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3 years ago

8

Is 4x positive or negative?

1 answer:

Nikolay [14]3 years ago

7 0

Answer:

if x is positive, the answer is positive. If x is negative, the answer is negative.

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M3+ 3 m3 + 4 m²<br><br>simplify

Kipish [7]

Answer:

${m}^{3} + {27}^{m} + 4 {m}^{2}$

Step-by-step explanation:

${m}^{3} + {3}^{m 3} + 4 {m}^{2}$

${m}^{3} + {3}^{m \times 3} + 4 {m}^{2}$

${m}^{3} + {3}^{3m} + 4 {m}^{2}$

${3}^{3m} = ({3}^{3} {)}^{m} = {27}^{m}$

➡️ ${m}^{3} + {27}^{m} + 4 {m}^{2}$

5 0

2 years ago

MATH HELP ASAP!!! MAKING BRAINLIST

Ulleksa [173]

Answer:

22.86

Step-by-step explanation:

2.54 x 9 will give you the answer.

7 0

3 years ago

Read 2 more answers

A game field is 360 feet long. The practice field is 2/5 shorter than the practice field. How long is the game field?feet

vlabodo [156]

Answer:

The practice field is 144 feet long.

Step-by-step explanation:

360 x 2/5 = 144

(You would multiply these numbers NOT divide!)

To check my math:

2/5 = 0.4

144/360 = 0.4

By doing the above check, I know the two equations are the same. By seeing this I can know that the answer is correct.

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

3 years ago

Read 2 more answers

Solve y = x2-5 for x.<br> O A. x = +/-5<br> B. X = y + 5<br> C. X = +/y+5<br> O D. x = y-5

OLEGan [10]

I got C b I’m not 100% sure

7 0

3 years ago