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3 years ago

Find the greatest common factor: 4ab^2 c^4-2a^2 b^3 c^2+6a^3 b^4 c

1 answer:

6 0

Well the GCF is the number that can be used to simplify, or that fits in all of your numbers for example this are all small numbers so 2 should fit in all of them 2 fits in 4 two times 2 fits in 2 one time and 2 fits in 6 three times now lets check our variable, to find the GCF of the variable first chekc if they all have the same if they don't u can't get none of taht variebles out, but if they repeat in all like the a u will take out the smallest amount out, for example the a as one in the degree of 1 the other in to the 2 degree and the last one to teh 3 degree well the smallest degree will be 1 so u will only take 1 a out so now ur GCF looks like 2a... Lets check the other variables, b is used in all of them and the smallest degree is b2 so we will take out 2 b's out so now my GCF looks like 2ab2 now lets check our last variable the, the c has the smallest degree of 1 so we will only take 1 c out
this means our final GCF is "2ab2c"

Hoep this helps

You might be interested in

-1/4b + 1/2 = 5/6<br>Find b

QveST [7]

Answer:

b = -4/3

Step-by-step explanation:

First, subtract the 1/2 over

-1/4b+1/2=5/6

5/6-1/2

To get a common denominator of 6, multipy 1/2 by 3

5/6-1/2(3)

5/6-3/6=2/6

-1/4b=2/6

Next, multiply by the reciprocal of -1/4, or -4

(-4)-1/4b=2/6(-4)

b=-4/3

7 0

3 years ago

A particle is projected with a velocity of <img src="https://tex.z-dn.net/?f=40ms%5E-%5E1" id="TexFormula1" title="40ms^-^1" alt

Katena32 [7]

Answer:

$2\sqrt{55}\text{ m/s or }\approx 14.8\text{m/s}$

Step-by-step explanation:

The vertical component of the initial launch can be found using basic trigonometry. In any right triangle, the sine of an angle is equal to its opposite side divided by the hypotenuse. Let the vertical component at launch be $y$ . The magnitude of $40\text{ m/s}$ will be the hypotenuse, and the relevant angle is the angle to the horizontal at launch. Since we're given that the angle of elevation is $60^{\circ}$ , we have:

$\sin 60^{\circ}=\frac{y}{40},\\y=40\sin 60^{\circ},\\y=20\sqrt{3}$ (Recall that $\sin 60^{\circ}=\frac{\sqrt{3}}{2}$ )

Now that we've found the vertical component of the velocity and launch, we can use kinematics equation $v_f^2=v_i^2+2a\Delta y$ to solve this problem, where $v_f/v_i$ is final and initial velocity, respectively, $a$ is acceleration, and $\Delta y$ is distance travelled. The only acceleration is acceleration due to gravity, which is approximately $9.8\:\mathrm{m/s^2}$ . However, since the projectile is moving up and gravity is pulling down, acceleration should be negative, represent the direction of the acceleration.

What we know:

$v_i=20\sqrt{3}\text{ m/s}$
$a=-9.8\:\mathrm{m/s^2}$
$\Delta y =50\text{ m}$

Solving for $v_f$ :

$v_f^2=(20\sqrt{3})^2+2(-9.8)(50),\\v_f^2=1200-980,\\v_f^2=220,\\v_f=\sqrt{220}=\boxed{2\sqrt{55}\text{ m/s}}$

3 0

3 years ago

I am being asked to calculate and plot residuals and I don't know how my x values are 13.8, 18, 16,7, 18, 0.7, 21.9, 9.2, 19.5,

Igoryamba

Answer:

(13.8,1.42) (18,3.7)(16.7,3.21) and so on REMEMBER ITS (X,Y)

Step-by-step explanation:

3 0

3 years ago

25 POINTS PLEASE PLEASE HELPPPP, ALSO THE QUESTION IS WORTH 4 MARKS SO PLEASEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEE HELPPPPP

patriot [66]

So u already know that the radius is half of the circle which in this case is 28 mm. And both circles at the top touch both sides of the rectangle.
Process:
28x4=112.
This is the top side, but because you could fit 2 circles at the top and you know the measurement, it is common sense to think that it would be the same on the other side. So the area of the square/ rectangle:
112x112= 12,534 mm

5 0

3 years ago

The price of a pair of Jordans was reduced from $200 to $160. By percentage was the price reduced

hichkok12 [17]

Answer:

20%

Step-by-step explanation:

3 0

2 years ago