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valentina_108 [34]
3 years ago
8

What are the solutions to the equation

Mathematics
1 answer:
sineoko [7]3 years ago
7 0

Answer:

3i , -3i

Step-by-step explanation:

This is correct, I don't see what's wrong.

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A scientist collected four water samples from local ponds. Each sample was the same size and he collected 2.6 liters in all. Wha
Temka [501]

Answer: 0.65 Liters

Steps:

Divide 2.6/4 to get 0.65 Liters

3 0
3 years ago
Please I need help! answer below.<br> Is A. B. C. D.?
Maksim231197 [3]

Answer:

it is B

Step-by-step explanation:

6 0
3 years ago
4/9x + 1/5x =58 need help solving this
iogann1982 [59]

Answer:

x=90

Step-by-step explanation:

First, it would help to make it so that the x's are not next to fractions. To do this, you can multiply all terms in the equation by 45. 45 is the closest common factor of 9 and 5, the denominators of the fractions. The answer would be;

20x+9x=2610

Then solve.

29x=2610

x=90

6 0
3 years ago
Elena receives ​$131 per year in simple interest from three investments totaling ​$3000. Part is invested at​ 3%, part at​ 4% an
ioda

Answer:

Elena invested $ 1,700 at 5%, $ 700 at 4%, and $ 600 at 3%.

Step-by-step explanation:

Given that Elena receives $ 131 per year in simple interest from three investments totaling $ 3000, and part is invested at 3%, part at 4% and part at 5%, and there is $ 1000 more invested at 5% than at 4%, to find the amount invested at each rate, the following calculations must be performed:

1500 x 0.05 + 500 x 0.04 + 1000 x 0.03 = 75 + 20 + 30 = 125

1600 x 0.05 + 600 x 0.04 + 800 x 0.03 = 80 + 24 + 24 = 128

1700 x 0.05 + 700 x 0.04 + 600 x 0.03 = 85 + 28 + 18 = 131

Therefore, Elena invested $ 1,700 at 5%, $ 700 at 4%, and $ 600 at 3%

7 0
3 years ago
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}
5 0
3 years ago
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