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2 years ago

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A map has the scale 2 inches equals 190 miles. The distance from Springfield to Glenview is 14 inches. How many actual miles sep

arate Springfield from Glenview? A. 140 B. 1,330 C. 2,800 D. 38

1 answer:

Kay [80]2 years ago

8 0

Lord help me hdbdbdbdbfbbfbfbrbbdbdbdbdnrjbdbdbdbdbdbdbhdhrbtbrbdbdbbrrbrgrggrgrhhrhr

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What are the correct answers to this geometry question?

Maru [420]

Answer:

Ti = 5

AT = 6

Kt = 10

Ai = 11

ke = 19

ae = 15

Step-by-step explanation:

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3 years ago

-1/2 divided by X=7/3

marshall27 [118]

The exact form is x= -3/14

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3 years ago

Find cotθ, cosθ, and secθ, where θ is the angle shown in the figure.

Darina [25.2K]

Answer:

$\cos( \theta) = \frac{5}{8} \\ {8}^{2} = {5}^{2} + {opp}^{2} \\ {opp}^{2} = 64 - 25 = 39 \\ opp = \sqrt{39} \\ \sin(\theta) = \frac{\sqrt{39}}{8} \\ \tan(\theta) = \frac{ \sqrt{39} }{5} \\ \cot(\theta) = \frac{1}{\tan(\theta)} = \frac{1}{\frac{ \sqrt{39} }{5}} \\ \cot(\theta) = \frac{5}{ \sqrt{39} } \\ \csc( \theta) = \frac{1}{\sin(\theta)} = \frac{1}{\frac{\sqrt{39}}{8}} \\ \csc( \theta) = \frac{8}{ \sqrt{39} } \\ \sec( \theta) = \frac{1}{\cos( \theta) } = \frac{1}{ \frac{5}{8} } \\ \sec( \theta) = \frac{8}{5}$

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2 years ago

Solve for XX. Assume XX is a 2×22×2 matrix and II denotes the 2×22×2 identity matrix. Do not use decimal numbers in your answer.

sveticcg [70]

The question is incomplete. The complete question is as follows:

Solve for X. Assume X is a 2x2 matrix and I denotes the 2x2 identity matrix. Do not use decimal numbers in your answer. If there are fractions, leave them unevaluated.

$\left[\begin{array}{cc}2&8\\-6&-9\end{array}\right]$ · X· $\left[\begin{array}{ccc}9&-3\\7&-6\end{array}\right]$ =<em>I</em>.

First, we have to identify the matrix <em>I. </em>As it was said, the matrix is the identiy matrix, which means

<em>I</em> = $\left[\begin{array}{ccc}1&0\\0&1\end{array}\right]$

So, $\left[\begin{array}{cc}2&8\\-6&-9\end{array}\right]$ · X· $\left[\begin{array}{ccc}9&-3\\7&-6\end{array}\right]$ = $\left[\begin{array}{ccc}1&0\\0&1\end{array}\right]$

Isolating the X, we have

X· $\left[\begin{array}{ccc}9&-3\\7&-6\end{array}\right]$ = $\left[\begin{array}{cc}2&8\\-6&-9\end{array}\right]$ - $\left[\begin{array}{ccc}1&0\\0&1\end{array}\right]$

Resolving:

X· $\left[\begin{array}{ccc}9&-3\\7&-6\end{array}\right]$ = $\left[\begin{array}{ccc}2-1&8-0\\-6-0&-9-1\end{array}\right]$

X· $\left[\begin{array}{ccc}9&-3\\7&-6\end{array}\right]$ = $\left[\begin{array}{ccc}1&8\\-6&-10\end{array}\right]$

Now, we have a problem similar to A.X=B. To solve it and because we don't divide matrices, we do X=A⁻¹·B. In this case,

X= $\left[\begin{array}{ccc}9&-3\\7&-6\end{array}\right]$ ⁻¹· $\left[\begin{array}{ccc}1&8\\-6&-10\end{array}\right]$

Now, a matrix with index -1 is called Inverse Matrix and is calculated as: A . A⁻¹ = I.

So,

$\left[\begin{array}{ccc}9&-3\\7&-6\end{array}\right]$ · $\left[\begin{array}{ccc}a&b\\c&d\end{array}\right]$ = $\left[\begin{array}{ccc}1&0\\0&1\end{array}\right]$

9a - 3b = 1

7a - 6b = 0

9c - 3d = 0

7c - 6d = 1

Resolving these equations, we have a= $\frac{2}{11}$ ; b= $\frac{7}{33}$ ; c= $\frac{-1}{11}$ and d= $\frac{-3}{11}$ . Substituting:

X= $\left[\begin{array}{ccc}\frac{2}{11} &\frac{-1}{11} \\\frac{7}{33}&\frac{-3}{11} \end{array}\right]$ · $\left[\begin{array}{ccc}1&8\\-6&-10\end{array}\right]$

Multiplying the matrices, we have

X= $\left[\begin{array}{ccc}\frac{8}{11} &\frac{26}{11} \\\frac{39}{11}&\frac{198}{11} \end{array}\right]$

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3 years ago

What is the length of the diameter of a circle inscribed in a right triangle with the length of hypotenuse c and the sum of the

jeyben [28]

Answer:

diameter = m - c

Step-by-step explanation:

In ΔABC, let ∠C be the right angle. The length of the tangents from point C to the inscribed circle are "r", the radius. Then the lengths of tangents from point A are (b-r), and those from point B have length (a-r).

The sum of the lengths of the tangents from points A and B on side "c" is ...

(b-r) +(a-r) = c

(a+b) -2r = c

Now, the problem statement defines the sum of side lengths as ...

a+b = m

and, of course, the diameter (d) is 2r, so we can rewrite the above equation as ...

m -d = c

m - c = d . . . . add d-c

The diameter of the inscribed circle is the difference between the sum of leg lengths and the hypotenuse.

5 0

3 years ago