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3 years ago

9) True or False: Great Britain officially recognized the Confederate States of America

1 answer:

6 0

It legally recognised the belligerent status of the Confederate States of America (CSA) but never recognised it as a nation and neither signed a treaty with it nor ever exchanged ambassadors.

And i could tell that its not math... :)

Its False

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N times 2+1 equals what need help for 35 points

kupik [55]

Answer:

6846

Step-by-step explanation:

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3 years ago

Find the solution of y = -x - 3 for x = -2.

Neko [114]

Answer:

(-2,-1)

Step-by-step explanation:

y = -x - 3

you replace the x with the -2

y = -(-2) - 3

y=2-3

y=-1

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3 years ago

Solve for the value of z: 90/15=6/z help asap !!

Anuta_ua [19.1K]

Answer:

Step-by-step explanation:

90/15=6/z

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3 years ago

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Help math question derivative!

atroni [7]

Let $f(x)=\sec^{-1}x$ . Then $\sec f(x)=x$ , and differentiating both sides with respect to $x$ gives

$(\sec f(x))'=\sec f(x)\tan f(x)\,f'(x)=1$
$f'(x)=\dfrac1{\sec f(x)\tan f(x)}$

Now, when $x=\sqrt2$ , you get

$(\sec^{-1})'(\sqrt2)=f'(\sqrt2)=\dfrac1{\sec\left(\sec^{-1}\sqrt2\right)\tan\left(\sec^{-1}\sqrt2\right)}$

You have $\sec^{-1}\sqrt2=\dfrac\pi4$ , so $\sec\left(\sec^{-1}\sqrt2\right)=\sqrt2$ and $\tan\left(\sec^{-1}\sqrt2\right)=1$ . So $(\sec^{-1})'(\sqrt2)=\dfrac1{\sqrt2\times1}=\dfrac1{\sqrt2}$

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3 years ago

What is the relationship among proportional relationships, lines, rates of change, and slope?

Genrish500 [490]

For each of the following relationships, graph the proportional relationship between the two quantities, write the equation representing the relationship, and describe how the unit rate, or slope is represented on the graph.

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3 years ago