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2 years ago

1.(01.01) Evaluate -6 - (-1). (1 point) 5 -5 6

Mathematics

2 answers:

Deffense [45]2 years ago

8 0

The answer is -5 have a great day

Elanso [62]2 years ago

3 0

Answer:

I don't understand the question. Write it again & briefly.

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<img src="https://tex.z-dn.net/?f=%20%5Chuge%20%20%5Cred%7B%5Cboxed%7B%20%5Ccolorbox%7Bblack%7D%7B%20%5Ccolor%7Bgold%7D%7Bquesti

timofeeve [1]

Answer:

3√2

Step-by-step explanation:

First, combine all the numbers in the square root, and then square root the number provided:

√(3 + 3 + 3 + 3 + 3 + 3) = √18

When square rooting, first, find prime factors of the number, 18. If there are pairs, they will be combined into a whole number:

√18 = √(3 * 3 * 2) = 3√2

3√2 is your answer.

3 0

2 years ago

Find the output, y, when the input, r, is – 7.<br><br> y=

Elodia [21]

Answer:

<h2>i dont know if im correct or not but i thincC its -1</h2>

Step-by-step explanation:

4 0

3 years ago

WILL G8VE BRAINLIEST IF YOU ARE CORRECT!!!

vaieri [72.5K]

The equivalent is 2.45 because 147/60 = 2 27/60 and 27/60 = .45

5 0

3 years ago

The image of a parallelogram is given. Use the translation (x,y) → (x+2, y-1) to find the coordinates of S.

MAVERICK [17]

So to do that, find the point of S then transform

in (x,y) form

we see that the point 'S' is 1 unit to the right and 2 units up therefor
the point is (1,2)
so we jsut apply the thingy translation to is
x=1
y=2
x+2=1+2=3
y-1=2-1=1
the new point is (3,1)

8 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