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Len [333]
3 years ago
13

How do you find y? ∣3y−1∣+∣2y+3∣>5

Mathematics
1 answer:
Alexandra [31]3 years ago
5 0
Let's solve your inequality step-by-step.

<span><span><span><span><span>3y</span>−1</span>+<span>2y</span></span>+3</span>>5

</span>Step 1: Simplify both sides of the inequality.<span><span><span>5y</span>+2</span>>5

</span>Step 2: Subtract 2 from both sides.<span><span><span><span>
5y</span>+2</span>−2</span>><span>5−2</span></span><span><span>
5y</span>>3</span>
Step 3: Divide both sides by 5.<span><span><span>
5y/</span>5</span>><span>3/5</span></span><span>
y><span>3/<span>5

hopefully this helps and here's a link that i used to use a lot in middle school 
https://www.symbolab.com/solver/solve-for-equation-calculator
</span></span></span>
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Several logs are stored in a pile with 22 logs on the bottom layer, 21 on the second layer, 20 on the
Oduvanchick [21]

Answer:

[22 x 21] / 2=231 or "c"

Step-by-step explanation:

4 0
2 years ago
Can anyone tell me why by direct substitution of x, the equation (circled ones) equals to the indeterminate form, 0/0? When you
Katena32 [7]

Answer:

See explanation and hopefully it answers your question.

Basically because the expression has a hole at x=3.

Step-by-step explanation:

Let h(x)=( x^2-k ) / ( hx-15 )

This function, h, has a hole in the curve at hx-15=0 if it also makes the numerator 0 for the same x value.

Solving for x in that equation:

Adding 15 on both sides:

hx=15

Dividing both sides by h:

x=15/h

For it be a hole, you also must have the numerator is zero at x=15/h.

x^2-k=0 at x=15/h gives:

(15/h)^2-k=0

225/h^2-k=0

k=225/h^2

So if we wanted to evaluate the following limit:

Lim x->15/h ( x^2-k ) / ( hx-15 )

Or

Lim x->15/h ( x^2-(225/h^2) ) / ( hx-15 ) you couldn't use direct substitution because of the hole at x=15/h.

We were ask to evaluate

Lim x->3 ( x^2-k ) / ( hx-15 )

Comparing the two limits h=5 and k=225/h^2=225/25=9.

3 0
2 years ago
Imagine a pond. In it sits one lilypad, which reproduces once a day. Each of its offspring also reproduces once a day, doubling
sergeinik [125]

Answer: On the 29th day

Step-by-step explanation:

According to this problem, no lilypad dies and the lilypads always reproduce, so we can apply the following reasoning.

On the first day there is only 1 lilypad in the pond. On the second day, the lilypad from the first reproduces, so there are 2 lilypads. On day 3, the 2 lilypads from the second day reproduce, so there are 2×2=4 lilypads. Similarly, on day 4 there are 8 lilypads. Following this pattern, on day 30 there are 2×N lilypads, where N is the number of lilypads on day 29.

The pond is full on the 30th day, when there are 2×N lilypads, so it is half-full when it has N lilypads, that is, on the 29th day. Actually, there are 2^{30} lilypads on the 30th, and 2^{29} lilypads on the 29th. This can be deduced multiplying succesively by 2.  

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