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3 years ago

Can you help with #21 please. I will mark Brainliest.

Mathematics

2 answers:

Leokris [45]3 years ago

7 0

Answer:

A, C, E

Step-by-step explanation:

By analyzing the graph, we can see that the roots are $-2$ and $-4$ .

We now want to find all equations of the form $(x-(-2))(x-(-4))=(x+2)(x+4)=0$ .

This immediately gives C as 1 answer.

We can expand our equation to find other possible solutions!

By the FOIL method: $(x+2)(x+4)=(x)(x)+(2)(x)+(4)(x)+(2)(4)=x^2+6x+8=0$

We can then divide both sides of the equation to find that $-x^2-6x-8=0$ , so A is another answer.

We can complete the square on $x^2+6x+8$ to find the last possible solution.

$x^2+6x+8=x^2+6x+9-9+8=x^2+6x+9-1=(x+3)^2-1=0$ .

We can then multiply both sides of the equation by $4$ to find that $4(x+3)^2-4=0$ , so E is the last answer.

The proof that the other answers don't work is left as an exercise.

So, the answers are A, C, and E, and we're done!

tamaranim1 [39]3 years ago

4 0

Answer:

its going to be letter b because 2 and 4 well the x intercepts those

Step-by-step explanation:

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Step-by-step explanation:

Use permutation to answer this question taking notice of the order.

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3 years ago

PLEASE HELP!!

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So: (j - 1/2j) - (a - 1/3a).

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4 years ago

Point C,D and E are collinear on CE, and CD:DE = 3/5. C is located at (1,8), D is located (4,5), and E is located at (x,y)

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3 years ago

Chloe has been offered a new job in the women’s shoe department at Macy’s. She has two salary options. She can receive a salary

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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

Can you help with #21 please. I will mark Brainliest.​

Can you help with #21 please. I will mark Brainliest.