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kiruha [24]
3 years ago
8

Find two integers whose sum is -10 and product is 16 TWO INTERGES PLEASE

Mathematics
1 answer:
Lady bird [3.3K]3 years ago
8 0

Answer:

-2 and -8

Step-by-step explanation:

-2 + (-8) = -10

-2 * -8 = 16

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Which function is graphed?
Alexandra [31]

Hey there! :)

Answer:

C. {x² + 4, x < 2

    {-x + 4   x ≥ 2

Step-by-step explanation:

This is a piecewise function, where the two equations are different. They are:

y = x²+ 4

y = -x + 4

The function x² + 4 is graphed where x < 2. (< is used because the circle is open)

The function -x + 4 is graphed where x ≥ 2. (≥ is used since the endpoint is closed)

Therefore, the correct answer is:

C. x² + 4, x < 2

   -x + 4   x ≥ 2

5 0
3 years ago
Please help me on this TIA!!
tamaranim1 [39]
50.24 hope it helps:)
8 0
3 years ago
PLEASE HELP!!!! I will mark brainliest
strojnjashka [21]

Answer:

I think B.

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
(1 point) consider the function f(t)=⎧⎩⎨⎪⎪⎪⎪0,−5,−6,6,t&lt;00≤t&lt;11≤t&lt;7t≥7;f(t)={0,t&lt;0−5,0≤t&lt;1−6,1≤t&lt;76,t≥7; 1. wr
sergij07 [2.7K]
f(t)=\begin{cases}0&\text{for }t

Recall that

u(t)=\begin{cases}0&\text{for }t

Take it one piece at a time. For t\ge0, we can scale u(t) by -5:

-5u(t)=\begin{cases}0&\text{for }t

If we shift the argument by 1 and scale by -5, we have

-5u(t-1)=\begin{cases}0&\text{for }t

so if we subtract this from -5u(t), we'll end up with

-5u(t)+5u(t-1)=\begin{cases}0&\text{for }t

For the next piece, we can add another scaled and shifted step like

-6u(t-1)+6u(t-7)=\begin{cases}0&\text{for }t

so that

-5u(t)+5u(t-1)-6u(t-1)+6u(t-7)=\begin{cases}0&\text{for }t

For the last piece, we add one more term:

6u(t-7)=\begin{cases}0&\text{for }t

and so putting everything together, we get f(t):

f(t)\equiv-5u(t)+5u(t-1)-6u(t-1)+6u(t-7)+6u(t-7)
f(t)\equiv-5u(t)-u(t-1)+12u(t-7)
5 0
3 years ago
A pencil box contains five
KonstantinChe [14]

Answer:

red pencil=5/24

yellow pencil=6/24

Blue pencil=8/24

Orange pencil=3/24

purple pencil=2/24

5 0
3 years ago
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