All categories

3 years ago

Help please will give brainliest

Mathematics

2 answers:

aleksley [76]3 years ago

8 0

Answer:

uhh either 4/4 1 or 2/2

Step-by-step explanation:

if the words covered by the light reflection says "the shaded part", then there are 4 total pieces, which means all four are shaded.

MAXImum [283]3 years ago

5 0

4/4 . hope this helped !

You might be interested in

What is the square root of 1078?

german

32.8329103188 is the square root

7 0

4 years ago

1/2 greater than or less than 3/15

Colt1911 [192]

Greater
If we get common denominators
1/2=15/30
3/15=6/30
15>6 therefore 1/2>3/15

5 0

3 years ago

Line segment CD begins at (−1,1) and ends at (4,1). The segment is translated 2 units down to form line segment C'D'. Line segme

dusya [7]

Answer:

D' = (4, -1)

Step-by-step explanation:

Translation down by 2 units subtracts 2 from the y-coordinate. The x-coordinate remains unchanged. You can see this in the descriptions of C and C':

C = (-1, 1)
C' = (-1, -1) . . . . . 2 is subtracted from y=1 to get y=-1

Likewise:

D = (4, 1)
D' = (4, -1)

The translated segment is still a horizontal segment, now at y=-1 instead of at its previous position of y=1.

6 0

4 years ago

Find the critical points of the function and use the First Derivative Test to determine whether the critical point is a local mi

Andre45 [30]

Answer:

Critical points are 1 and -1

Maximum at x=1

Minimum at x=-1

Step-by-step explanation:

We are given that a function

$f(x)=3tan^{-1}(x)-\frac{3}{2} x+5$ on ( $-\infty,\infty$ )

We have to find the critical points of the function.

To find the critical point we will differentiate function w.r.t x and then substitute f'(x)=0

$f'(x)=\frac{3}{1+x^2}-\frac{3}{2}$

$\frac{d(tan^{-1}x)}{dx}=\frac{1}{1+x^2}$

$f'(x)=0$

$\frac{3}{1+x^2}-\frac{3}{2}=0$

$\frac{3}{1+x^2}=\frac{3}{2}$

$1+x^2=2$

$x^2=2-1=1$

$x=\pm1$

Therefore, the critical points of the given function are 1 and -1.

$f(0)=3-\frac{3}{2}=\frac{3}{2}$

$f'(1)=0$

$f'(2)=\frac{3}{5}-\frac{3}{2}=-\frac{9}{10}$

When we goes from 0 to 2 then the sign of derivative change from positive to negative .Therefore, function has local maximum at x=1.

$f(-2)=\frac{3}{5}-\frac{3}{2}=-\frac{9}{10}$

$f(-1)=0$

$f(0)=\frac{3}{2}$

When we goes form -2 to 0 then the sign of derivative change from negative to positive .Hence , function has local minimum at x=-1

Hence, critical points are local maximum and local minimum .

3 0

3 years ago

I need help with this

julsineya [31]

The answer would be the third number line

7 0

3 years ago

Read 2 more answers