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3 years ago

Opdojwjdojodwojwdojwdjdowojdojojddowjdojdwojdojwdo

Mathematics

2 answers:

s344n2d4d5 [400]3 years ago

8 0

Answer:

elflaerkhg/lndfklnds/lfknas.dgsnf.raekjhg.re;ihnvdltnh45

Step-by-step explanation:

yeah <3

MaRussiya [10]3 years ago

3 0

Answer:

I provided an attachment

Step-by-step explanation:

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Evaluate the expression if x=2 and y=4 4x + 3y 1. 13 2. 15 3. 20 4. 21

zimovet [89]

4(2) + 3(4) =

8 + 12 =

and your final answer is 20

8 0

3 years ago

What is the area shaded in the region?

solong [7]

Answer:

Option C

The Area Of Shaded Region is 50.2 cm²

Step-by-step explanation:

For A given Circle,

For Small (Non-Shaded) Circle:

Diameter (d) = 6 cm

Radius = d ÷ 2 = 6/2 = 3 cm

For Area of Small (Non-Shaded) Circle

<h3>Formula:</h3>

A = πr²

A = 3.14 × (3)² cm

A = 3.14 × 9 cm

A = 28.26 cm²

For Area of Big (Shaded) Circle:

Diameter (d) = 10 cm

Radius = d ÷ 2 = 10/2 = 5 cm

For Area of Small (Non-Shaded) Circle

<h3>Formula:</h3>

A = πr²

A = 3.14 × (5)² cm

A = 3.14 × 25 cm

A = 78.5 cm²

For The Area Of Shaded Region

Area of Big Circle - Area of Small Circle

78.5 cm² - 28.26 cm² = 50.24 cm²

Thus, The Area Of Shaded Region is

50.2 cm²

-TheUnknownScientist

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3 years ago

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How do you solve the function 5x-4=-2(3x+2)

NemiM [27]

5x-4=-2(3x+2)
5x-4= -6x-4
5x+6x= -4+4
11x=0
X=0/11

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4 years ago

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The number of messages that arrive at a Web site is a Poisson distributed random variable with a mean of 5 messages per hour. Ro

BaLLatris [955]

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Step-by-step explanation:

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3 years ago

Jacob solves the system of equations by forming a matrix equation.

Lyrx [107]

Simultaneous equations can be solved using inverse matrix operation.

The complete steps of Jacob's solution are:

$\left[\begin{array}{cc}4&1\\-2&3\end{array}\right]^{-1} \cdot \left[\begin{array}{cc}4&1\\-2&3\end{array}\right]\left[\begin{array}{c}x&y\end{array}\right] = \frac{1}{14}\left[\begin{array}{cc}3&-1\\2&4\end{array}\right] \cdot \left[\begin{array}{c}2&-22\end{array}\right]$

$\left[\begin{array}{c}x&y\end{array}\right] = \left[\begin{array}{cc}4&1\\-2&3\end{array}\right] \cdot \left[\begin{array}{c}2&-22\end{array}\right]$

$\left[\begin{array}{c}x&y\end{array}\right] = \frac{1}{14} \left[\begin{array}{c}28&-84\end{array}\right]$

$\left[\begin{array}{c}x&y\end{array}\right] = \left[\begin{array}{c}2&-6\end{array}\right]$

We have:

$4x + y = 2$

$-2x + 4y = -22$

Calculate the determinant of $\left[\begin{array}{cc}4&1\\-2&3\end{array}\right]$

$|A| = 4 \times 3 -1 \times -2$

$|A| = 12 +2$

$|A| = 14$

So, the inverse matrix becomes

$A = \frac{1}{14}\left[\begin{array}{cc}4&1\\-2&3\end{array}\right]$

Replace the first column with $\left[\begin{array}{c}2&-22\end{array}\right]$ to calculate the value of x

$x = \frac{1}{14}\left[\begin{array}{cc}2&1\\-22&3\end{array}\right]$

So, we have:

$x = \frac{1}{14}(2 \times 3 - 1 \times -22)$

$x = \frac{1}{14}(6 +22)$

$x = \frac{1}{14}(28)$

$x = 2$

Replace the second column with $\left[\begin{array}{c}2&-22\end{array}\right]$ to calculate the value of y

$y = \frac{1}{14}\left[\begin{array}{cc}4&2\\-2&-22\end{array}\right]$

So, we have:

$y = \frac{1}{14}(4 \times -22 - 2 \times -2)$

$y = \frac{1}{14}(-88 +4)$

$y = \frac{1}{14}(-84)$

$y = -6$

Hence, the complete process is:

$\left[\begin{array}{c}x&y\end{array}\right] = \left[\begin{array}{cc}4&1\\-2&3\end{array}\right] \cdot \left[\begin{array}{c}2&-22\end{array}\right]$

$\left[\begin{array}{c}x&y\end{array}\right] = \frac{1}{14} \left[\begin{array}{c}28&-84\end{array}\right]$

$\left[\begin{array}{c}x&y\end{array}\right] = \left[\begin{array}{c}2&-6\end{array}\right]$