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3 years ago

What are some of the diseases that Natterson-Horowitz sees in both human and animal patients?

Mathematics

1 answer:

german3 years ago

3 0

Answer:

Some the disease that Natterson-Horowitz sees in both human and animal patients are congestive heart failure, brain tumors, leukemia, diabetes, arthritis and breast cancer.

Step-by-step explanation:

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. What is our rule about how heat is transferred?

V125BC [204]

Answer:

hope this helps

Step-by-step explanation:

Heat can be transferred from one place to another by three methods: conduction in solids, convection of fluids (liquids or gases), and radiation through anything that will allow radiation to pass. The method used to transfer heat is usually the one that is the most efficient.

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2 years ago

HELP PLZ!!!!!!!!!!!!!

ArbitrLikvidat [17]

I think it is C it is like a right angle

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3 years ago

Fiona has $18 to spend. She spent $4.25, including tax, to buy a notebook. She needs to save $9.75, but she wants to buy a snack

Sati [7]

18 - 4.25 = 13.75.
13.75 - 9.75 = what is left to spend, $4.00

This means she can buy no more than 8 cracker packs.

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3 years ago

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Which system of equations can you use to find the roots of the equation?<br> x3 – 10x = x2 – 6

vivado [14]

$\bf \stackrel{y}{x^3-10x}=\stackrel{y}{x^2-6}\implies \begin{cases} y=x^3-10x\\ y=x^2-6 \end{cases}$

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2 years ago

Read 2 more answers

Use a matrix equation to solve the system of linear equations. left brace Start 2 By 1 Matrix 1st Row 1st Column 2nd Row 1st Col

natali 33 [55]

Answer:

$\left[\begin{array}{ccc}3&-1&\\-1&1/2\\\end{array}\right]$

Step-by-step explanation:

The matrix system for the linear equations: x + 2y = 8, 2x + 6y = 9

$\left[\begin{array}{ccc}1&2&\\2&6\\\end{array}\right] \left[\begin{array}{ccc}x\\y\end{array}\right] = \left[\begin{array}{ccc}8\\9\end{array}\right]$

To get the coefficient of x and y, the inverse of the first matrix (let the first matrix be A) must be known.

$A^{-1}$ = (1 / determinant of A) x Adjoint of A

the determinant of A = (1 x 6) - (2 x 2) = 6 - 4 = 2

Adjoint of A = $\left[\begin{array}{ccc}6&-2&\\-2&1\\\end{array}\right]$

$A^{-1}$ = $\frac{1}{2} \left[\begin{array}{ccc}6&-2\\-2&1\end{array}\right]$ = $\left[\begin{array}{ccc}3&-1&\\-1&1/2\\\end{array}\right]$

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2 years ago