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omeli [17]
3 years ago
5

98 PTS HELP!!! Please show work! ExPlAiN PlEaSe.

Mathematics
2 answers:
cupoosta [38]3 years ago
5 0

Answer:

2x ≥ 4

x ≥ 2

-x+7<-6

-x < - 13

x > 13

Region towards right of x = 13, with x = 13 being dotted

y > -x + 3

Draw a dotted line passing through (0,3) and (3,0)

Shade the region above this line which is towards right of x = 13

ludmilkaskok [199]3 years ago
4 0

Answer:

see below

Step-by-step explanation:

2x ≥ 4

Divide by 2

2x/2 ≥ 4/2

x ≥ 2

-x+7<-6

Subtract 7 from each side

-x+7-7<-6-7

-x < -13

Divide each side by -1, remembering to flip the inequality

x > 13

y>-x+3

This is a line with slope -1 and y intercept 3

This is a dotted line since there is no equals sign and graph above

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Can someone plz help me find the answer to 2x=4 y-16. and 2x-y=-7
telo118 [61]

Answer:

So, y = 2x + 7;

Then, 2x = 4( 2x + 7 ) - 16;

2x = 8x + 28 - 16;

2x = 8x + 12;

- 6x = 12;

x = - 2;

Then, y = 2· (-2) + 7;

y = - 4 + 7;

y = 3.

Step-by-step explanation:


8 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
A rectangle has a side length of 3 feet and an area of 24 squared feet. A larger similar triangle has a corresponding length of
Zinaida [17]
3*w=24
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Answer:

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2 years ago
What are a domain and range?
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Answer:

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5 0
2 years ago
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