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3 years ago

One kilometer equals 0.62 miles. How many kilometers equal 16 miles? Round the answer to the nearest tenth.

Mathematics

2 answers:

OlgaM077 [116]3 years ago

7 0

Answer: 25.7495 full: 25.75 rounded

Step-by-step explanation: Multiply 16*0.62, simple. From that you get 25.7495 Then, since there is a 9 after the four, you would round the 4 up to a 5.

Nitella [24]3 years ago

6 0

Answer:

25.8

Step-by-step explanation:

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So I don’t have enough points for asking two answer so please anyone help me thnaks

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3 years ago

Please help asap asap

Ivanshal [37]

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No, it's not because 6 x 4 = 24 and 2 x 3 = 6.

Step-by-step explanation:

In order for it to be equivalent, the simplified ratio needs to be multiplied by the same number, meaning because 6 x 4 = 24, the next number needs to be the product of 2 x 4, not 2 x 3.

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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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tamaranim1 [39]

Answer: D.

Step-by-step explanation: Apply the distributive property.

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