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saw5 [17]
3 years ago
10

What is the solution and the inequality ?

Mathematics
1 answer:
Nikolay [14]3 years ago
4 0
Inequality : 19.50n+24≤$110

Solution: Alyssa can buy 4 DVDs

Work: 110-24=86
           86/19.50 = 4.4

You can't have a fraction of a DVD, so you have to round down to 4. 
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PLEEEEEEASE HELP MEEEEE!! ill give you 20 points
Nana76 [90]
331.3%
Convert 53/16 to simplest form, 3 5/16, convert 5/16 into numbers —> 0.3125
Take the whole number and add it into the ones place: 3.3125
Move the decimal into two places then round to the nearest tenth: 331.3
8 0
3 years ago
Find the Antilog of 547.840​
bija089 [108]

Answer:

It's impossible because the figure is greater than 10

Step-by-step explanation:

{ \boxed{ \bf{antilog \: of \: x =  \frac{x}{ log} =  {10}^{x}  }}}

Therefore:

{ \sf{anti(547.840) =  {10}^{547.840} }} \\ { \tt{ \red{math \: error \: !}}}

3 0
3 years ago
Se tiene en una caja de canicas 10, 4 blancas y 6 negras
fgiga [73]

Answer:

Step-by-step explanation:

look at it

8 0
3 years ago
Is floor(x) periodic?
diamong [38]

If you look at the graph of y = floor(x), you'll see a stairstep pattern that climbs up as you read from left to right. There are no vertical components to the graph. There are only horizontal components.

The graph is not periodic because the y values do not repeat themselves after a certain x value is passed. For instance, start at x = 0 and go to x = 3. You'll see the y values dont repeat themselves as if a sine function would. If you wanted the function to be periodic, the steps would have to go downhill at some point; however, this does not happen.

Conclusion: The function floor(x) is <u>not</u> periodic.

4 0
3 years ago
Read 2 more answers
Help me pleaseee.....
Lana71 [14]

Answer:

titutex=cos\alp,\alp∈[0:;π]

\displaystyle Then\; |x+\sqrt{1-x^2}|=\sqrt{2}(2x^2-1)\Leftright |cos\alp +sin\alp |=\sqrt{2}(2cos^2\alp -1)Then∣x+

1−x

2

∣=

2

(2x

2

−1)\Leftright∣cos\alp+sin\alp∣=

2

(2cos

2

\alp−1)

\displaystyle |\N {\sqrt{2}}cos(\alp-\frac{\pi}{4})|=\N {\sqrt{2}}cos(2\alp )\Right \alp\in[0\: ;\: \frac{\pi}{4}]\cup [\frac{3\pi}{4}\: ;\: \pi]∣N

2

cos(\alp−

4

π

)∣=N

2

cos(2\alp)\Right\alp∈[0;

4

π

]∪[

4

3π

;π]

1) \displaystyle \alp \in [0\: ;\: \frac{\pi}{4}]\alp∈[0;

4

π

]

\displaystyle cos(\alp -\frac{\pi}{4})=cos(2\alp )\dotscos(\alp−

4

π

)=cos(2\alp)…

2. \displaystyle \alp\in [\frac{3\pi}{4}\: ;\: \pi]\alp∈[

4

3π

;π]

\displaystyle -cos(\alp -\frac{\pi}{4})=cos(2\alp )\dots−cos(\alp−

4

π

)=cos(2\alp)…

1

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Display

6 0
3 years ago
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