(2,3) so x=2, y=3 and h=(2^2+3^2)^(1/2)=√13
sina=3/√13, cosa=2/√13, tana=3/2
Afterwards multiply sina and cosa by √13/√13 and get sina=(3√13)/13 and cosa=(2√13)/13
Answer: you multiply then add.
Step-by-step explanation:
Answer:
y= -3
Step-by-step explanation:
<u>Multiply all terms by the same value to eliminate fraction denominators:</u>
2/y +3 = 3/3y + 9
(3y+9) (y+3) (2/y+3) = (3y +9) (3/3y+9)
<u>Simplify:</u>
6
+
1
8
=
3
+
9
<u>Subtract 18 from both sides of the equation:</u>
6y +18 =3y +9
6y+18 -18 =3y +9 -18
<u>Simplify Again: </u>
6y +3y-9
<u>Subtract 3y from both sides of the equation:</u>
6y =3y -9
6y-3y = 3y -9 -3y
<u>Answer:</u>
y= -3
Step-by-step explanation:
We have
87+86+x
T is the total scores
n is number of tests done = 3
T/n is greater than or equal to 90
To get minimum test
T/n = 90
T = 90*n
= 90n
We solve for x
Remember n = 3
90x3 = 270
270 = 87+86+x
270 = 173+x
X = 270-173
X = 97
1.
X = 97
2.
You didn't add a value in your part b question. I used 80 though.
T/n < 80
T<80n
80x3= 240
T<240
87+86+x<240
X<240-173
X<67
X has to be less than 67 so we conclude that if x <=67 b grade will be lost.
Answer:
I think the answer would be −329/8 or −41.125 .
Step-by-step explanation:
Hope this helped .
