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zvonat [6]
3 years ago
10

12 POINT HURRY HURYRJXNXNDJDJDNDN

Mathematics
1 answer:
padilas [110]3 years ago
8 0

Answer:

The fourth choice, Pink, is the correct answer.

Step-by-step explanation:

The first two options intersect, therefore the intersection is the solution. The third set of lines may seem like one line, however since we are dealing with two linear equations, it can be presumed that those are two lines on top of one another, therefore it has infinite solutions. The last option remaining is Pink, in which the lines are parallel and therefore will never intersect, making them have no solution.

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A cylinder has a radius of 4 centimeters and a height of 12 centimeters. A smaller cylinder has linear dimensions that are one-f
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A2=A1/16 is the answer for this one
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3 years ago
HELP me on this please
Rzqust [24]

Answer:

angle FEA is 100°

angle AEC is 20°

angle CEG is 60°

Step-by-step explanation:

angle AEC=angle DEB=20°

angle AEG=angle FED=80°

80°-20°=60°; angle CEG=60°

angle FEA+angle AEC+ angle CEG=180°

angle FEA=100°

6 0
3 years ago
18 take away negative 10 is wha
Ilya [14]
18--10= 
the two negative signs make a plus so you simply have 18 + 10 which is 28
3 0
3 years ago
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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
I NEED HELP WITH THIS PROBLEM
coldgirl [10]

Answer:

3 1/2, 3

Step-by-step explanation:

Good luck!

3 0
3 years ago
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