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2 years ago

Perform the operation (5x^2 + 4) - (-2y^2 + 4)

Mathematics

1 answer:

KIM [24]2 years ago

6 0

5x^2+2y^2 is the answer to you problem.

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I could use some more help

kherson [118]

Answer:

$\left[\begin{array}{ccc}-30&22&8\\-4&-4&16\\30&-28&18\end{array}\right]$

Step-by-step explanation:

Given

$A = \left[\begin{array}{ccc}-1&9&2\\10&-10&2\\-5&6&-5\end{array}\right]$

$B = \left[\begin{array}{ccc}7&-1&-1\\6&-4&-3\\-10&10&-7\end{array}\right]$

Required

2A - 4B

To solve 2A - 4B, we first multiply matrix A by 2 and matrix B by 4

So, if

$A = \left[\begin{array}{ccc}-1&9&2\\10&-10&2\\-5&6&-5\end{array}\right]$

$2A = 2 *\left[\begin{array}{ccc}-1&9&2\\10&-10&2\\-5&6&-5\end{array}\right]$

$2A = \left[\begin{array}{ccc}-2&18&4\\20&-20&4\\-10&12&-10\end{array}\right]$

$B = \left[\begin{array}{ccc}7&-1&-1\\6&-4&-3\\-10&10&-7\end{array}\right]$

then

$4B = 4*\left[\begin{array}{ccc}7&-1&-1\\6&-4&-3\\-10&10&-7\end{array}\right]$

$4B = \left[\begin{array}{ccc}28&-4&-4\\24&-16&-12\\-40&40&-28\end{array}\right]$

So; 2A - 4B becomes

$\left[\begin{array}{ccc}-2&18&4\\20&-20&4\\-10&12&-10\end{array}\right] - \left[\begin{array}{ccc}28&-4&-4\\24&-16&-12\\-40&40&-28\end{array}\right]$

$\left[\begin{array}{ccc}-2-28&18-(-4)&4-(-4)\\20-24&-20-(-16)&4-(-12)\\-10-(-40)&12-40&-10-(-28)\end{array}\right]$

$\left[\begin{array}{ccc}-30&18+4&4+4\\20-24&-20+16&4+12\\-10+40&12-40&-10+28\end{array}\right]$

$\left[\begin{array}{ccc}-30&22&8\\-4&-4&16\\30&-28&18\end{array}\right]$

Hence, 2A - 4B is equivalent to

$\left[\begin{array}{ccc}-30&22&8\\-4&-4&16\\30&-28&18\end{array}\right]$

3 0

2 years ago

Name five fractions whose values are between 1/2and<br> 7/8

pantera1 [17]

Answer:

9/16 10/16 11/16 12/16 13/16

Step-by-step explanation:

7 0

2 years ago

Help me i need the answer

Korolek [52]

Answer:

4 solutions

Step-by-step explanation:

remember that y=mx+b

So 2/5x would be you m.And -3/7 is your b

6 0

1 year ago

Determine the longest interval in which the given initialvalue problemis certainto have a unique twicedifferentiable solution. D

AlekseyPX

Answer:

t_o = 3, so solution exists on (0,4).

Step-by-step explanation:

Use Theorem

Divide equation with t(t — 4).

y''+[3/(t-4)]*y'+ [4/t(t-4)]*y=2/t(t-4)

p(t)=3/t-4—> continuous on (-∞, 4) and (4,∞)

q(t) = 4/t(t-4) —> continuous on (-∞,0), (0,4) and (4, ∞)

g(t) = 2/t(t-4)—> continuous on (-∞, 0), (0,4) and (4,∞)

t_o = 3, so solution exists on (0,4).

6 0

2 years ago

PLEASE HELP ASAP I WILL MARK BRILLIANT WHOEVER ANSWERS THANKS (:

Mila [183]

The answer is 38.5 cm^2

5 0

2 years ago

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