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Anarel [89]
3 years ago
15

Solve and graph -4 + 2t - 14 - 18t > -6 - 100t

Mathematics
1 answer:
irina [24]3 years ago
7 0
Okay so first u just add the terms on each side 
2t-18>-6-100t  
2(t-9)>2(-3-50t)
t-9>-3-50t
51t>6
t>6/51
t>2/17
so just take a number line put an empty circle like this : O on 2/17 and draw a line showing it continues to the right infinitely. Sorry if my math is wrong lol i am functioning on one hour of sleep. 
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Find the x- and y- intercepts of the line. 2X+3y=-18
Anna71 [15]
To find the x-intercept, plug zero into y and solve for x.

2x+3(0)=-18

2x=-18
/2     /2

x=-9
To solve for the y-intercept, do the same thing, but instead plug zero into the x-intercept to solve for y.

2(0)+3y=-18
3y=-18
/3     /3

y=-6

So, your answer to your question is the y-intercept=-6, and the x-intercept=-9
Hope this helps!:)
6 0
3 years ago
What is the mean and mode of the data set shown below? {2, 4, 5, 6, 8, 2, 5, 6}
likoan [24]
The average is calculated by adding all numbers and dividing by the count:

(2+4+5+6+8+2+5+6)/8 = 38/8 = 4.75

The mode is the number that occurs most often, however, this is the case for 2, 5 and 6. A data set can have multiple modes. 2, 5 and 6 are modes.
4 0
3 years ago
Simplfy 4y-6(4z-5y)-4z
victus00 [196]

The correct answer: −18y−40




6 0
3 years ago
(sinA÷secA)÷(cosA÷sinA) =tan A​
Sloan [31]

(Sin A/ secA) × ( sec A / cos A) = Sin A / cos A = tan A

Question should be cos A ÷ sec A

4 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
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