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3 years ago

9

Which factorization is equivalent to this expression? −42k − 54

1 answer:

Elodia [21]3 years ago

4 0

Answer:

-6(7k +9)

Step-by-step explanation:

−42k − 54

Factor out a -6

-6(7k +9)

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Find sin3A, cos<br> 3A and tan 3A when sin=1/2<br>

Slav-nsk [51]

Answer:

Sin 3A = 1

cos 3A = 0

tan 3 A = undefined

Step-by-step explanation:

$\sin A = \frac{1}{2} \\ \therefore \: \sin A = \sin \: 30 \degree.. \bigg( \because \: \sin \: 30 \degree = \frac{1}{2} \bigg) \\ \therefore \: A = 30 \degree \\ \\ now \\ \sin 3A =\sin (3 \times 30 \degree)= \sin \: 90 \degree = 1 \\ \\ \cos 3A =\cos (3 \times 30 \degree)= \cos \: 90 \degree = 0\\\\ \tan 3A =\tan (3 \times 30 \degree)= \tan \: 90 \degree = \infty$

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4 years ago

what is the distance between the first integer that is less than square root 40 amd the first integer that is greater than squar

ki77a [65]

Answer:

6

Step-by-step explanation:

$\sqrt{40} = 6.32 so 6 \\\sqrt{125} = 11.18 so 12\\12 - 6 = 6$

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4 years ago

Sam needs to take out a personal loan for $8,900 to pay for a trip to Europe with his classmates. His bank has offered him the f

saul85 [17]

Answer:

The Answer is D

Step-by-step explanation:

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3 years ago

Need it solved correctly for khan academy

RUDIKE [14]

The tiger population loses 3/5 of its size every 2.94 decades

<h3>Rate of change using differential calculus</h3>

The given equation is:

$N(t)=710(\frac{8}{125} )^t$

Find the derivative of the given function

$\frac{dN}{dt} =710(0.064)^tln(0.064)\\\\\frac{dN}{dt} =-1951.7(0.064)^t$

When the tiger loses 3/5 of its population

dN/dt = 3/5

Solve for t

$\frac{3}{5} =-1951.7(0.064)^t\\\\-0.0003=(0.064)^t$

Take the natural logarithm of both sides

$ln(-0.0003)=t(ln0.064)\\\\-8.087=-2.75t\\\\t=\frac{-8.087}{-2.75} \\\\t=2.94$

The tiger population loses 3/5 of its size every 2.94 decades

Learn more on rate of change using calculus here: brainly.com/question/96116

#SPJ1

6 0

2 years ago

Find the general form, Ax2+Ay2+Dx+Ey+F=0, by identifying the coefficients A,D,E,&F center:(0,1); r=1 (x-0)2+(y-1)2=1 x2+y2-2

siniylev [52]

Answer:

The value of A is 1, D is 0, E is -2 and F is 0.

Step-by-step explanation:

The given equation is

$Ax^2+Ay^2+Dx+Ey+F=0$ ...(1)

The standard form of the circle is

$(x-h)^2+(y-k)^2=r^2$

Where, (h,k) is the center of the circle and r is the radius.

$(x-0)^2+(y-1)^2=1$

$x^2+y^2-2y+1-1=0$

$x^2+y^2-2y=0$

It can be written as

$x^2+y^2+0x-2y+0=0$ ....(2)

On comparing (1) and (2) we get.

$A=1$

$D=0$

$E=-2$

$F=0$

Therefore the value of A is 1, D is 0, E is -2 and F is 0.

6 0

4 years ago

Read 2 more answers