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gayaneshka [121]
3 years ago
12

The greater of two numbers is 1 more than twice the smaller. Three times the greater exceeds 5 times the smaller by 10. Find the

numbers
Mathematics
1 answer:
earnstyle [38]3 years ago
5 0
X<y
y is 1 more than 2 times the smaler
y=1+2 times x
y=1+2x

3 times y is bigger than 5 times x by 10
3y=5x+10
subsitute
y=1+2x
3(1+2x)=5x+10
3+6x=5x+10
subtract 3 from both sides
6x=5x+7
subtract 5x from both sides
x=7
subsitute
y=1+2x
y=1+2(7)
y=1+14
y=15

x=7
y=15

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Which of the following points would fall on the line produced by the point-slope form equation y - 2 = 3/2(x + 4) when graphed?
Zielflug [23.3K]

Answer:

c. (-2, 5)

Step-by-step explanation:

y - 2 = 3/2(x + 4)

to know which is the correct option we have to replace the equation x and y with what is in the point

a.

y - 2 = 3/2(x + 4)

(-1) - 2 = 3/2((-2) + 4)

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b.

y - 2 = 3/2(x + 4)

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Wrong

c.

y - 2 = 3/2(x + 4)

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3 = 3

Correct

d.

y - 2 = 3/2(x + 4)

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8 0
3 years ago
Can someone tell me what 8 x 1 1/2 equals
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7 0
4 years ago
Read 2 more answers
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
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