All categories

3 years ago

More help with math guys, I need all the help i can get

Mathematics

1 answer:

Fiesta28 [93]3 years ago

5 0

I think it’s A pretty sire

You might be interested in

Triangle PQR has vertices at the following coordinates : P(0,1) Q(3,2) R(5,-4) . Determine wether or not triangle PQR is a right

lord [1]

Here is my answer. Hope this help.

8 0

2 years ago

HELP AGAIN THESE ARE CONFUSING WHEN THEY SAY OF I WILL GIVE BRAINLEST

trasher [3.6K]

Answer:

1 18/35

Step-by-step explanation:

3 0

2 years ago

6x^2 - 19x + 10 <br> Factor

3241004551 [841]

Answer:

(2x-5)(3x-2)

Step-by-step explanation:

1)When factoring quadratic equations in for ax²+bx+c you need to separate the b term in a way that the two addends you separate it by should equal a•c. Just do trial and error. In this case you should get -4 and -15. Your separated equation should be:

6x²-4x-15x+10

2)now factor out a common factor from the first two terms and one from the last two terms you should have:

2x(3x-2)-5(3x-2)

3)finally rewrite this equation into two separate factors and you have your answer.

5 0

3 years ago

Find a, b , and h so that f(x)= a sin (b(x-h))<br> f(x)=5cosx+2sinx

TiliK225 [7]

$a\sin(b(x-h))=a\sin(bx)\cos(bh)-a\cos(bx)\sin(bh)$

For this to be equivalent to $5\cos x+2\sin x$ , you require $b=1$ and

$\begin{cases}a\cos h=2\\-a\sin h=5\end{cases}$

Dividing the second equation by the first gives

$\dfrac{-a\sin h}{a\cos h}=-\tan h=\dfrac52\implies h=\arctan\left(-\dfrac52\right)=-\arctan\dfrac52$

Meanwhile, you also get

$a\cos\left(-\arctan\dfrac52\right)=2\implies a\cos\left(\arctan\dfrac52\right)=\dfrac{2a}{\sqrt{29}}=2\implies a=\sqrt{29}$

So,

$f(x)=5\cos x+2\sin x=\sqrt{29}\sin\left(x+\arctan\dfrac52\right)$

8 0

2 years ago

Si la edad actual de Federico es la

Zina [86]

Answer:

can u say it in eglish pls

Step-by-step explanation:

7 0

2 years ago