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2 years ago

Will mark brainliest for whoever answers correctly

Mathematics

1 answer:

Veronika [31]2 years ago

4 0

Answer:

The answer is 2

Step-by-step explanation:

x+y = (1+i) + (1-i)

open the bracket...

1+i + 1-i

collect like terms

=1+1+i-i

= 2-0

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Please help me i need this answer asap

xxTIMURxx [149]

Answer:

Step-by-step explanation:

d = (x_2 - x_1)^2 + (y_2-y_1)^2

If any two points on both triangles have the same measure then they are congruent.

3 0

2 years ago

Factored 8x2y3-5= <br> I need the answer

ANTONII [103]

Answer: Reformatting the input :

Changes made to your input should not affect the solution:

(1): "y3" was replaced by "y^3". 1 more similar replacement(s).

STEP

Equation at the end of step 1

(23x2 • y3) - 5

STEP

Trying to factor as a Difference of Cubes

2.1 Factoring: 8x2y3-5

Theory : A difference of two perfect cubes, a3 - b3 can be factored into

(a-b) • (a2 +ab +b2)

Proof : (a-b)•(a2+ab+b2) =

a3+a2b+ab2-ba2-b2a-b3 =

a3+(a2b-ba2)+(ab2-b2a)-b3 =

a3+0+0+b3 =

a3+b3

Check : 8 is the cube of 2

Check : 5 is not a cube !!

Ruling : Binomial can not be factored as the difference of two perfect cubes

Final result :

8x2y3 - 5

5 0

3 years ago

Please select the best answer from the choices provided<br><br> T<br> F

bogdanovich [222]

Answer:

Step-by-step explanation:

The ordered pair (-2,-1) can also be represented by (x,y). To check the inequality we can substitute the values into the given inequality.

$y>-\sqrt{x+11+1}$

$-1>-\sqrt{-2+11+1}$

$-1>-\sqrt{10}$

$-1>-3.1622$

After substituting the values, we then have a clear view that -1 is Greater than -3.1622.

8 0

3 years ago

Question 4: 10 pts

UkoKoshka [18]

Answer:2

Step-by-step explanation: but it was real

7 0

2 years ago

Find the points on the curve where the tangent is horizontal or vertical. If you have a graphing device, graph the curve to chec

klasskru [66]

Answer:

There is a horizontal tangent at (0,-4)

The tangent is vertical at (-2,-3) and (2,-3).

Step-by-step explanation:

The given function is defined parametrically by the equations:

$x=t^3-3t$

and

$y=t^2-4$

The tangent function is given by:

$\frac{dy}{dx}=\frac{\frac{dy}{dt} }{\frac{dx}{dt} }$

$\implies \frac{dy}{dx}=\frac{2t}{3t^2-3}$

The tangent is vertical at when $\frac{dx}{dt}=0$

$\implies \frac{3t^2-3}{2t}=0$

$\implies 3t^2-3=0$

$\implies 3t^2=3$

$\implies t^2=1$

$\implies t=\pm1$

When t=1,

$x=1^3-3(1)=-2$ and $y=1^2-4=-3$

When t=-1,

$x=(-1)^3-3(-1)=2$ and $y=(-1)^2-4=-3$

The tangent is vertical at (-2,-3) and (2,-3).

The tangent is horizontal, when $\frac{dy}{dx}=0$ or $\frac{dy}{dt}=0$

$\implies 2t=0$

$\implies t=0$

When t=0,

$x=0^3-3(0)=0$ and $y=0^2-4=-4$

There is a horizontal tangent at (0,-4)

5 0

3 years ago