Ask question

Ask question

All categories

borishaifa [10]

2 years ago

5

Pls hurry and helppp NO LINKS

2 answers:

yuradex [85]2 years ago

8 0

Answer:

-x = 14

is not equivalent to x + 10 = 2x - 4

hope it helps

natita [175]2 years ago

8 0

THE OTHER GUY WAS CORRECT!!!but i hate links so much like ppl really spend time out of their day just to ruin ppls day that really need help on their work. so sad.

You might be interested in

500g of an item costs $12. Use the unitary method to find the equivalent cost of 320g.

zepelin [54]

A) 500g=$12 -> 100g=12:5 100g=$2,40
20g=2,40:5=$0,58
320g=3x2,40+0,58
320g=7,20+0,58
320g=7,78$

5 0

2 years ago

Trignometry problem help plz<br>no 3, 5 and 6

polet [3.4K]

3. First factor $\cos^6A-\sin^6A$ as a difference of cubes:

$\cos^6A-\sin^6A=\underbrace{(\cos^2A-\sin^2A)}_{\cos2A}(\cos^4A+\cos^2A\sin^2A+\sin^4A)$

For the remaining group, apply the double angle identity.

$\cos^2A=\dfrac{1+\cos2A}2$

$\sin^2A=\dfrac{1-\cos2A}2$

$\implies\begin{cases}\cos^4A=\left(\dfrac{1+\cos2A}2\right)^2=\dfrac{1+2\cos2A+\cos^22A}4\\\\\cos^2A\sin^2A=\dfrac{(1+\cos2A)(1-\cos2A)}4=\dfrac{1-\cos^22A}4\\\\\sin^4A=\left(\dfrac{1-\cos2A}2\right)^2=\dfrac{1-2\cos2A+\cos^22A}4\end{cases}$

$\implies4(\cos^6A-\sin^6A)=\cos2A[(1+2\cos2A+\cos^22A)+(1-\cos^22A)+(1-2\cos2A+\cos^22A)]$

$=\cos2A(3+\cos^22A)=\cos^32A+3\cos2A$

5. seems rather tricky. You might want to post another question for that problem alone...

6. Factorize the left side as a sum of cubes:

$\cos^320^\circ+\sin^310^\circ=(\cos20^\circ+\sin10^\circ)(\cos^220^\circ-\cos20^\circ\sin10^\circ+\sin^210^\circ)$

From here we have to prove that

$\cos^220^\circ-\cos20^\circ\sin10^\circ+\sin^210^\circ=\dfrac34$

We can write everything in terms of sine:

$\cos^220^\circ=(1-2\sin^210^\circ)^2=1-4\sin^210^\circ+4\sin^410^\circ$ (double angle identity)

$\cos20^\circ\sin10^\circ=\dfrac{\sin30^\circ-\sin10^\circ}2=\dfrac14-\dfrac12\sin10^\circ$ (angle sum identity)

After some simplifying, we're left with showing that

$4\sin^410^\circ-3\sin^210^\circ+\dfrac12\sin10^\circ=0$

or

$4\sin^310^\circ-3\sin10^\circ+\dfrac12=0$

This last equality follows from what you could the triple angle identity for sine,

$\sin3x=3\sin x-4\sin^3x$

3 0

3 years ago

(3 + 12) + 25 = 3 + (12 + 25)

Whitepunk [10]

Answer:Commutative property of addition

Step-by-step explanation: It’s because you’re adding stuff

6 0

2 years ago

Limit of this equation as h approaches 0

Vladimir [108]

Answer:

the 8

Step-by-step explanation:

4 0

2 years ago

What does "A" equal.

Mila [183]

A/6= -3.3
a/6*6 = -3.3*6 (6's cancel out)
a= -19.8

Hope it helps!

5 0

2 years ago