Help with that for the crown if you cant see it then just zoom in i guess.

Triangle A'B'C' is the image of triangle ABC under a rotation about point Q.<br><br>

yan [13]

9514 1404 393

Answer:

True; about 120° CCW rotation about Q

Step-by-step explanation:

Triangle A'B'C' appears to be congruent to triangle ABC and about the same distance from point Q. The angle of rotation from ABC to A'B'C' appears to be about 120° counterclockwise when measured using a geometry program.

Triangle A'B'C' is the image of triangle ABC under rotation about 120° CCW about point Q.

_____

<em>Additional comment</em>

If choices of rotation angle are ±135° or ±165°, the best choice would be ...

135°

3 0

2 years ago

If 180° < α < 270°, cos⁡ α = −817, 270° < β < 360°, and sin⁡ β = −45, what is cos⁡ (α + β)?

eduard

Answer:

$cos(\alpha+\beta)=-\frac{84}{85}$

Step-by-step explanation:

we know that

$cos(\alpha+\beta)=cos(\alpha)*cos(\beta)-sin(\alpha)*sin(\beta)$

Remember the identity

$cos^{2} (x)+sin^2(x)=1$

step 1

Find the value of $sin(\alpha)$

we have that

The angle alpha lie on the III Quadrant

so

The values of sine and cosine are negative

$cos(\alpha)=-\frac{8}{17}$

Find the value of sine

$cos^{2} (\alpha)+sin^2(\alpha)=1$

substitute

$(-\frac{8}{17})^{2}+sin^2(\alpha)=1$

$sin^2(\alpha)=1-\frac{64}{289}$

$sin^2(\alpha)=\frac{225}{289}$

$sin(\alpha)=-\frac{15}{17}$

step 2

Find the value of $cos(\beta)$

we have that

The angle beta lie on the IV Quadrant

so

The value of the cosine is positive and the value of the sine is negative

$sin(\beta)=-\frac{4}{5}$

Find the value of cosine

$cos^{2} (\beta)+sin^2(\beta)=1$

substitute

$(-\frac{4}{5})^{2}+cos^2(\beta)=1$

$cos^2(\beta)=1-\frac{16}{25}$

$cos^2(\beta)=\frac{9}{25}$

$cos(\beta)=\frac{3}{5}$

step 3

Find cos⁡ (α + β)

$cos(\alpha+\beta)=cos(\alpha)*cos(\beta)-sin(\alpha)*sin(\beta)$

we have

$cos(\alpha)=-\frac{8}{17}$

$sin(\alpha)=-\frac{15}{17}$

$sin(\beta)=-\frac{4}{5}$

$cos(\beta)=\frac{3}{5}$

substitute

$cos(\alpha+\beta)=-\frac{8}{17}*\frac{3}{5}-(-\frac{15}{17})*(-\frac{4}{5})$

$cos(\alpha+\beta)=-\frac{24}{85}-\frac{60}{85}$

$cos(\alpha+\beta)=-\frac{84}{85}$

4 0

3 years ago

What equations is equivalent to y = log x

saw5 [17]

"log" standing alone actually means "logarithm to the base 10."

Thus, y = log x <=> y = log x
10
y
Stated another way (inverse functions), x = 10

4 0

3 years ago

Cuánto dinero me gasto en colocarle baldosas a una sala comedor que tiene 9.30 de largo por 4.20 metros de ancho

Svetlanka [38]

La pregunta está incompleta ya que no se da el costo de la colocación de baldosas por m².

Suponga que el costo de los mosaicos por m² = c

Respuesta:

39.06c

Explicación paso a paso:

El costo del embaldosado será:

El costo por m² * área total a embaldosar

Dado que :

La dimensión de la habitación a embaldosar es:

Longitud = 9,30 metros

Ancho = 4.20 metros

El área total de la habitación a embaldosar es = Largo * ancho

Área total de la habitación a embaldosar = 9,30 m * 4,20 m

Superficie total de la habitación a embaldosar = 39,06 m²

Si el costo del mosaico por m² = c

El costo de embaldosar la habitación será:

39,06 * c = 39,06c

4 0

3 years ago

A cylinder and a cone have congruent heights and radii.what is the ratio of the volume of the cone to the volume of the cylinder

NemiM [27]

Answer:

1: 3 ratio of the volume of the cone to the volume of the cylinder

Step-by-step explanation:

Volume of cone(V) is given by:

$V = \frac{1}{3} \pi r^2h$

where, r is the radius and h is the height of the cone.

Volume of cylinder(V') is given by:

$V' = \pi r'^2h'$

where, r' is the radius and h' is the height of the cylinder.

As per the statement:

A cylinder and a cone have congruent heights and radii.

⇒r = r' and h = h'

then;

$\frac{V}{V'} = \frac{ \frac{1}{3} \pi r^2h}{\pi r'^2h'} = \frac{ \frac{1}{3} \pi r'^2h'}{\pi r'^2h'}$

⇒ $\frac{V}{V'} =\frac{1}{3} = 1 : 3$

Therefore, the ratio of the volume of the cone to the volume of the cylinder is, 1 : 3

4 0

3 years ago

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