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2 years ago

Will offer 40 points for answer Help with math question it’s in the image

Mathematics

1 answer:

dmitriy555 [2]2 years ago

7 0

I think it’s the first one because you mirror ABC from the m-axis and then rotate ABC 90 degrees clockwise from point B’ to get to the position of triangle A”B’C”

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(x5 + y5) ÷ (x + y) please answer ASAP

Shkiper50 [21]

Answer:

$\frac{x^5+y^5}{x+y}=x^4-x^3y+x^2y^2-xy^3+y^4$

Step-by-step explanation:

Given the expression

$\left(x^5+y^5\right)\div \left(x+y\right)$

$\mathrm{Apply\:factoring\:rule:\:}x^n+y^n=\left(x+y\right)\left(x^{n-1}-x^{n-2}y+\:\dots \:-\:xy^{n-2}\:+\:y^{n-1}\right)\:\quad \quad \mathrm{n\:is\:odd}$

$x^5+y^5=\left(x+y\right)\left(x^4-x^3y+x^2y^2-xy^3+y^4\right)$

$=\frac{\left(x+y\right)\left(x^4-x^3y+x^2y^2-xy^3+y^4\right)}{x+y}$

Cancel the common factor: x+y

$=x^4-x^3y+x^2y^2-xy^3+y^4$

Thus,

$\frac{x^5+y^5}{x+y}=x^4-x^3y+x^2y^2-xy^3+y^4$

3 0

3 years ago

4. An airplane descends at a rate of 1,400 feet per minute. Once the airplane starts its descent, what will be its change in ele

Evgesh-ka [11]

Answer:

5600 feet

Step-by-step explanation:

If it is dropping 1400 every minute then to find the elevation at 4 minutes, you have to multiply 1400 with 4.

5 0

3 years ago

For the love of God help me !! I'm desperate for it tomorrow

Eduardwww [97]

Try to relax. Your desperation has surely progressed to the point where
you're unable to think clearly, and to agonize over it any further would only
cause you more pain and frustration.
I've never seen this kind of problem before. But I arrived here in a calm state,
having just finished my dinner and spent a few minutes rubbing my dogs, and
I believe I've been able to crack the case.

Consider this: (2)^a negative power = (1/2)^the same power but positive.

So:
Whatever power (2) must be raised to, in order to reach some number 'N',
the same number 'N' can be reached by raising (1/2) to the same power
but negative.

What I just said in that paragraph was: log₂ of(N) = - log(base 1/2) of (N) .
I think that's the big breakthrough here.
The rest is just turning the crank.

Now let's look at the problem:

log₂(x-1) + log(base 1/2) (x-2) = log₂(x)

Subtract log₂(x) from each side:

log₂(x-1) - log₂(x) + log(base 1/2) (x-2) = 0

Subtract log(base 1/2) (x-2) from each side:

log₂(x-1) - log₂(x) = - log(base 1/2) (x-2) Notice the negative on the right.

The left side is the same as log₂[ (x-1)/x ]

==> The right side is the same as +log₂(x-2)

Now you have: log₂[ (x-1)/x ] = +log₂(x-2)

And that ugly [ log to the base of 1/2 ] is gone.

Take the antilog of each side:

(x-1)/x = x-2

Multiply each side by 'x' : x - 1 = x² - 2x

Subtract (x-1) from each side:

x² - 2x - (x-1) = 0

x² - 3x + 1 = 0

Using the quadratic equation, the solutions to that are
x = 2.618
and
x = 0.382 .

I think you have to say that x=2.618 is the solution to the original
log problem, and 0.382 has to be discarded, because there's an
(x-2) in the original problem, and (0.382 - 2) is negative, and
there's no such thing as the log of a negative number.

There,now. Doesn't that feel better.

4 0

3 years ago

25 Points ! Write a paragraph proof. Given: ∠T and ∠V are right angles. Prove: ∆TUW ∆VWU

vladimir2022 [97]

Answer:

Δ TUW ≅ ΔVWU ⇒ by AAS case

Step-by-step explanation:

* Lets revise the cases of congruent for triangles

- SSS ⇒ 3 sides in the 1st Δ ≅ 3 sides in the 2nd Δ

- SAS ⇒ 2 sides and including angle in the 1st Δ ≅ 2 sides and

including angle in the 2nd Δ

- ASA ⇒ 2 angles and the side whose joining them in the 1st Δ

≅ 2 angles and the side whose joining them in the 2nd Δ

- AAS ⇒ 2 angles and one side in the first triangle ≅ 2 angles

and one side in the 2ndΔ

- HL ⇒ hypotenuse leg of the first right angle triangle ≅ hypotenuse

leg of the 2nd right angle Δ

* Lets solve the problem

- There are two triangles TUW and VWU

- ∠T and ∠V are right angles

- LINE TW is parallel to line VU

∵ TW // VU and UW is a transversal

∴ m∠VUW = m∠TWU ⇒ alternate angles (Z shape)

- Now we have in the two triangles two pairs of angle equal each

other and one common side, so we can use the case AAS

- In Δ TUW and ΔVWU

∵ m∠T = m∠V ⇒ given (right angles)

∵ m∠TWU = m∠VUW ⇒ proved

∵ UW = WU ⇒ (common side in the 2 Δ)

∴ Δ TUW ≅ ΔVWU ⇒ by AAS case

7 0

3 years ago

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An oblique hexagonal prism has a base area of 42 square cm. The prism is 4 cm tall and has an edge length of 5 cm. An oblique he

Anna [14]

Answer:

Step-by-step explanation:

8 0

3 years ago

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