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11111nata11111 [884]

2 years ago

5

I need help 7th grade math

2 answers:

Digiron [165]2 years ago

6 0

The answer is D :) hope that help so

victus00 [196]2 years ago

4 0

Answer:

D

Step-by-step explanation:

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Im sorry it’s blurry! can someone please help me out it would mean a lot thank you!

alekssr [168]

Answer: B. -3, -2, 1, 2, 5

Step-by-step explanation:

Please give brainliest

3 0

2 years ago

Help me please!!! I need this answered fast!!!

denpristay [2]

Answer:

I think that the answer is A.

5 0

2 years ago

If angle 1 = angle 2 and angle 2 = angle 4, what property allows you to state that angle 1 = angle 4

Gnesinka [82]

Answer:

Transitive Property

Step-by-step explanation:

You can use it for example: a = b. b = c. So a = c (Transitive Property).

6 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

Question 7 of 10

Ganezh [65]

Answer:

Option B

Step-by-step explanation:

Given Equation :

$\qquad \sf \dashrightarrow \: 3(4x+3) = 2x - 5(3 - x) + 2$

Using distribute property:

$\qquad \sf \dashrightarrow \: 12x + 9 = 2x - 15 + 5x + 2$

Adding the like terms we get :

$\qquad \sf \dashrightarrow \: 12x + 9 = 2x + 5x - 15 + 2$

$\qquad \sf \dashrightarrow \: 12x + 9 = 7x - 13$

Transposing the variables on the right side and constant terms on the left side :

$\qquad \sf \dashrightarrow \: 12x - 7x = - 13 - 9$

$\qquad \sf \dashrightarrow \: 5x = - 22$

Dividing both sides by 5 :

$\qquad \sf \dashrightarrow \: \dfrac{5x}{5} = \dfrac{ - 22}{5}$

$\qquad \bf \dashrightarrow \: x = \dfrac{ - 22}{5}$

3 0

2 years ago