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3 years ago

Which equation is represented by the table of values below?

Mathematics

1 answer:

Kaylis [27]3 years ago

8 0

Answer:

Step-by-step explanation:

It is the only one that has a y-intercept on the table of values.

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Suppose that a person with a push mower can mow a large lawn in 5 hours, whereas the lawn can be mowed with a riding lawn mower

Reil [10]

------------------------------------------------------------------
Push Mower
------------------------------------------------------------------
It takes 5 hours to complete the job
1 hour = 1/5 of the job

------------------------------------------------------------------
Riding Lawn Mower
------------------------------------------------------------------
It takes 3 hours to complete the job
1 hour = 1/3 of the job

------------------------------------------------------------------
Push Mower + Riding Lawn Mower
------------------------------------------------------------------
1 hour = 1/5 + 1/3
1 hour = 3/15 + 5/15
1 hour = 8/15 of the job.

------------------------------------------------------------------
Calculate time needed to complete
------------------------------------------------------------------
8/15 of the job takes 1 hour
8/15 of the job = 1 hour

------------------------------------------------------------------
Divide by 8/15 on both side
------------------------------------------------------------------
8/15 ÷ 8/15 of the job = 1 ÷ 8/15 hour
Whole of the job = 1 x 15/8 hours
Whole of the job = 15/8 hours
Whole of the job = 1 7/8 hours

------------------------------------------------------------------
Answer: 1 7/8 hours
------------------------------------------------------------------

6 0

3 years ago

I need help on this and the first person who answer correctly gets a BRANILIST

True [87]

Answer:

first you need to set up the equation which is 20+2x=44 then isolate 2x by doing the inverse operation for 20 which is -20 and add them to both sides which would now give you 2x=24 so divde 24 by 2 which is 12, x=12 answer 12ml

Step-by-step explanation:

7 0

3 years ago

Can you solve: T-6=-4

Andreyy89

T-6=-4
Just simply add 6 to both sides, it will cancel out the -6 and leave T alone, and it will become T=2, There's your answer.

8 0

3 years ago

A graphing calculator is recommended. A crystal growth furnace is used in research to determine how best to manufacture crystals

Sophie [7]

Answer:

a) $w_1 = \frac{-2.156 -\sqrt{2.156^2 -4(0.1)(-183)}}{2*0.1}=-54.896$

$w_2 = \frac{-2.156 +\sqrt{2.156^2 -4(0.1)(-183)}}{2*0.1}=33.336$

And since the power can't be negative then the solution would be w = 33.34 watts.

b) $202= 0.1w^2 +2.156 w +20$

$0.1w^2 +2.156 w-182=$

$w_2 = \frac{-2.156 +\sqrt{2.156^2 -4(0.1)(-182)}}{2*0.1}=33.222$

$204= 0.1w^2 +2.156 w +20$

$0.1w^2 +2.156 w-184=$

$w_2 = \frac{-2.156 +\sqrt{2.156^2 -4(0.1)(-184)}}{2*0.1}=33.449$

So then the range of voltage would be between 33.22 W and 33.45 W.

c)For this case $\epsilon = \pm 1$ since that's the tolerance 1C

d) $\delta_1 =|33.222-33.336|=0.116$

$\delta_2 =|33.449-33.336|=0.113$

So then we select the smalles value on this case $\delta =0.113$

e) For this case if we assume a tolerance of $\epsilon=\pm 1C$ for the temperature and a tolerance for the power input $\delta =0.113$ we see that:

$lim_{x \to a} f(x) =L$

Where a = 33.34 W, f(x) represent the temperature, x represent the input power and L = 203C

Step-by-step explanation:

For this case we have the following function

$T(w)= 0.1 w^2 +2.156 w +20$

Where T represent the temperature in Celsius and w the power input in watts.

Part a

For this case we need to find the value of w that makes the temperature 203C, so we can set the following equation:

$203= 0.1w^2 +2.156 w +20$

And we can rewrite the expression like this:

$0.1w^2 +2.156 w-183=$

And we can solve this using the quadratic formula given by:

$w =\frac{-b \pm \sqrt{b^2 -4ac}}{2a}$

Where a =0.1, b =2.156 and c=-183. If we replace we got:

$w_1 = \frac{-2.156 -\sqrt{2.156^2 -4(0.1)(-183)}}{2*0.1}=-54.896$

$w_2 = \frac{-2.156 +\sqrt{2.156^2 -4(0.1)(-183)}}{2*0.1}=33.336$

And since the power can't be negative then the solution would be w = 33.34 watts.

Part b

For this case we can find the values of w for the temperatures 203-1= 202C and 203+1 = 204 C. And we got this:

$202= 0.1w^2 +2.156 w +20$

$0.1w^2 +2.156 w-182=$

$w_2 = \frac{-2.156 +\sqrt{2.156^2 -4(0.1)(-182)}}{2*0.1}=33.222$

$204= 0.1w^2 +2.156 w +20$

$0.1w^2 +2.156 w-184=$

$w_2 = \frac{-2.156 +\sqrt{2.156^2 -4(0.1)(-184)}}{2*0.1}=33.449$

So then the range of voltage would be between 33.22 W and 33.45 W.

Part c

For this case $\epsilon = \pm 1$ since that's the tolerance 1C

Part d

For this case we can do this:

$\delta_1 =|33.222-33.336|=0.116$

$\delta_2 =|33.449-33.336|=0.113$

So then we select the smallest value on this case $\delta =0.113$

Part e

For this case if we assume a tolerance of $\epsilon=\pm 1C$ for the temperature and a tolerance for the power input $\delta =0.113$ we see that:

$lim_{x \to a} f(x) =L$

Where a = 33.34 W, f(x) represent the temperature, x represent the input power and L = 203C

8 0

4 years ago

HELP PLSSS ASAP pls

lara31 [8.8K]

Answer:

first choice - it is dilated by 1/3

Step-by-step explanation:

write out your triangles points

A (-12,12) A' (4, -4)

B (-3, 12) . B' (1, -4)

C (-9,6) . C' (3, -2)

3 0

3 years ago