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3 years ago

13

How do I round decimals?

2 answers:

Sergio [31]3 years ago

8 0

Ok heres how you round decimals example if you have 1.5 you can increase it to 2 let me teach you a trick to do it easy 5 or more raise the score example
1.7 can round up to 2 and 4 or less let it rest example 1.3 you cant round that so it stays the same

Ray Of Light [21]3 years ago

3 0

So its pretty much like if your rounding reguraly but the only diffrience is that it had a decimal in it or a dot and you always have to remeber to round the big one first

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What are the coefficients in this expression? 2x +9y+(-4) Select all that apply.

zalisa [80]

A) 2 and B) 9
Coefficients are the numbers in front of a variable. The x and y wouldn’t be considered coefficients.

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3 years ago

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How many times does X squared minus 4X -12 cross the X axis

kolbaska11 [484]

We can solve this problem using discriminant.

x^2-4x-12's discriminant is

(-4)^2-4*-12=16+48 which is clearly larger than 0

This means that it crosses over the axes 2 times.

In case you don't know what discriminant is, its in equation ax^2+bx+c

the discriminant is b^2-4ac.

If its positive it has 2 crosses with x axis, if negative then 0 crosses, if 0 then 1 cross.

Hope this helped at least a little bit :D

4 0

3 years ago

Please help me, i will give you brainliest

True [87]

Answer:

4

Step-by-step explanation:

(segment piece) x (segment piece) = (segment piece) x (segment piece)

JN* NK = LN * NM

3x = 2*6

3x = 12

Divide by 3

3x/3 =12/3

x =4

6 0

3 years ago

Solve x4 – 17x2 + 16 = 0.<br> Let u =<br> Select all of the zeroes of the function. i i

Soloha48 [4]

$Solve \ x^4 - 17x^2 + 16 = 0$

To solve this equation , we need to write it in quadratic form

$ax^2+bx+c=0$

To get the equation in quadratic form we replace x^2 with u

$x^2 = u$

$x^4 - 17x^2 + 16 = 0$ can be written as

$(x^2)^2 - 17x^2 + 16 = 0$ , Replace u for x^2

So equation becomes

$u^2 - 17u + 16 = 0$

Now we factor the left hand side

-16 and -1 are the two factors whose product is +16 and sum is -17

(u-16) (u-1) = 0

u -16 = 0 so u=16

u-1 =0 so u=1

WE assume u = x^2, Now we replace u with x^2

$u = 16 so \ it \ becomes \ x^2 =16$

Now take square root on both sides , x= +4 and x=-4

$u = 1 so \ it \ becomes \ x^2 =1$

Now take square root on both sides , x= +1 and x=-1

So zeros of the function are -4, -1, 1, 4

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3 years ago

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I need someone to explain what I did wrong

egoroff_w [7]

Answer:

g(f(4)) = 9

Step-by-step explanation:

Fill in gf(x) for where x is in g(x): (x-7)²

Then, fill in 4 for where x is: (4-7)²

Square it: (-3)²

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3 years ago