Answer:
Step-by-step explanation:
A(-6, 2), B(6,-3) and C(-6, -3)
AB² = (x₂ - x₁)² + (y₂ - y₁)²
= ( 6 -[-6])² + ( -3 -2)²
= ( 6 + 6)² + ( -3 -2)² = 12² + (-5)² = 144 + 25 =169
AB = √169 = 13 units
BC² = ( -6 -6)² + ( -3 - [-3])² = (-6-6)² + (-3 +3)²
= (-12)² + 0 = 144
BC = √144 = 12 unis
CA² = (-6 - [-6])² +(-3-2)² = (-6 + 6)² + ( -3-2)²
= 0 + (-5)² = 25
CA =√25 = 5 units
length of the hypotenuse of a right triangle = 13units
The number of quarters she have 11.
<h3>What is a system of equations?</h3>
A system of equations is two or more equations that can be solved to get a unique solution. the power of the equation must be in one degree.
Let consider y = quarters; x = dimes
We know that there are 3 times as many dimes as quarters. So we can state that x = 3y.
Then, we say that 25y + 10x = 605
(Value of coin * amount of coins)
Then we substitute x=3y into the equation, yielding:
25y + 10(3y) = 605
25y + 30y = 605
55y = 605
605/55 = 11 = y
Therefore, the number of quarters she have 11.
Learn more about equations here;
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The expression equivalent to -7(y - 2) is -7y + 14
<h3>Which expression is equivalent to -7(y - 2)?</h3>
Given the expression; -7(y - 2)
We expand the expression by multiplying each element inside the parentheses the element out i.e -7
-7(y - 2)
[ -7 × y and -7 × -2 ]
-7y + 14
Therefore, the expression equivalent to -7(y - 2) is -7y + 14.
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The calculation uses the accumulated daily balance method (ADB).
We assume the statement is based on calendar month (rare!).
George owes $500 from beginning to end of June, so 30 days out of 30.
Interest accrued is 500*0.013*30/30=$6.50.
He also owes $2000 from June 12 to June 30, so 19 days inclusively.
Interest accrued is $2000*.013*(19/30)=16.47
Total interest at the end of the month=$6.50+$16.47=$22.97
40√3+432/7 is the answer
Hope this helps :D