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allsm [11]
3 years ago
11

Evaluate |6| - |-6|. -36 0 6 12

Mathematics
2 answers:
Daniel [21]3 years ago
7 0

I believe that it is 0. Sorry if I´m wrong.

lora16 [44]3 years ago
6 0
0 bcz modulas function every negative no. in positive
therefore it becomes 6 - 6 I.e equal to 0
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Which type of graph is typically not used for quantitative data?
Verdich [7]

Answer:

A histogram often looks similar to a bar graph, but they are different because of the level of measurement of the data. Bar graphs measure the frequency of categorical data. A categorical variable is one that has two or more categories, such as gender or hair color

Step-by-step explanation:

4 0
3 years ago
Helpppp me ASAP!!!!!!!!!
abruzzese [7]

Answer:

11/1 or 11

Step-by-step explanation:

The scale of the y-axis is 11 and the scale of the x-axis is 1.  As said before the Unit Rate is y/x

If you pick one point, let's say (2,22), that will give you 22/2. If we simplify that fraction you get 11/1 or 11.

5 0
3 years ago
As part of a screening process, computer chips must be operated in an oven at 145 °C. Ten minutes after starting, the temperatur
Novosadov [1.4K]

Answer:

Step-by-step explanation:

I solved this using initial conditions and calculus, so I hope that's what you are doing in math.  It's actually NOT calculus, just a concept that is taught in calculus.

The initial condition formula we need is

y=Ce^{kt}

Filling in our formula with the 2 conditions we are given:

65=Ce^{10k}   and   85=Ce^{15k}

With those 2 equations, we have 2 unknowns, the C (initial value) and the k (the constant). We know that the initial value (or starting temp) for both conditions is the same, so we solve for C in one equation, sub it into the other equation and solve for k.  If

65=Ce^{10k} then

\frac{65}{e^{10k}}=C which, by exponential rules is the same as

C=65e^{-10k}

Since that value of C is the same as the value of C in the other equation, we sub it in:

85=65e^{-10k}(e^{15k})

Divide both sides by 65 and use the rules of exponents again to get

\frac{85}{65}=e^{-10k+15k} which simplifies down to

\frac{85}{65}=e^{5k}

Take the natural log of both sides to get

ln(\frac{85}{65})=5k

Do the log thing on your calculator to get

.2682639866 = 5k and divide both sides by 5 to find k:

k = .0536527973

Now that we have k, we sub THAT value in to one of the original equations to find C:

65=Ce^{10(.0536527973)}

which simplifies down to

65=Ce^{.536527973}

Raise e to that power on your calculator to get

65 = C(1.710059171) and divide to solve for C:

C = 38.01038064

Now sub in k and C to the final problem when t = 23:

y=38.01038064e^{(.0536527973)(23)} which simplifies a bit to

y=38.01038064e^{1.234014338}

Raise e to that power on your calculator to get

y = 38.01038064(3.434991111) and

finally, the temp at 23 minutes is

130.565

6 0
3 years ago
Using the formula y=ab/2c, express: the variable c in terms of a, b, and y.
adelina 88 [10]

Answer:

aby/2 = c

Step-by-step explanation:

1) ab * y = ab/2c

2) aby/2 = 2c/2

Final answer: aby/2 = c

4 0
3 years ago
I wanted to knwo what two to the third power was please help
nikitadnepr [17]
Eight is the answer.
8 0
3 years ago
Read 2 more answers
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