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3 years ago

How do I find the value of each variable?

Mathematics

1 answer:

cricket20 [7]3 years ago

6 0

I believe how you would find the variable z is by seeing that next to it is an exterior angle of 110 and (hopefully) you know that an exterior and exterior angle should add up to 180 so it would be z=70

I have not done geometry in a long time so sorry if I am wrong and I don't quite remember how to find y. I still hope this was helpful!

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Solve the eqaution -12+5k=15-4k

Veronika [31]

Simple....

you have:

-12+5k=15-4k

isolate the variable...

-12+5k=15-4k
+12 +12

5k=27-4k

5k=27-4k
+4k +4k

9k=27

k=3

Thus, your answer.

7 0

4 years ago

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The sum of two times a number and 5 is 11

Alika [10]

So it can be written as 2x + 5 = 11, and x = 3

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3 years ago

Solve for the roots in simplest form using the quadratic formula:<br>4x2 + 21<br>20x

Artyom0805 [142]

Answer:

(0,0) and (-530,0)

Step-by-step explanation:

If you have to use the quadratic formula, you need to know your A, B, and C values. Your A value is 4, your B is 2120, and your C value, or constant, appears to be zero. Then you need to plug these values into the quadratic formula to get 0 and -530. This would make your roots to be at points (0,0) and (-530,0).

6 0

3 years ago

Twenty men can cut thirty trees in four hours. If four men leave the job, how many

Mice21 [21]

Answer:

As in 4 hours , 20 men can cut 30 trees. So, in 1 hour , 20 men can cut 30 trees / 4. So, in 6 hours, 20 men can cut 30 * 6 trees / 4 = 45 trees.

Step-by-step explanation:

done

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