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egoroff_w [7]
3 years ago
8

PLEASE I NEED HELP!!!!!!!!

Mathematics
1 answer:
Yanka [14]3 years ago
7 0

Answer:

1) x= 3i, x= - 3i

2) x= +i\sqrt{7}, x = - i

Step-by-step explanation:

1) x^2+9=0

(x+3i)(x-3i)=0

x= 3i, x=-3i

2) x^2-4=-11

x^2=-7

x= ±i\sqrt{7}

x= +i\sqrt{7}, x = - i

hope this helps!! :))

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The National Assessment of Educational Progress (NAEP) gave a test of basic arithmetic and the ability to apply it in everyday l
mylen [45]

Answer:

a) 2.0702

b) 2.07

Step-by-step explanation:

Given:

Sample, n = 840

standard deviation, σ = 60

a) The standard deviation of the sampling distribution, will be:

= \frac{\sigma}{\sqrt{n}}

= \frac{60}{\sqrt{840}}

= 2.0702

Standard deviation of the sampling distribution is 2.0702

b) According to the 68 part of the 68-95-99.7 rule, 68% of all values of x fall within 1 standard deviation on either side of the unknown mean μ.

Therefore, missing number will be:

1 * [\frac{\sigma}{\sqrt{n}}]

= 1 * 2.0702

= 2.0702 ≈ 2.07

6 0
3 years ago
Nancy bought 2 packs of basketball cards for 3.78 each and 4 decks of basketballs cards for 7.38 each with 2 20 bills. How much
love history [14]

Answer:

3.78*2=7.56

7.38*4=29.52

29.52+7.56=37.08

220-37.08=182.92

8 0
3 years ago
-7b&lt; -35<br> =less than or equal two
timama [110]

Answer:

b>5

=greater than or equal too

Step-by-step explanation:

divide -35 by -5, and since you are dividing by a negative, the sign flips

6 0
3 years ago
Show that the line integral is independent of path by finding a function f such that ?f = f. c 2xe?ydx (2y ? x2e?ydy, c is any p
Juli2301 [7.4K]
I'm reading this as

\displaystyle\int_C2xe^{-y}\,\mathrm dx+(2y-x^2e^{-y})\,\mathrm dy

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The value of the integral will be independent of the path if we can find a function f(x,y) that satisfies the gradient equation above.

You have

\begin{cases}\dfrac{\partial f}{\partial x}=2xe^{-y}\\\\\dfrac{\partial f}{\partial y}=2y-x^2e^{-y}\end{cases}

Integrate \dfrac{\partial f}{\partial x} with respect to x. You get

\displaystyle\int\dfrac{\partial f}{\partial x}\,\mathrm dx=\int2xe^{-y}\,\mathrm dx
f=x^2e^{-y}+g(y)

Differentiate with respect to y. You get

\dfrac{\partial f}{\partial y}=\dfrac{\partial}{\partial y}[x^2e^{-y}+g(y)]
2y-x^2e^{-y}=-x^2e^{-y}+g'(y)
2y=g'(y)

Integrate both sides with respect to y to arrive at

\displaystyle\int2y\,\mathrm dy=\int g'(y)\,\mathrm dy
y^2=g(y)+C
g(y)=y^2+C

So you have

f(x,y)=x^2e^{-y}+y^2+C

The gradient is continuous for all x,y, so the fundamental theorem of calculus applies, and so the value of the integral, regardless of the path taken, is

\displaystyle\int_C2xe^{-y}\,\mathrm dx+(2y-x^2e^{-y})\,\mathrm dy=f(4,1)-f(1,0)=\frac9e
8 0
3 years ago
Which two equations would be most appropriately solved by using the zero product property?
anzhelika [568]
For the zero product property questions it's choices 2 and 3 since they're quadratics factored and set equal to zero.

for the quadratic formula questions it would be answer choices 2 and 3 also because they're unfactored trinomial quadratic expressions
6 0
3 years ago
Read 2 more answers
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