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JulijaS [17]
3 years ago
15

Which of the following values are solutions to the inequality 9 > - 3 - 4x

Mathematics
2 answers:
IceJOKER [234]3 years ago
8 0

Answer:

x>−3

Step-by-step explanation:

zvonat [6]3 years ago
3 0

Answer:

the answer is 12

Step-by-step explanation:

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padilas [110]

Answer:

its choice 1 can i get brainly

Step-by-step explanation:

8 0
3 years ago
Please help I really need it
Musya8 [376]

Answer:

X = -2, Y = 13

X = 0, Y = 9

X = 2, Y = 5

X = 4, Y = 1

Step-by-step explanation:

Take the value given for X and plug it into the equation. For example, when X = -4, you will make your equation look as follows:

y = -2(-4) + 9

Simplify it:

y = 8 + 9

y = 17

Repeat this process with the rest of the X values to get your Y values.

6 0
3 years ago
I need help with internet’s I am not doing good in math and I really need help
sweet [91]

Answer:

Then get some help

5 0
4 years ago
Because of a problem in the program, the timer in a video player did not begin counting until the video had been playing for sev
kotykmax [81]

Answer:

105 frames

Step-by-step explanation:

Given that 25 frames are played per second

25 frames= 1 sec

The video already played 3 and 2/5 seconds before the player started to count 0

Write 3 and 2/5 seconds as an improper fraction

=(5*3)+2 / 5 = 17/5 seconds

Multiply by 25 frames

17/5 *25 =85 frames

So according to the video counter,after 17/5 seconds,it should count 85 frames.However,at 0 seconds,it indicated a count of 190 frames.Thus,to get the number of frames that were already in count you subtract 85 frames from the 190 frames.

190-85=105 frames.

7 0
4 years ago
Read 2 more answers
Using sum or difference formulas, find the exact value of sin(165∘)
Nikitich [7]
\bf \textit{Sum and Difference Identities}
\\ \quad \\
sin({{ \alpha}} + {{ \beta}})=sin({{ \alpha}})cos({{ \beta}}) + cos({{ \alpha}})sin({{ \beta}})
\\ \quad \\
sin({{ \alpha}} - {{ \beta}})=sin({{ \alpha}})cos({{ \beta}})- cos({{ \alpha}})sin({{ \beta}})\\\\
-------------------------------\\\\

\bf sin(165^o)\implies sin(120^o+45^o)
\\\\\\
sin(120^o)cos(45^o)~+~cos(120^o)sin(45^o)\implies \cfrac{\sqrt{3}}{2}\cdot \cfrac{\sqrt{2}}{2}~+~\cfrac{-1}{2}\cdot \cfrac{\sqrt{2}}{2}
\\\\\\
\cfrac{\sqrt{6}}{4}~-~\cfrac{\sqrt{2}}{4}\implies \cfrac{\sqrt{6}-\sqrt{2}}{4}
6 0
3 years ago
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