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3 years ago

11

Choose the correct graph of the following condition. {(x, y) : x = 4) o [(x,y): y = 4)

1 answer:

Studentka2010 [4]3 years ago

7 0

Answer:

There is no solution

Step-by-step explanation:

You might be interested in

The price ( p) of a school lunch is 8:50 The restaurant increases the price by 12% which statement is true for this situation

professor190 [17]

Answer:

B

Step-by-step explanation:

1 is 100% so increasing it by 12% means adding 0.12 to it

4 0

3 years ago

Help!! 50 points and brainliest!

Viktor [21]

Answer:

Second choice:

$x=2t$

$y=4t^2+4t-3$

Fifth choice:

$x=t+1$

$y=t^2+4t$

Step-by-step explanation:

Let's look at choice 1.

$x=t+1$

$y=t^2+2t$

I'm going to subtract 1 on both sides for the first equation giving me $x-1=t$ . I will replace the $t$ in the second equation with this substitution from equation 1.

$y=(x-1)^2+2(x-1)$

Expand using the distributive property and the identity $(u+v)^2=u^2+2uv+v^2$ :

$y=(x^2-2x+1)+(2x-2)$

$y=x^2+(-2x+2x)+(1-2)$

$y=x^2+0+-1$

$y=x^2$

So this not the desired result.

Let's look at choice 2.

$x=2t$

$y=4t^2+4t-3$

Solve the first equation for $t$ by dividing both sides by 2:

$t=\frac{x}{2}$ .

Let's plug this into equation 2:

$y=4(\frac{x}{2})^2+4(\frac{x}{2})-3$

$y=4(\frac{x^2}{4})+2x-3$

$y=x^2+2x-3$

This is the desired result.

Choice 3:

$x=t-3$

$y=t^2+2t$

Solve the first equation for $t$ by adding 3 on both sides:

$x+3=t$ .

Plug into second equation:

$y=(x+3)^2+2(x+3)$

Expanding using the distributive property and the earlier identity mentioned to expand the binomial square:

$y=(x^2+6x+9)+(2x+6)$

$y=(x^2)+(6x+2x)+(9+6)$

$y=x^2+8x+15$

Not the desired result.

Choice 4:

$x=t^2$

$y=2t-3$

I'm going to solve the bottom equation for $t$ since I don't want to deal with square roots.

Add 3 on both sides:

$y+3=2t$

Divide both sides by 2:

$\frac{y+3}{2}=t$

Plug into equation 1:

$x=(\frac{y+3}{2})^2$

This is not the desired result because the $y$ variable will be squared now instead of the $x$ variable.

Choice 5:

$x=t+1$

$y=t^2+4t$

Solve the first equation for $t$ by subtracting 1 on both sides:

$x-1=t$ .

Plug into equation 2:

$y=(x-1)^2+4(x-1)$

Distribute and use the binomial square identity used earlier:

$y=(x^2-2x+1)+(4x-4)$

$y=(x^2)+(-2x+4x)+(1-4)$

$y=x^2+2x+-3$

$y=x^2+2x-3$ .

This is the desired result.

3 0

4 years ago

Read 2 more answers

317 multiplied by d<br> 317*d

Goshia [24]

The answer would be 317d or you could
do 317(d)

3 0

3 years ago

Help me answer these questions please 15, 19 and 20 please

SpyIntel [72]

Answer:

15) K'(t) = 5[5^(t)•In 5] - 2[3^(t)•In 3]

19) P'(w) = 2e^(w) - (1/5)[2^(w)•In 2]

20) Q'(w) = -6w^(-3) - (2/5)w^(-7/5) - ¼w^(-¾)

Step-by-step explanation:

We are to find the derivative of the questions pointed out.

15) K(t) = 5(5^(t)) - 2(3^(t))

Using implicit differentiation, we have;

K'(t) = 5[5^(t)•In 5] - 2[3^(t)•In 3]

19) P(w) = 2e^(w) - (2^(w))/5

P'(w) = 2e^(w) - (1/5)[2^(w)•In 2]

20) Q(W) = 3w^(-2) + w^(-2/5) - w^(¼)

Q'(w) = -6w^(-2 - 1) + (-2/5)w^(-2/5 - 1) - ¼w^(¼ - 1)

Q'(w) = -6w^(-3) - (2/5)w^(-7/5) - ¼w^(-¾)

7 0

3 years ago

Will someone please help asap!<br> Thank you

kolezko [41]

Answer:

35

Step-by-step explanation:

7 0

3 years ago