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3 years ago

I need help on all of these

Mathematics

1 answer:

lord [1]3 years ago

5 0

well #7 is a solution.

y= 9x (4,36)

9 times 4 = 36

36/9=4

so the ordered pair is a solution.

y=3x-5 (7,16)

3 times 7 -5= 16

so once again the ordered pair is a solution.

Hoped this helped.

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Please help me and please write the answer out

Vitek1552 [10]

15 degrees is the answer

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

4 years ago

Read 2 more answers

Question 9 (3 points)

Lilit [14]

Answer:

Precambrian period

Step-by-step explanation:

Since the Great oxidation period = 2.4 Billion years ago era = Precambrian period.

Micro-fossils attributed to cyanobacteria are abundant during the Proterozoic that came later 2,5 billion years ago and ended 540 million years ago,

7 0

3 years ago

Factorise x^2/4-y^2/4

Nuetrik [128]

Answer:

$\large\boxed{\left(\dfrac{x}{2}-\dfrac{y}{2}\right)\left(\dfrac{x}{2}+\dfrac{y}{2}\right)=\dfrac{x-y}{2}\cdot\dfrac{x+y}{2}}$

Step-by-step explanation:

$\dfrac{x^2}{4}-\dfrac{y^2}{4}=\dfrac{x^2}{2^2}-\dfrac{y^2}{2^2}\qquad\text{use}\ \left(\dfrac{a}{b}\right)^n=\dfrac{a^n}{b^n}\\\\=\left(\dfrac{x}{2}\right)^2-\left(\dfrac{y}{2}\right)^2\qquad\text{use}\ (a-b)(a+b)=a^2-b^2\\\\=\left(\dfrac{x}{2}-\dfrac{y}{2}\right)\left(\dfrac{x}{2}+\dfrac{y}{2}\right)=\dfrac{x-y}{2}\cdot\dfrac{x+y}{2}$