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2 years ago

Find the LCM for 6 and 4.

Mathematics

2 answers:

DENIUS [597]2 years ago

5 0

The LCM of 6 and 4, is 12.

nlexa [21]2 years ago

3 0

The LCM for 6 and 4 is 12. (6, |12| 18/4, 8, |12|)

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Fatima evaluated the expression when m = -2 and n = 4. Her work is shown below.

Kruka [31]

Answer:

The answer to your question is: third option is correct.

Step-by-step explanation:

The third option is correct

4(-2)⁻²(4)⁻³

$\frac{4}{(2)^{2}(4)^{3}}}$

4 0

3 years ago

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Mr. Toshio lent out 11 rulers at the beginning of class, collected 4 rulers in the middle of class, and gave out 7 at the end of

Drupady [299]

Mr. Toshio had 32 rulers. 32-11 = 21 rulers. 21+4 = 25 rulers. and 25-7 = 18 rulers

8 0

3 years ago

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Describe how to transform the graph of f(x) = x² to obtain

lord [1]

Answer:

The functions given are:

f(x) = x²

g(x) = f(-4x-3) + 1

First, find f(-4x-3):

f(x) = x²

f(-4x-3) = (-4x-3)²

Find g(x):

g(x) = f(-4x-3) + 1

g(x) = (-4x-3)² + 1

g(x) = (-1)² (4x+3)² + 1

g(x) = (4x+3)² + 1

First take

y = (x)²

Compress the graph along x axis by multiplying x with 4

y = (4x)²

Shift the graph left by 0.75 units, by adding 3 to x term.

y = (4x+3)²

Shift the graph up by 1 unit by adding 1 to the total terms.

y = (4x+3)² +1

4 0

3 years ago

For every 1 litre of water used to make medicine, 200ml of sucrose and 300ml of saline solution are used. Express the amount of

DiKsa [7]

Answer:

taste of your own medicine jk

Step-by-step explanation:

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3 years ago

HELP ASAP I WILL GIVE BRAINLIEST!!

Alexxx [7]

Consider the function F(x)= x³ +½x² −2x +3. Use calc and algebra to find any and all stationary points on the graph of the function.Precalc need help ASAP

Solution:

the above function has more than one root or more than one solution. The derivative of the function represents the slope of the tangent line to the curve

F(x)= x³ +½x² −2x +3, F'(x) = 3(x3-1)+(1/2)(2)(x2-1)-2(x1-1)+0

F'(x)=3x2+x-2 is the slope of the tangent line

setting the slope =0 and solving the quadratic equations, you find all the

critical points of interest

3x2+x-2=0, solve by factoring (3x-2)(x+1)=0, then 3x-2=0, x= 2/3

and x+1=0, x=-1 substituting these values of x in the original equation

you get the corresponding values of F(x) such that;

F(x)= x³ +½x² −2x +3, F(x)=(2/3)3+1/2(2/3)2-2(2/3)+3

= 8/27+4/18-4/3+3 = 8/27+2/9-4/3+3/1 = 59/27 and the first point is

(2/3, 59/27)

F(x)= x³ +½x² −2x +3, F(x)=(-1)3+1/2(-1)2-2(-1)+3

= -1+1/2+2+3 = 9/2 and the 2nd point is ( -1, 9/2)

these two points represents points where the slope of the tangent line to the curve is 0 and could be either maximum or minimum points

You can also find the y intercepts where x = 0 F(x)= x³ +½x² −2x +3, setting x = 0, y= F(x)=3 and the point ( 0,3) is a y intercept setting y = 0, then x intercepts can be found upon solving the equation x³ +½x² −2x +3 =0

3 0

2 years ago