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skad [1K]
3 years ago
5

Which equation is the inverse of y = 7x2 – 10? y = StartFraction plus-or-minus StartRoot x + 10 EndRoot Over 7 EndFraction y = p

lus-or-minus StartRoot StartFraction x + 10 Over 7 EndFraction EndRoot x = plus-or-minus StartRoot StartFraction x Over 7 EndFraction + 10 EndRoot y = StartFraction plus-or-minus StartRoot x EndRoot Over 7 EndFraction plus-or-minus StartFraction StartRoot 10 EndRoot Over 7 EndFraction
Mathematics
2 answers:
lbvjy [14]3 years ago
8 0

Answer:

Letter C

y = plus-or-minus StartRoot StartFraction x + 10 Over 7 EndFraction EndRoot

Step-by-step explanation:

Just took the test.

Juli2301 [7.4K]3 years ago
7 0

Answer:

C

Step-by-step explanation:

I took the test on E2020.

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Whats the slope of the line?
TiliK225 [7]

Answer:

m = rise/run = 4/5.

Step-by-step explanation:

This line goes throught the "exact" grid intersections (-2, 0) and (3, 4).  As we move from the first point to the second, we see that x increases by 5 (this is the 'run') and y increases by 4 (this is the 'rise').  Thus, the slope of this line is m = rise/run = 4/5.

6 0
3 years ago
Derivative of tan(2x+3) using first principle
kodGreya [7K]
f(x)=\tan(2x+3)

The derivative is given by the limit

f'(x)=\displaystyle\lim_{h\to0}\frac{f(x+h)-f(x)}h

You have

\displaystyle\lim_{h\to0}\frac{\tan(2(x+h)+3)-\tan(2x+3)}h
\displaystyle\lim_{h\to0}\frac{\tan((2x+3)+2h)-\tan(2x+3)}h

Use the angle sum identity for tangent. I don't remember it off the top of my head, but I do remember the ones for (co)sine.

\tan(a+b)=\dfrac{\sin(a+b)}{\cos(a+b)}=\dfrac{\sin a\cos b+\cos a\sin b}{\cos a\cos b-\sin a\sin b}=\dfrac{\tan a+\tan b}{1-\tan a\tan b}

By this identity, you have

\tan((2x+3)+2h)=\dfrac{\tan(2x+3)+\tan2h}{1-\tan(2x+3)\tan2h}

So in the limit you get

\displaystyle\lim_{h\to0}\frac{\dfrac{\tan(2x+3)+\tan2h}{1-\tan(2x+3)\tan2h}-\tan(2x+3)}h
\displaystyle\lim_{h\to0}\frac{\tan(2x+3)+\tan2h-\tan(2x+3)(1-\tan(2x+3)\tan2h)}{h(1-\tan(2x+3)\tan2h)}
\displaystyle\lim_{h\to0}\frac{\tan2h+\tan^2(2x+3)\tan2h}{h(1-\tan(2x+3)\tan2h)}
\displaystyle\lim_{h\to0}\frac{\tan2h}h\times\lim_{h\to0}\frac{1+\tan^2(2x+3)}{1-\tan(2x+3)\tan2h}
\displaystyle\frac12\lim_{h\to0}\frac1{\cos2h}\times\lim_{h\to0}\frac{\sin2h}{2h}\times\lim_{h\to0}\frac{\sec^2(2x+3)}{1-\tan(2x+3)\tan2h}

The first two limits are both 1, and the single term in the last limit approaches 0 as h\to0, so you're left with

f'(x)=\dfrac12\sec^2(2x+3)

which agrees with the result you get from applying the chain rule.
7 0
3 years ago
Question 16
Vadim26 [7]

Answer: 175

Step-by-step explanation: total = x + 5y dollars and dollars = 300 so

x + 5y = 300 then since we know y =  200 we can substitute that

x + 5(200) = 300 then subtract x from 200

x + 5(200 - x) = 300 now solve for x

x + 5(200) + 5(-x) = 300 we know 5 times 200 is 1000

x + 1000 - 5x = 300 then subtract 5x - x = -4x

-4x + 1000 = 300 switch around 1000 and 300

-4x = 300 - 1000 subtract 300 - 1000 = -700

-4x = -700 multiply

x = -700/(-4) then finish up and solve

x = 175

hope this helps please mark brainliest if you can

4 0
4 years ago
dianna made several cakes yesterday. each cake required 2 2/3 cups of flour. all together, she used 13 1/3 cups of flours. how m
Kay [80]

13 1/3 divided by 2 2/3

= 40/3 /8/3= 5

Hope that helps!


3 0
3 years ago
Read 2 more answers
A few probability questions. Quickly please :) I have tried to do them but seem to be unable to figure these 2 out.
Zepler [3.9K]
1.choosing a red marble since the probability would be 7/12 but the blue would be 5/12 so red is most possible
2.the possibility of a red being chosen is 2/20=1/10
4 0
3 years ago
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