Round to the nearest hundredth. I need to be walked through the steps :)

Mathematics

2 answers:

Mamont248 [21]3 years ago

6 0

It's 45 because they are congruent. If the vertex is in the center of a circle then the angle and the arc are congruent

devlian [24]3 years ago

6 0

To solve this, you'll need the arc length formula for degrees:
2 $\pi$ *radius(C/360)
(C is "central angle")

2 $\pi$ *radius is the circumference of the entire circle, 360 is a full angle, so this formula gives you the ratio of full circumference/full angle to the angle and arc length you have.

2 $\pi$ *7(45/360) = 5.497
= 5.50 to the nearest hundred.

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The greatest integer that satisfies the inequality 1.6−(3−2y)<5

Cerrena [4.2K]

Solve the inequality 1.6-(3-2y)<5.

1. Rewrite this inequality without brackets:

1.6-3+2y<5.

2. Separate terms with y and without y in different sides of inequality:

2y<5-1.6+3,

2y<6.4.

3. Divide this inequality by 2:

y<3.2

4. The greatest integer that satisfies this inequality is 3.

Answer: 3.

6 0

3 years ago

Consider two vectors A and B A=14i and B= -4i+8j

DaniilM [7]

Assuming $\mathbf a,\mathbf b\in\mathbb R^3$ , you have

$\mathbf a\cdot\mathbf b=(14)(-4)+(0)(8)=-56$

so

$((\mathbf a\cdot\mathbf b)\,\mathbf i)\cdot\mathbf a=(-56\,\mathbf i)\cdot(14\,\mathbf i)=(-56)(14)=-784$

Next,

$\mathbf a+\mathbf b=(14\,\mathbf i)+(-4\,\mathbf i+8\,\mathbf j)=10\,\mathbf i+8\,\mathbf j$

Then

$(\mathbf a+\mathbf b)\times\mathbf b=\begin{vmatrix}\mathbf i&\mathbf j&\mathbf k\\10&8&0\\-4&8&0\end{vmatrix}=112\,\mathbf k$

$((\mathbf a+\mathbf b)\times\mathbf b)\times\mathbf k=\begin{vmatrix}\mathbf i&\mathbf j&\mathbf k\\0&0&112\\0&0&1\end{vmatrix}=\mathbf 0$