Under a dilation, triangle XYZ where X(4, 7), Y(-3, 8), and Z(-2, -1) becomes triangle X’Y’Z’ where X’(12, 21), Y’(-9, 24), and
-BARSIC- [3]
It is 3, the points are all 3 times greater
<h3>
Answer: Largest value is a = 9</h3>
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Work Shown:
b = 5
(2b)^2 = (2*5)^2 = 100
So we want the expression a^2+3b to be less than (2b)^2 = 100
We need to solve a^2 + 3b < 100 which turns into
a^2 + 3b < 100
a^2 + 3(5) < 100
a^2 + 15 < 100
after substituting in b = 5.
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Let's isolate 'a'
a^2 + 15 < 100
a^2 < 100-15
a^2 < 85
a < sqrt(85)
a < 9.2195
'a' is an integer, so we round down to the nearest whole number to get 
So the greatest integer possible for 'a' is a = 9.
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Check:
plug in a = 9 and b = 5
a^2 + 3b < 100
9^2 + 3(5) < 100
81 + 15 < 100
96 < 100 .... true statement
now try a = 10 and b = 5
a^2 + 3b < 100
10^2 + 3(5) < 100
100 + 15 < 100 ... you can probably already see the issue
115 < 100 ... this is false, so a = 10 doesn't work
Answer: no
Step-by-step explanation: no
Answer:
No solution
Step-by-step explanation:
-3x + 3y = 4
-x + y = 3
<u>Solve the equation for x:</u>
<u>Move the variable to the right side and change its sign</u>
-x + y = 3
-x = 3 - y
<u>Change the signs on both sides of the equation</u>
-x = 3 - y
x = -3 + y
<u>Substitute the given value of x into the equation -3x + 3y = 4</u>
-3x + 3y = 4
x = -3 + y
-3(-3+y)+3y=4
<u>Solve the equation for y</u>
-3(-3+y)+3y=4
y = 0
There is no solution for y.
And since there's no solution for y, the system has no solution.