find t1 and t2?

Question

875=2(t1 * t2)+t1^2
90= t1+t2

find t1 and t2?

Please help and show steps.

Answer 1

875=2(t1 * t2)+t1^2 ............eq 1
90= t1+t2 .............................eq 2

Square both sides of eq2
t1^2 + 2*t1*t2 + t2^2 = 8100
Substitute from eq 2 :
875 + t2^2 = 8100
t2 = 85

Put in 2
t1 = 5

Hope this helps.

Answer 2

first expand
875 = 2t1t2 + t1^2
875 = t1(2t2 + t1)

we know t2 = 90 - t1

875 = t1( 2(90-t1) + t1)
875 = t1(180 - 2t1 + t1)
875 = t1(180 - t1)
-t1^2 + 180t1 - 875 = 0
(-t1 + 5 )(t1 - 175) = 0

so t1 = 5 or t1 = 175

if t1 = 5, t2 = 85
t2 = 175, t2 = -85

Answer 3

875 = 2(t1 * t2) + t1^2
90 = t1 + t2
t2 = 90 - t1
875 = 2t1(90 - t1) + t1^2
875 = 180t1 - 2t1^2 + t1^2
875 = 180t1 - t1^2
t1^2 - 180t1 + 875 = 0
t1^2 - 180t1 + 8100 - 8100 + 875 = 0
(t1 - 90)^2 - 7,225 = 0
(t1 - 90 + 85)(t1 - 90 - 85) = 0
(t1 - 5)(t1 - 175) = 0
t1 = 5, 175
t2 = 85, - 85
(t1,t2) = (5,85), (175,-85)

Answer 4

875 = 2(t1 * t2) + t1^2
90 = t1 + t2

90 = t1 + t2
t2 = 90 - t1

875 = 2(t1(90 - t1)) + t1^2
875 = 2(90t1 - t1^2) + t1^2
875 = 180t1 - 2t1^2 + t1^2
875 = -t1^2 + 180t1
t1^2 - 180t1 + 875 = 0

t1 = (-b ± sqrt(b^2 - 4ac))/(2a)

t1 = (-(-180) ± sqrt((-180)^2 - 4(1)(875)))/(2(1))
t1 = (180 ± sqrt(32400 - 3500))/2
t1 = (180 ± sqrt(28900))/2
t1 = (180 ± 170)/2
t1 = 10/2 or 350/2
t1 = 5 or 175

t2 = 90 - t1
t2 = 90 - (5 or 175)
t2 = 85 or -85

ANS :
t1 = 5 and t2 = 85
or
t1 = 175 and t2 = -85

Answer 5

CAN BE EXPRESSED AS:

x^2 + 2xy = 875
x + y = 90

WHERE, x = t1 AND y = t2

EQUATE FOR X:

x = 90 - y

SUBSTITUTE X:

(90 - y)^2 + 2 (90 - y) y = 875

(90 + y)(90 - y) = 875

8100 - y^2 = 875

-y^2 = 875 - 8100

-y^2 = -7225

sqrt(-y^2) = sqrt(-7225)

y = 85

OR

y = -85

SOLVE FOR X:

x + y = 90

x + 85 = 90

x = 90 - 85

x = 5

OR

x + y = 90

x + -85 = 90

x = 90 + 85

x = 175

THEREFORE:


t1, t2 = {5,85}

OR

t1, t2 = {175,-85}

Answer 6

t1+t2=90
t2=90-t1
substitute in other eqn
t1^2+2((90*t1)-t1^2)-875=0
-t1^2+180*t1-875=0
t1=175 or t1=5
t2=-85 or t2=85

Answer 7

875 = 2*t1*t2 + t1^2
90 = t1+t2 => t1 = 90-t2

875 = 2*(90-t2)*t2 + (90-t2)^2
875 = 2*(90t2-t2^2) + 8100-180t2+t2^2
875 = 180t2 -2t2^2 + 8100 -180t2 + t2^2
875 = -2t2^2 + 8100 + t2^2
875 = 8100 - t2^2 | -875
0 = 7225 - t2^2 | + t2^2
t2^2 = 7225 | square root
t2 = 85 or t2 = -85

if t2 = 85 then t1 = 90-85 = 5
if t2 = -85 then t1 = 90-(-85) = 90+85 = 175

Answer 8

lets take the first equation

875=2(t1*t2)+t1^2

we can take t1 as common

875= t1(2t2+t1) ------name it as 3

from 2nd equation

t1= 90 - t2

sub t1= 90 - t2 in 3

875 = (90 - t2 ) (2t2 + 90 - t2)

875 = (90 - t2) (t2 + 90)
875= (90+t2) (90 - t2) it is of the form (a+b)(a-b)
875 = 90^2 - t2^2
t2^2= 8100 - 875
t2 = 85 or -85

if we sub t2=85 in 2nd equation , we will get t1= 5

if we sub t2= -85 in 2nd equation , we will get t1=175

(t1,t2)=(5,85) or (175,-85)

Answer 9

I am assuming this is just a normal algebra question in which case 't' would equal the same thing... and you should be working this out on ur own!!

Answer 10

sorry, don't get what the problem is.... you increase the pressure (maintaing the volume constant) and therefore the temperature should increase (from 25C to 67.7C).....