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4 years ago

15

there is a rectangle built with 12 tiles in this rectangle half of the tiles are red one fourth of the tiles are yellow and one

fourth of the tiles are green how many tiles are red how many tiles are yellow how many tiles are green

1 answer:

yarga [219]4 years ago

6 0

There are 6 red tiles since 6 is half of 12. 1/4 of 12 is 3 (divide 12 by 4), so there are 3 green tiles and 3 yellow tiles

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Can someone help me with math I need help I will attempt to give y’all help too!

lukranit [14]

Answer:

A, 52.5

Step-by-step explanation:

We see that the 42 side corresponds to the side with side length 16, so the factor is 42/16 = 21/8. Now, we are looking for the side that corresponds to x, which is 20. Multiplying by the scale factor, we have 20 x 21/8 = 52.5

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2 years ago

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Which set of line segments could create a right triangle?

Maslowich

Its not d it is b or a

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3 years ago

guajiro [1.7K]

$▪▪▪▪▪▪▪▪▪▪▪▪▪ {\huge\mathfrak{Answer}}▪▪▪▪▪▪▪▪▪▪▪▪▪▪$

Simplification ~

$- 8(13 - 5c + 12d)$

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3 years ago

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A baseball player leads off the game and hits a long home run. the ball leaves the bat at a height of 4ft above the ground and a

shepuryov [24]

Answer:

time = 4.4 s

horizontal distance = 488.8 m

Step-by-step explanation:

Height, h = 4 ft

Angle, A = 32 degree above X axis

velocity, u = 131 ft/s

Let the time taken is t.

Use second equation of motion

$h = u t + 0.5 gt^2\\\\- 4 = 131 sin 32 t - 16 t^2\\\\16t^2 - 69.4t - 4 = 0\\\\t=\frac{69.4\pm \sqrt{4816.36 +256}}{32}\\\\t = \frac{69.4\pm71.2}{32}\\\\t = - 0.05625 s, 4.4 s$

So, time is 4.4 s.

The horizontal distance is

d = u cos A t

d = 131 cos 32 x 4.4 = 488.8 m

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3 years ago

Let [v1,v2,v3] be a set of nonzero vectors in r^m such that the (transpose of vi)*vj = 0 when i is not equal to j. show that the

babunello [35]

Let $\mathbf V$ be the $m\times3$ matrix whose columns are $\mathbf v_1,\mathbf v_2,\mathbf v_3$ , and let $\mathbf c$ be the vector whose components are the constants $c_1,c_2,c_3$ . Now consider the matrix equation

$\mathbf V\mathbf c=\mathbf 0$

Multiplying both sides by $\mathbf V^\top$ , we have

$\mathbf V^\top(\mathbf V\mathbf c)=(\mathbf V^\top\mathbf V)\mathbf c=\mathbf 0$

More explicitly, we're writing

$\mathbf V=\begin{bmatrix}\mathbf v_1&\mathbf v_2&\mathbf v_3\end{bmatrix}$

Multiply both sides by $\mathbf V^\top$ and the left hand side can be written as

$\mathbf V^\top\mathbf V=\begin{bmatrix}{\mathbf v_1}^\top\\{\mathbf v_2}^\top\\{\mathbf v_3}^\top\end{bmatrix}\begin{bmatrix}\mathbf v_1&\mathbf v_2&\mathbf v_3\end{bmatrix}=\begin{bmatrix}{\mathbf v_1}^\top\mathbf v_1&{\mathbf v_1}^\top\mathbf v_2&{\mathbf v_1}^\top\mathbf v_3\\{\mathbf v_2}^\top\mathbf v_1&{\mathbf v_2}^\top\mathbf v_2&{\mathbf v_2}^\top\mathbf v_3\\{\mathbf v_3}^\top\mathbf v_1&{\mathbf v_3}^\top\mathbf v_2&{\mathbf v_3}^\top\mathbf v_3\end{bmatrix}$

We're told that ${\mathbf v_i}^\top\mathbf v_j=0$ whenever $i\neq j$ , so we're left with

$\mathbf V^\top\mathbf V=\begin{bmatrix}\|\mathbf v_1\|^2&0&0\\0&\|\mathbf v_2\|^2&0\\0&0&\|\mathbf v_3\|^2\end{bmatrix}$

Each of $\mathbf v_1,\mathbf v_2,\mathbf v_3$ are nonzero, which means their norms are nonzero, which necessarily implies that $\mathbf c=0$ , and so the vectors $\mathbf v_1,\mathbf v_2,\mathbf v_3$ must necessarily be linearly independent.

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3 years ago