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2007-10-15 14:54:27 · 11 answers · asked by CECILIA G 1 in Science & Mathematics Mathematics

11 answers

Just plug in x= -2 to evaluate the function at that value.

f(-2) = -5(-2) - 1
f(-2) = 10 - 1
f(-2) = 9

2007-10-15 14:56:48 · answer #1 · answered by Anonymous · 0 0

f (- 2 ) = (- 5 x - 2) - 1 = 10 - 1 = 9

2007-10-15 19:53:41 · answer #2 · answered by Como 7 · 0 0

9

2007-10-15 14:58:09 · answer #3 · answered by john 4 · 0 0

given this function, you just have to substitute -2 to the value of x...

f(-2) = -5x - 1
= -5(-2) - 1
= 10 - 1
= 9

There you have it!!

2007-10-15 15:00:18 · answer #4 · answered by Anonymous · 0 0

u just substitute it -2 in for x and solve so it would be
-5(-2)-1 and the answer is 9

2007-10-15 14:58:49 · answer #5 · answered by Hippo 1 · 0 0

What you are able to desire to do is to set the equation equivalent to sixteen, ex : sixteen = 16x2-64x+sixty 4. this might in essence mean f(x) = sixteen, considering f(x) = 6x2-64x+sixty 4. very properly, next you are able to simplify by using subtracting sixteen on the two aspects, ex : 16x2-64x+40 8=0. by using dividing this by using the utmost elementary factor of sixteen,sixty 4 and 40 8, particularly sixteen, you will get x2-4x+3=0. What this implies is that for you in looking a x such that f(x) = sixteen is comparable to looking a x so as that x2-4x+3=0. you are able to now factorize to get (x-a million)(x-3)=0. The values of x that should make this equation genuine is a million and 3. the two those solutions would be maximum suitable, on condition that there are actually not any barriers on the area.

2016-12-18 08:40:27 · answer #6 · answered by Anonymous · 0 0

plug in -2 for x so its -5(-2)-1=9

2007-10-15 14:58:34 · answer #7 · answered by johnnydepp2010 2 · 0 0

f(x)= -5x-1
then
f(-2)= 5(-2)-1
= -10-1
= -11

2007-10-15 14:58:34 · answer #8 · answered by Philip O 2 · 0 1

you substitute -2 for x

so -5*-2 -1

f(-2)=9

2007-10-15 14:57:10 · answer #9 · answered by 11swim11 3 · 0 0

-14

really dont beleive me im a 7th grader
and a stupid on at that.....

2007-10-15 14:58:10 · answer #10 · answered by Peyton(: 2 · 0 0

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