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this is an algebra equation which the answer to it is -9 but i cant seem to solve it. i keep gettin the wrong answer

b to the power of 2 + 5b - 3 when b = -2

2006-09-12 01:10:42 · 8 answers · asked by ▲▼ßððĝiз▼▲ 4 in Education & Reference Primary & Secondary Education

8 answers

-2 times -2 equals 4, then 5 times -2 is -10..you add them together and get -6...then -6 minus -3 is -9

2006-09-12 01:17:30 · answer #1 · answered by Michael D 5 · 1 0

Sure,whynot?

you said that, the equation is, b to the power of 2+5b - 3
while the result is -9
and the value of b is -2.

ok,Lets start

b raised 2+5b-3=-9 b=-2
and substitute every letter by -2
{(-2raised 2)+[5(-2)]-3}=-9
{(-2x-2)+(-10)-3}=-9
{4+(-10)-3}=-9
{(4-10)-3}=-9
{-6-3}=-9
-9=-9

hope it helps

2006-09-12 01:43:16 · answer #2 · answered by Anonymous · 0 0

Step 1 2 + 5b - 3 = ? and b = 2

Step 2 2 + 5 (2) - 3 = ?

Step 3 2 + 10 - 3 = ?

Step 4 12 - 3 = 9

2006-09-12 01:38:51 · answer #3 · answered by Wijekumara 4 · 0 0

whats wrong in that !!!
b2+5b-3//when b=-2
the sqaure of -2 is =4
thus 4+5b-3
4+5(-2)-3
4-10-3
4-13
= -9

2006-09-12 01:19:50 · answer #4 · answered by Anonymous · 0 0

b to the power of two would = 4 ... (-2)*(-2) = 4
+5b = -10
so, with the three terms as numerical values, you are left with
4 -10 -3 = -9

2006-09-12 01:19:48 · answer #5 · answered by metatron 4 · 0 0

I will be glad to help.

first solve the initial problem by substituting -2 for b
-2^2 + 5(-2) -3 = neg 2 squared is +4
4 + (-10) - 3 = then solve
4 + -13 = -9

when subtracting, add the opposite;
in other words make it an addition problem by adding a negative.

I hope this helps!

2006-09-12 01:23:21 · answer #6 · answered by grrlgenius5173 2 · 0 0

dude, that was simple...I did it in my mind in 5 seconds. Just always remember that 2 negatives muliplied become a postive and a negative and a postive mulitplied becomes a negative answer. That's probably where you messed up.

2006-09-12 01:58:34 · answer #7 · answered by boz4425 4 · 0 0

i am not knowing

2006-09-12 01:34:22 · answer #8 · answered by R.kumar s 1 · 0 0

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