Answer:
Step-by-step explanation:
22) J(2,-1); K(2,5)
Distance = √(x₂-x₁)² + (y₂-y₁)²
= √( 2- 2)² + (5 - [-1] )²
= √0 + (5 + 1 )²
=√ ( 6 )² = 6
25) A(0,3) ; B (0,12)
distance =√(0-0)² + (12 - 3 )²
=√( 9)² = 9
28) Q(12,-12) ; T(5,12)
distance = √(5 - 12)² + (12 - [-12] )²
= √( -7)² + (12 + 12] )²
= √ 49 + (24)²
= √ 49 + 576
= √625
=25
Answer:
(3,-4)
Step-by-step explanation:
I have seen this question before and I think you meant 10 more dimes than nickels.
We can use substitution to answer this question. The value of a nickel is 5 cents, and we can use the variable n to represent the number of nickels. The value of a dime is 10 cents, and we can use the variable d to represent the number of dimes.
First lets figure out the equations.
.10d+.5n=2.80 (the number of nickels (n) multiplied by .5 will tell us their money value. Same thing for the dimes)
d-n=10 (since there are 13 more dimes than nickels, the number of dimes value (d) minus the number of nickels value (n) will give us 10)
Now lets isolate a variable in one of the equations, preferably the second one because it doesn't have any visible coefficients,
d-n=10
-n=10-d (subtracted the d from both sides)
n=-10+d (made the n positive)
Now that we have the value of n, we can plug it into the other equation.
.10d+.05n=2.80
.10d+.05(-10+d)=2.80 (we replaced the n with the value that we previously got)
.10d-.5+.05d=2.80 (did the multiplication)
.15d-.5=2.80 (combined like terms)
.15d=3.30 (added the .5 to both sides)
d=22 (divided both sides by the .15)
Now that we know that there are 22 dimes and we also know that there are 10 less nickels than dimes, so we can subtract 10 from 22 to get the number of nickels. 22-10=12
d=22
n=12
Answer:
(-3, 3)
Step-by-step explanation:
Substitute for y:
x+6 = -2x -3
3x = -9 . . . . . . . add 2x -6 to both sides
x = -3 . . . . . . . . divide by 3
y = (-3) +6 = 3 . . substitute for x in the first equation
The solution is ...
(x, y) = (-3, 3)
The volume of a rectangular prism is its length times width times height, or algebraically,
. You may be used to computing volume with numbers, but remember, a variable is a stand-in for a number. So you can solve this in the same way. Substitute
into the formula for volume. You get
, and you multiply these factors together. As you would with ordinary fractions, multiply the numerators and denominators across. You get
. It seems that the book wants you to simplify by bringing the 6 up to the denominator. Recall the rule
, if n is non-negative. The opposite applies so that
. For your final answer, you write
. This corresponds to
answer choice B.
