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2 years ago

14

Relationship B has a greater rate than Relationship A This Graph Represents Relationship A Which equations could represent relat

ionship b

1 answer:

Anarel [89]2 years ago

5 0

Answer:

Step-by-step explanation:

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4. As shown in Figure 5.111 to the right if the

Juli2301 [7.4K]

Answer:

Area of the shaded part is 3.14 square unit.

Step-by-step explanation:

Area of the shaded part = Area of large semicircle - (Area of two small semi circles)

Area of large semicircle with center O = $\frac{1}{2}(\pi r^2)$

= $\frac{1}{2}\pi (2)^{2}$

= 2π

Area of semicircle with center O' = $\frac{1}{2}\pi (1)^2$

= $\frac{\pi }{2}$

Area of semicircle with center O" = $\frac{\pi }{2}$

Now substitute these values in the formula,

Area of shaded part = $2\pi - (\frac{\pi }{2}+\frac{\pi }{2})$

= $\pi$

= 3.14 square unit

Area of the shaded part is 3.14 square unit.

7 0

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

6 0

3 years ago

Please HELP I will give Brainliest!

ankoles [38]

The answer is the first page, first graph

5 0

2 years ago

Montano1993 [528]

Taking this example into account, we can see that setting the first value equal to 1, we obtain that $F(x)=0.5x+1=4$ and $x=6$ . Using this information, we find that $F(x+1)=0.5(x+1)+1=0.5(6+1)+1=4.5$ . It shows that when x is positive, the succussive terms are increasing.

Referring to that finding, if we set initial value less than zero, which means that we are solving 0.5x+1<0 and taking a number in the interval of the solution, which means x ∈ (- ∞, -20). Setting x=-19, we find that $F(x)=0.5x+1=-19$ and x=-40. In the next iteration, $F(x+1)=0.5(x+1)+1=0.5(1-40)+1=-18.5$ . In the next iteration, $F(x+2)=0.5(x+2)+1=0.5(2-40)+1=-18$ . By this way, we find that even if the initial value is less than zero, value of the successive iterations is increasing.

Using the function $g(x)=-x+2$ and taking the initial value equal to 4, we find that $g(x)=-x+2=4$ and x=-2. In the next iteration, $g(x+1)=-(x+1)+2=-(-2+1)+2=3$ . If we continue the iterations we'll see that they are decreasing.
Setting the initial value equal to 2, we find that $g(x)=-x+2=2$ and x=0. The next iteration is $g(x+1)=-(x+1)+2=1$ . In this case, the interations are also decreasing.
If we set the initial value equal to 1, we find that $g(x)=-x+2=1$ and x=1. In the next iteration, $g(x+1)=-(x+1)+2=0$ and the iterations are decreasing.

8 0

3 years ago

On december 1, jonas jewelry wholesale center started with an inventory of 1,000 watches. shipments of 168, 231, and 411 were se

denis-greek [22]

Answer: Third option is correct.

Step-by-step explanation:

Since we have given that

Total watches in an inventory = 1000

Number of shipments sent out = 168, 231, 411

So, Remaining inventory of watches would be :

$1000-(168+231+411)\\\\=1000-810\\\\=190$

New receipts = 21, 145 and 11

So, Number of watches on hand becomes :

$190+(21+145+11)\\\\=367$

Hence, Third option is correct.

5 0

3 years ago