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Vlada [557]
2 years ago
14

Choose the most convenient method to graph the line x - y = 5.

Mathematics
2 answers:
givi [52]2 years ago
7 0
The most convenient method I would choose is slope-intercept form, which is y=mx+b. So, make this equation equal in terms of y.

x-y=5
subtract x from both sides
-y = 5 - x
multiply both sides by -1 so y isn't negative
y = 5 + x
5 +x is the same thing as x+5, so
y = x+5

Now, you can easily sub in any number for x and you can easily get y. Here are some points you can plot:

(0, 5), (1, 6), (2, 7), (3, 8) and the list goes on. I hope this helped!
dmitriy555 [2]2 years ago
7 0

Answer:

Step-by-step explanation:

x - y = 5

Write in slope-intercept form: y = mx +b

-y = -x + 5

Multiply the entire equation by(-1)

y = x - 5

When x = 0 ; y = 0 - 5 = -5    ;              (0,-5)

When x = 1  ; y = 1 - 5 = - 4     ;              (1 , -4)

When x = 4 ; y =  4 - 5 = -1      ;              (4 , -1)

When x = -1 ; y = -1 - 5 = - 6      ;           (-1,-6)

Plot the points and join the points

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Can someone please help me, Thanks - Alyssa T.
Oxana [17]
A=LW
192=(13+x)((9+x)
192=117+22x+x^2 (subtract 192 from both sides)
x^2+22x-75+0 (factor)
(x+25)(x-3)=0
so x=-25 or 3
Check work
since we are adding x it can only be the positive solution so x=3 so the new dimensions are:
L=13+3=16m
W=9+3=12
check: 16*12=192 so the answer is correct hope this helped
7 0
3 years ago
The solution x=7 is a solution to which of the following equations? Check all that apply
IRISSAK [1]
The correct answer is D.
X=7
8=7+1
8=8
3 0
3 years ago
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I need help.. i really want to go sleep.. thank you so much...
Effectus [21]

Answer:

1) True 2) False

Step-by-step explanation:

1) Given  \sum\limits_{k=0}^8\frac{1}{k+3}=\sum\limits_{i=3}^{11}\frac{1}{i}

To verify that the above equality is true or false:

Now find \sum\limits_{k=0}^8\frac{1}{k+3}

Expanding the summation we get

\sum\limits_{k=0}^8\frac{1}{k+3}=\frac{1}{0+3}+\frac{1}{1+3}+\frac{1}{2+3}+\frac{1}{3+3}+\frac{1}{4+3}+\frac{1}{5+3}+\frac{1}{6+3}+\frac{1}{7+3}+\frac{1}{8+3} \sum\limits_{k=0}^8\frac{1}{k+3}=\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10}+\frac{1}{11}

Now find \sum\limits_{i=3}^{11}\frac{1}{i}

Expanding the summation we get

\sum\limits_{i=3}^{11}\frac{1}{i}=\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10}+\frac{1}{11}

 Comparing the two series  we get,

\sum\limits_{k=0}^8\frac{1}{k+3}=\sum\limits_{i=3}^{11}\frac{1}{i} so the given equality is true.

2) Given \sum\limits_{k=0}^4\frac{3k+3}{k+6}=\sum\limits_{i=1}^3\frac{3i}{i+5}

Verify the above equality is true or false

Now find \sum\limits_{k=0}^4\frac{3k+3}{k+6}

Expanding the summation we get

\sum\limits_{k=0}^4\frac{3k+3}{k+6}=\frac{3(0)+3}{0+6}+\frac{3(1)+3}{1+6}+\frac{3(2)+3}{2+6}+\frac{3(3)+4}{3+6}+\frac{3(4)+3}{4+6}

\sum\limits_{k=0}^4\frac{3k+3}{k+6}=\frac{3}{6}+\frac{6}{7}+\frac{9}{8}+\frac{12}{8}+\frac{15}{10}

now find \sum\limits_{i=1}^3\frac{3i}{i+5}

Expanding the summation we get

\sum\limits_{i=1}^3\frac{3i}{i+5}=\frac{3(0)}{0+5}+\frac{3(1)}{1+5}+\frac{3(2)}{2+5}+\frac{3(3)}{3+5}

\sum\limits_{i=1}^3\frac{3i}{i+5}=\frac{3}{6}+\frac{6}{7}+\frac{9}{8}

Comparing the series we get that the given equality is false.

ie, \sum\limits_{k=0}^4\frac{3k+3}{k+6}\neq\sum\limits_{i=1}^3\frac{3i}{i+5}

6 0
3 years ago
Which is greater, 12/17 or 5/8?
Vilka [71]

Answer:

5/8

Step-by-step explanation:

12/17 is 0.70

5/8 is 0.62

However, 12/17 had a lot more numbers than that so

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3 0
3 years ago
Read 2 more answers
Please help me .......
Misha Larkins [42]
First we have to know the Pythagoreans theorem which is

a^2+b^2=c^2
7^2+8^2=c^2
49+64= 10.63
Hope this helps
3 0
3 years ago
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