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3 years ago

How do i do translation

Mathematics

2 answers:

igor_vitrenko [27]3 years ago

6 0

Search on google “how do I do translation” an it depends on the subject.

MA_775_DIABLO [31]3 years ago

4 0

You can use google translate

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Which describes the calculations that could be used to solve this problem?

Darya [45]

Answer:

Subtract 37 – 13. Then divide the difference by 4.

Step-by-step explanation:

After the box:

37 - 13 left

No. of bags:

(37 - 13)/4

4 0

3 years ago

Which of the following is not part of the Triangle Proportionality Theorem?

natka813 [3]

Answer:

A. AA Similarity Theorem . True

B. Two sides divided proportionally. True

C. A parallel line to one side of a triangle. True

D. A perpendicular line to one side of a triangle. False

Step-by-step explanation:

The Triangle Proportionality Theorem says that if a line is parallel to one side of a triangle, then it splits the other two sides into proportional sections.

Now, here the given statements are:

A. AA Similarity Theorem

TRUE, as the angles opposite to equal sides are equal.

And if lines are divided proportionally, both the triangles are Similar by AA similarity theorem.

B. Two sides divided proportionally.

TRUE,as the other two sides of the triangle are divided proportionally by the parallel line.

C. A parallel line to one side of a triangle.

TRUE, as the parallel line divides the other two non- parallel sides in equal proportions.

D. A perpendicular line to one side of a triangle.

FALSE, this part is not required anywhere in the given theorem result.

3 0

2 years ago

CAn you please explain thist to me?

boyakko [2]

Find the volume of the cone Vc
Find volume of sphere Vs
Volume of shaded figure Vsh=Vc-Vs

8 0

2 years ago

Let alpha and beta be conjugate complex numbers such that frac{\alpha}{\beta^2} is a real number and alpha - \beta| = 2 \sqrt{3}

miskamm [114]

Answer:

$-3+i\sqrt{3} , 1+\sqrt{3}$

Step-by-step explanation:

Given that alpha and beta be conjugate complex numbers

such that frac{\alpha}{\beta^2} is a real number and alpha - \beta| = 2 \sqrt{3}.

Let

$\alpha = x+iy\\\beta = x-iy$

since they are conjugates

$\alpha-\beta = x+iy-(x-iy)\\= 2iy= 2i\sqrt{3} \\y =\sqrt{3}$

$\frac{\alpha}{\beta^2} }\\=\frac{x+i\sqrt{3} }{(x-i\sqrt{3})^2} \\=\frac{x+i\sqrt{3}}{x^2-3-2i\sqrt{3}} \\=\frac{x+i\sqrt{3}((x^2-3+2i\sqrt{3}) }{(x^2-3-2i\sqrt{3)}(x^2-3-2i\sqrt{3})}$