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3 years ago

Correct Answers only please!

Mathematics

1 answer:

elixir [45]3 years ago

5 0

Answer:

Step-by-step explanation:

so basically batting average is hits with no strikeouts or catches. add all the hits 50+10+5+11 which is 76. take 76/260 (76 hits out of 260 tries) to get .29

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Use the given inverse to solve the system of equations. left brace Start 3 By 3 Matrix 1st Row 1st Column x minus y plus z 2nd C

natka813 [3]

The interpretation of the given question is as follows:

Use the given inverse to solve the system of equations

$x- y - z = -6 \\ \\ 2y + z = -6 \\ \\ 3x -8 y = - \dfrac{1}{2}$

The inverse of $\left[\begin{array}{ccc}1&-1&1\\0&2&1\\3&-8&0\end{array}\right] is \left[\begin{array}{ccc}-8&8&3\\-3&3&1\\6&-5&-2\end{array}\right]$

x =

y =

z =

Answer:

x = - 1.5

y = - 0.5

z = - 5

Step-by-step explanation:

Using the correlation of inverse of matrix AX = B to solve the question above;

AX = B

⇒ A⁻¹(AX) = A⁻¹ B

X = A⁻¹ B

So ;

X = A⁻¹ B

$\left[\begin{array}{c}x\\y\\z\end{array}\right] =$ $\left[\begin{array}{ccc}-8&8&3\\-3&3&1\\6&-5&-2\end{array}\right] =$ $\left[\begin{array}{ccc}-6\\ -6\\- \dfrac{1}{2}\end{array}\right]$

$\left[\begin{array}{c}x\\y\\z\end{array}\right] =\left[\begin{array}{ccc}(-8*-6)+(8*-6)+(3*-\dfrac{1}{2})\\(-3*-6)+(3*-6)+(1*-\dfrac{1}{2})\\(6*-6)+(5*-6)+(-2* - \dfrac{1}{2})\end{array}\right]$

$\left[\begin{array}{c}x\\y\\z\end{array}\right] =\left[\begin{array}{ccc}(48)+(-48)+(\dfrac{-3}{2})\\(18)+(-18)+(\dfrac{-1}{2})\\(-36)+(30)+(1)\end{array}\right]$

$\left[\begin{array}{c}x\\y\\z\end{array}\right] =\left[\begin{array}{ccc}(\dfrac{-3}{2})\\(\dfrac{-1}{2})\\(-5)\end{array}\right]$

$\left[\begin{array}{c}x\\y\\z\end{array}\right] =\left[\begin{array}{ccc}-1.5\\-0.5\\ -5\end{array}\right]$

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3 years ago

What is the following product?

devlian [24]

Answer: Last Option

$4x^5\sqrt[3]{3x}$

Step-by-step explanation:

To make the product of these expressions you must use the property of multiplication of roots:

$\sqrt[n]{x^m}*\sqrt[n]{x^b} = \sqrt[n]{x^{m+b}}$

we also know that:

$\sqrt[3]{x^3} = x$

$\sqrt[3]{16x^7}*\sqrt[3]{12x^9}\\\\\sqrt[3]{16x^3x^3x}*\sqrt[3]{12(x^3)^3}\\\\x^2\sqrt[3]{16x}*x^3\sqrt[3]{12}\\\\x^5\sqrt[3]{16x*12}\\\\x^5\sqrt[3]{2^4x*2^2*3}\\\\x^5\sqrt[3]{2^6x*3}\\\\4x^5\sqrt[3]{3x}$

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lions [1.4K]

The two negetive points of the block add thise to you answer

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Paraphin [41]

Answer:

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Step-by-step explanation:

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Step-by-step explanation:

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