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emmasim [6.3K]
3 years ago
12

Rational number between -1 and 1​

Mathematics
1 answer:
Nonamiya [84]3 years ago
8 0

Answer:

\dfrac{-1}{2}

Step-by-step explanation:

We need to find ration numbers between -1 and 1.

Multiplying and dividing both -1 and 1 by 2.

\dfrac{-1\times 2}{2}=\dfrac{-2}{2}

and \dfrac{1\times 2}{2}=\dfrac{2}{2}

Now, the rational number between -2/2 and 2/2 is \dfrac{-1}{2}.

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Place the decimal point in the answer below to make it correct. Explain your reasoning
sladkih [1.3K]

Answer:

AC is greater than BC because segment AC is the hypotenuse of right triangle ABC, and the hypotenuse is the longest side of a right triangle.



5 0
3 years ago
Which of the following inequalities have the same solution set? Select all that apply (>= means greater than or equal to and
Alex_Xolod [135]
<span>7n + 4 - 5n <= 2(n+2) Let's simplify these expression first 2n + 4 <= 2(n + 2) 2(n +4) = 2n + 4 The LHS and RHS expressions are the same but the inequality says contrary Hence 2n + 4 is not less than but equal to 2(n + 2). 7+ 6a >= -4(2-a) - 2a -4(2-a) - 2a = -8 + 4a - 2a = -8 + 2a. It should be obvious that 7 +6a is greater than the RHS expression. Suppose a= 2 then we have 7 + 6(2) = 19 while -8 + 2(2) = - 8 + 4 = -4 So it follows that 7 + 6a >= -4(2-4) -2a 4x >= 8. The expression is always true so the equal to option is out.</span>
5 0
3 years ago
A contaminant is leaking into a lake at a rate of R(t) = 1700e^0.06t gal/h. Enzymes that neutralize the contaminant have been ad
olasank [31]

Answer:

16,460 gallons

Step-by-step explanation:

This is a differential equation problem, we have a constant flow of contaminant into the lake, but also we know that only a fraction of that quantity of contaminant remains because of the enzymes. For that reason, the differential equation of contaminant's flow into the lake would be:

\frac{dQ}{dt} =1700exp(0.06t)*exp(-0.32t)\\\frac{dQ}{dt} =1700exp(-0.26t)\\

Then, we have to integrate in order to find the equation for Q(t), as the quantity of contaminant in the lake, in function of time.

\int\limits^0_t {dQ}=\int\limits^0_t {1700exp(-0.26t)dt}\\Q(t)=\frac{1700}{-0.26} exp(-0.26t)+C \\

Now, we use the given conditions to replace them in the equation, in order to solve for C

t_{0} =0\\Q_{0}=10,000\\Q_{0}=-6538exp(-0.26*0)+C\\C=10,000+6538=16538

Then, we reorganize the equation and we replace t for 17 hours, in order to determine the quantity of contaminant at that time:

Q_{t} =-6538exp(-0.26t)+16538\\Q_{17} =-6538exp(-0.26*17)+16538\\Q_{17} =16460 gallons

3 0
3 years ago
2x-2=2(2x-11) solve for x
Mkey [24]

Instead of distributing 2, we can actually divide both sides by 2:

x - 1 = 2x - 11

Isolate x:

x = 10

7 0
3 years ago
Read 2 more answers
Which should I choose
maw [93]

Answer:

Now thats you've had time to think about it.

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