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3 years ago

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Please help I’m stuck.

1 answer:

Sholpan [36]3 years ago

7 0

What are you stuck on ??????

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Megan has 40$ in her wallet. She wants to buy a dress. The regular price of the dress is 45$. The dress is sold at a discount of

devlian [24]

Answer:

yes

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

What would my GPA be? Here are my grades.

marta [7]

That is approximately, in order, a C, a C, a B, a D, an A, a B, and an A.

In grade points that is a 2, a 2, a 3, a 1, a 4, a 3, and a 4.

The average of those numbers is about 2.7, so you have a 2.7.

You can raise that by bringing up the D as it is an outlier here.

4 0

3 years ago

Plz help ASAP!! Explain your answer! I will mark at brainliest!!! And don’t copy anybody else’s answer

Klio2033 [76]

Answer:

No, it is not a square

Step-by-step explanation:

If one wall is 19", that would mean the wall perpendicular to this wall is also 19" (in fact all of the walls would be 19"!) If this was a square, then the diagonal we draw at 20.62" would serve as the hypotenuse of a right triangle. One wall would serve as a leg, and another wall as another leg. If this is a square, then the Pythagorean's Theorem would be satisfied when we plug in the 2 wall measures for a and b, and the diagonal for c:

$19^2+19^2=20.62^2$

We need to see if this is a true statement. If the left side equals the right side, then the 2 legs of the right triangle are the same length, and the room, then is a square.

361 + 361 = 425.1844

Is this true? Does 722 = 425.1844? Definitely not. That means that the room is not a square.

8 0

3 years ago

Solve the following system

scZoUnD [109]

Answer:

{x = -4 , y = 2 , z = 1

Step-by-step explanation:

Solve the following system:

{-2 x + y + 2 z = 12 | (equation 1)

2 x - 4 y + z = -15 | (equation 2)

y + 4 z = 6 | (equation 3)

Add equation 1 to equation 2:

{-(2 x) + y + 2 z = 12 | (equation 1)

0 x - 3 y + 3 z = -3 | (equation 2)

0 x+y + 4 z = 6 | (equation 3)

Divide equation 2 by 3:

{-(2 x) + y + 2 z = 12 | (equation 1)

0 x - y + z = -1 | (equation 2)

0 x+y + 4 z = 6 | (equation 3)

Add equation 2 to equation 3:

{-(2 x) + y + 2 z = 12 | (equation 1)

0 x - y + z = -1 | (equation 2)

0 x+0 y+5 z = 5 | (equation 3)

Divide equation 3 by 5:

{-(2 x) + y + 2 z = 12 | (equation 1)

0 x - y + z = -1 | (equation 2)

0 x+0 y+z = 1 | (equation 3)

Subtract equation 3 from equation 2:

{-(2 x) + y + 2 z = 12 | (equation 1)

0 x - y+0 z = -2 | (equation 2)

0 x+0 y+z = 1 | (equation 3)

Multiply equation 2 by -1:

{-(2 x) + y + 2 z = 12 | (equation 1)

0 x+y+0 z = 2 | (equation 2)

0 x+0 y+z = 1 | (equation 3)

Subtract equation 2 from equation 1:

{-(2 x) + 0 y+2 z = 10 | (equation 1)

0 x+y+0 z = 2 | (equation 2)

0 x+0 y+z = 1 | (equation 3)

Subtract 2 × (equation 3) from equation 1:

{-(2 x)+0 y+0 z = 8 | (equation 1)

0 x+y+0 z = 2 | (equation 2)

0 x+0 y+z = 1 | (equation 3)

Divide equation 1 by -2:

{x+0 y+0 z = -4 | (equation 1)

0 x+y+0 z = 2 | (equation 2)

0 x+0 y+z = 1 | (equation 3)

Collect results:

Answer: {x = -4 , y = 2 , z = 1

4 0

3 years ago

Can somebody explain how these would be done? The selected answer is incorrect, and I was told "Nice try...express the product b

trapecia [35]

Answer:

Solution ( Second Attachment ) : - 2.017 + 0.656i

Solution ( First Attachment ) : 16.140 - 5.244i

Step-by-step explanation:

Second Attachment : The quotient of the two expressions would be the following,

$6\left[\cos \left(\frac{2\pi }{5}\right)+i\sin \left(\frac{2\pi \:}{5}\right)\right]$ ÷ $2\sqrt{2}\left[\cos \left(\frac{-\pi }{2}\right)+i\sin \left(\frac{-\pi \:}{2}\right)\right]$

So if we want to determine this expression in standard complex form, we can first convert it into trigonometric form, then apply trivial identities. Either that, or we can straight away apply the following identities and substitute,

( 1 ) cos(x) = sin(π / 2 - x)

( 2 ) sin(x) = cos(π / 2 - x)

If cos(x) = sin(π / 2 - x), then cos(2π / 5) = sin(π / 2 - 2π / 5) = sin(π / 10). Respectively sin(2π / 5) = cos(π / 2 - 2π / 5) = cos(π / 10). Let's simplify sin(π / 10) and cos(π / 10) with two more identities,

( 1 ) $\cos \left(\frac{x}{2}\right)=\sqrt{\frac{1+\cos \left(x\right)}{2}}$

( 2 ) $\sin \left(\frac{x}{2}\right)=\sqrt{\frac{1-\cos \left(x\right)}{2}}$

These two identities makes sin(π / 10) = $\frac{\sqrt{2}\sqrt{3-\sqrt{5}}}{4}$ , and cos(π / 10) = $\frac{\sqrt{2}\sqrt{5+\sqrt{5}}}{4}$ .

Therefore cos(2π / 5) = $\frac{\sqrt{2}\sqrt{3-\sqrt{5}}}{4}$ , and sin(2π / 5) = $\frac{\sqrt{2}\sqrt{5+\sqrt{5}}}{4}$ . Substitute,

$6\left[ \left\frac{\sqrt{2}\sqrt{3-\sqrt{5}}}{4}+i\left\frac{\sqrt{2}\sqrt{5+\sqrt{5}}}{4}\right]$ ÷ $2\sqrt{2}\left[\cos \left(\frac{-\pi }{2}\right)+i\sin \left(\frac{-\pi \:}{2}\right)\right]$

Remember that cos(- π / 2) = 0, and sin(- π / 2) = - 1. Substituting those values,

$6\left[ \left\frac{\sqrt{2}\sqrt{3-\sqrt{5}}}{4}+i\left\frac{\sqrt{2}\sqrt{5+\sqrt{5}}}{4}\right]$ ÷ $2\sqrt{2}\left[0-i\right]$

And now simplify this expression to receive our answer,

$6\left[ \left\frac{\sqrt{2}\sqrt{3-\sqrt{5}}}{4}+i\left\frac{\sqrt{2}\sqrt{5+\sqrt{5}}}{4}\right]$ ÷ $2\sqrt{2}\left[0-i\right]$ = $-\frac{3\sqrt{5+\sqrt{5}}}{4}+\frac{3\sqrt{3-\sqrt{5}}}{4}i$ ,

$-\frac{3\sqrt{5+\sqrt{5}}}{4}$ = $-2.01749\dots$ and $\:\frac{3\sqrt{3-\sqrt{5}}}{4}$ = $0.65552\dots$

= $-2.01749+0.65552i$

As you can see our solution is option c. - 2.01749 was rounded to - 2.017, and 0.65552 was rounded to 0.656.

________________________________________

First Attachment : We know from the previous problem that cos(2π / 5) = $\frac{\sqrt{2}\sqrt{3-\sqrt{5}}}{4}$ , sin(2π / 5) = $\frac{\sqrt{2}\sqrt{5+\sqrt{5}}}{4}$ , cos(- π / 2) = 0, and sin(- π / 2) = - 1. Substituting we receive a simplified expression,

$6\sqrt{5+\sqrt{5}}-6i\sqrt{3-\sqrt{5}}$

We know that $6\sqrt{5+\sqrt{5}} = 16.13996\dots$ and $-\:6\sqrt{3-\sqrt{5}} = -5.24419\dots$ . Therefore,

Solution : $16.13996 - 5.24419i$

Which rounds to about option b.

7 0

3 years ago