All categories

3 years ago

GEOMETRY HELP PLEASE!!! FAST!!!

Mathematics

1 answer:

Lubov Fominskaja [6]3 years ago

7 0

The <em><u>correct answer</u></em> is:

Rotation of 90° counterclockwise about the origin and a translation 2 units right

Explanation:

A rotation 90° counterclockwise maps each point (x, y) to (-y, x). This means our points would be:

A(-4,3)→(-3, 4); B(-1,3)→(-3, -1); C(-2,1)→(-1, -2)

A translation 2 units right will add 2 units to the new x-coordinates; this gives us

(-3, 4)→(-1, 4); (-3, -1)→(-1, -1); and (-1, -2)→(1, -2)

These are the points in the image, so this is the correct set of transformations.

You might be interested in

Which of these is not a possible r-value?

Tems11 [23]

Answer:

Step-by-step explanation:

4 0

4 years ago

If for two non-zero vectors a,b : ∣a+b∣=∣a−b∣ then angle between them is:______a. 0b. 45c. 60d. 90

Eddi Din [679]

Answer:

$90^\circ$

Step-by-step explanation:

Given two non zero vectors, $\vec a, \vec b$ .

Let the angle between the two vectors = $\theta$

Given that:

$|\vec a+\vec b|=|\vec a-\vec b|$

Let us have a look at the formula for magnitude of addition of two vectors:

$|\vec x+\vec y|=\sqrt{x^2+y^2+2xycos\theta}$

Where $\theta$ is the angle between the two vectors.

formula for magnitude of subtraction of two vectors:

$|\vec x-\vec y|=\sqrt{x^2+y^2-2xycos\theta}$

As per the given condition:

$\sqrt{a^2+b^2+2abcos\theta}=\sqrt{a^2+b^2-2abcos\theta}$

Squaring both sides:

$a^2+b^2+2abcos\theta=a^2+b^2-2abcos\theta\\\Rightarrow 2abcos\theta=-2abcos\theta\\\Rightarrow cos\theta = -cos\theta\\\Rightarrow 2cos\theta = 0\\\Rightarrow cos\theta = 0\\\Rightarrow \theta = 90^\circ$

So, the angle between the two vectors is: $90^\circ$

7 0

3 years ago

Given △GHI and △JKL, GI = 5, HI = 4, JK = 4, and JL = 5, what else do you need to know to prove the two triangles are congruent

blsea [12.9K]

Answer:

△ GHI ≅ △ JKL. (Proved)

Step-by-step explanation:

In triangle Δ GHI, GI = 5 and HI = 4.

If we consider the triangle is a right triangle having hypotenuse = 5 and any other leg = 4.

Then this will follow the Pythagoras theorem as 3² + 4² = 5², where 3 is the other leg.

Therefore, Δ GHI is a right triangle having hypotenuse 5 and one leg 4.

Similarly, we can prove that △ JKL is also a right triangle(Since JK = 4 and JL = 5), having hypotenuse 5 and one leg 4.

Therefore, applying HL rule, we cam conclude △ GHI ≅ △ JKL. (Proved)

7 0

3 years ago

What is the percent of $31 to $35

il63 [147K]

The percent is %89. Might I suggest using a calculator?

3 0

4 years ago

What is the quotient (6x4 − 15x3 − 2x2 − 10x − 4) ÷ (3x2 + 2)?

wariber [46]

we are given

$(6x^4-15x^3-2x^2-10x-4)$ ÷ $(3x^2+2)$

we can divide it by using long division method

we will get

$Quotient=2x^2 -5x-2$

$Remainder=0$

$(6x^4-15x^3-2x^2-10x-4)$ ÷ $(3x^2+2)$ = $2x^2 -5x-2$

..............Answer

5 0

3 years ago