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2 years ago

Look at the diagram. Which of the following is another name ∠2?

Mathematics

1 answer:

____ [38]2 years ago

8 0

The other name for angle 2 is ∠DBC

The correct option is (D)

<h3>What is Angle?</h3>

A figure which is formed by two rays or lines that shares a common endpoint is called an angle.

We know that while naming the angle we have to consider the point where the angle is made should be at center.

Hence, the angle 2 can also be written as ∠DCB.

Learn more about angle here:

brainly.com/question/13954458

#SPJ1

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What is the answer to (6+4)tothe3power-1tothe8power

Ipatiy [6.2K]

Hey there!

$\huge\boxed{(6 + 4)^3 - 1^8}$

$\huge\boxed{= 10^3 - 1^8}$

$\huge\boxed{10^3}$

$\huge\boxed{= 10\times10\times10}$

$\huge\boxed{= 100\times10 }$

$\huge\boxed{= 1,000}$

$\huge\boxed{1^{8}}$

$\huge\boxed{= 1\times1\times1 \times 1 \times 1 \times 1 \times 1 \times 1}$

$\huge\boxed{= 1 \times 1 \times 1 \times 1}$

$\huge\boxed{= 1\times1}$

$\huge\boxed{= 1}$

$\huge\boxed{1,000 - 1}$

$\huge\boxed{= 999}$

$\huge\text{Answer: \textsf{999}}$

$\huge\text{Good luck on your assignment \& enjoy your day!}$

~ $\huge\boxed{\frak{Amphitrite1040:)}}$

8 0

2 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.

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3 years ago

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mezya [45]

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3 years ago

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Vera_Pavlovna [14]

Answer:

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3 years ago

A credit card statement showed these transactions during October.

AURORKA [14]

Kshdnxnnxnxnxnxnnxnxnxnxnxnxnxn =5

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3 years ago