0:\quad k=-3\\

\mathrm{Solve\:}\:k-5=0:\quad k=5\\\\k =5 , k=-3

7 0
3 years ago
Read 2 more answers
Y = x + 2 and y = 2x + 3
Reil [10]
Im confused as to what you are asking!
8 0
3 years ago
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