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2 years ago

Describe the transformation that moves ABC to its image A'B'C'.

Mathematics

1 answer:

kap26 [50]2 years ago

8 0

All it did was it counter clock wise rotated to make the shape look like it’s a mirror

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This is probley easy...

Yakvenalex [24]

Answer:

Step-by-step explanation:

divide 450 and 75

4 0

3 years ago

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Find the measure of the missing angles

Ipatiy [6.2K]

1- 38
2-90
3-38
hopefully i’m right i’m not 100% sure though :)

6 0

3 years ago

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Mind explaining anyone?

8090 [49]

Do 45/225 and simplify and that’s your answer

7 0

3 years ago

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Find the value of x, y, and z in the rhombus below.

Yuki888 [10]

Solution:

It should be noted:

Opposite sides of a rhombus are always equal.
Opposite angles of a rhombus are always equal.

Thus:

(-y - 10) = 90°
3z - 3 = 90°
4x - 2 = 90°

Finding x:

4x - 2 = 90°
=> 4x = 90 + 2
=> 4x = 92
=> x = 23

Finding y:

(-y - 10) = 90°
=> -y - 10 = 90°
=> -y = 100
=> y = -100

Finding z:

3z - 3 = 90°
=> 3z = 90 + 3
=> 3z = 93
=> z = 31

4 0

2 years ago

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Find the solution of the differential equation that satisfies the given initial condition. dy/dx=x/y , y(0)=-1

Charra [1.4K]

Separating variables, we have

$\dfrac{dy}{dx} = \dfrac xy \implies y\,dy = x\,dx$

Integrate both sides.

$\displaystyle \int y\,dy = \int x\,dx$

$\dfrac12 y^2 = \dfrac12 x^2 + C$

Given that $y(0)=-1$ , we find

$\dfrac12 (-1)^2 = \dfrac12 0^2 + C \implies C = \dfrac12$

Then the particular solution is

$\dfrac12 y^2 = \dfrac12 x^2 + \dfrac12$

$y^2 = x^2 + 1$

$y = \pm\sqrt{x^2 + 1}$

and because $y(0)=-1$ , we take the negative solution to accommodate this initial value.

$\boxed{y(x) = -\sqrt{x^2+1}}$

7 0

2 years ago