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frutty [35]
3 years ago
6

is the product if 19 negative numbers and 22 positive numbers a postive number or a negative number?​

Mathematics
1 answer:
kvv77 [185]3 years ago
4 0
The answer to this question is I don’t know
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Using all my points for this. Sorry for making somebody have to do this but I have to do some other homework on the side that I
givi [52]

Answer:

I am busy but all the questions from 1 to 6 just add the measure of 2 angles which are given and then subtract it from 180, you will get the measure of the remaining angle.

7.

8.: 180/3

9. x = 5

10. i dont know

i dont know = no time....

Step-by-step explanation:

8 0
3 years ago
The roots of the quadratic equation $z^2 + bz + c = 0$ are $-7 + 2i$ and $-7 - 2i$. What is $b+c$?
Alexxx [7]

Answer:

67

Step-by-step explanation: Given the quadratic equation $z^2 + bz + c = 0$, Vieta's formulas tell us the sum of the roots is $-b$, and the product of the roots is $c$. Thus,

\[-b = (-7 + 2i) + (-7 - 2i) = -14,\]so $b = 14.$

Also,

\[c = (-7 + 2i)(-7 - 2i) = (-7)^2 - (2i)^2 = 49 + 4 = 53.\]Therefore, we have $b+c = \boxed{67}$.

There are many other solutions to this problem. You might have started with the factored form $(z - (-7 + 2i))(z - (-7 - 2i)),$ or even thought about the quadratic formula.

This is the aops answer :)

8 0
3 years ago
What is wrong with the following equation: 8 + (6/2) = 17 - 5
Sidana [21]
8+6/2=17-5

We can start by reducing the fraction using basic division.

Remember, numerator divided by denominator.

6 \ 2 = 3.

Rewrite the equation-

8+3=17-5

It doesn’t matter which side of the equal sign you start with, but in this case I started with the right hand.

8+3= 11

Rewrite-

11=17-5

Now subtract.


17-5=12

Write the equality.

11=12.

This is false.

11≠12.
6 0
3 years ago
Need help due tonight!!
blagie [28]

Answer:

h = 80

Step-by-step explanation:

Remark

The general Volume formula for a pyramid is V = 1/3 * B * h

Givens

base edge = 3 m

V = 240 m^2

Solution

V = 1/3 * B * h

B = s^2

B = base edge^2

B = 3^2

B = 9

240 = 1/3  * 9 * h             multiply by 3

240 * 3 = 9 h                   Divide by 9

80  = h

8 0
3 years ago
Read 2 more answers
Find the counterclockwise circulation and outward flux of the field F=7xyi+5y^2j around and over the boundary of the region C en
dezoksy [38]

Split up the boundary of <em>C</em> (which I denote ∂<em>C</em> throughout) into the parabolic segment from (1, 1) to (0, 0) (the part corresponding to <em>y</em> = <em>x</em> ²), and the line segment from (1, 1) to (0, 0) (the part of ∂<em>C</em> on the line <em>y</em> = <em>x</em>).

Parameterize these pieces respectively by

<em>r</em><em>(t)</em> = <em>x(t)</em> <em>i</em> + <em>y(t)</em> <em>j</em> = <em>t</em> <em>i</em> + <em>t</em> ² <em>j</em>

and

<em>s</em><em>(t)</em> = <em>x(t)</em> <em>i</em> + <em>y(t)</em> <em>j</em> = (1 - <em>t</em> ) <em>i</em> + (1 - <em>t</em> ) <em>j</em>

both with 0 ≤ <em>t</em> ≤ 1.

The circulation of <em>F</em> around ∂<em>C</em> is given by the line integral with respect to arc length,

\displaystyle \int_{\partial C}\mathbf F\cdot\mathbf T \,\mathrm ds

where <em>T</em> denotes the <em>tangent</em> vector to ∂<em>C</em>. Split up the integral over each piece of ∂<em>C</em> :

• on the parabolic segment, we have

<em>T</em> = d<em>r</em>/d<em>t</em> = <em>i</em> + 2<em>t</em> <em>j</em>

• on the line segment,

<em>T</em> = d<em>s</em>/d<em>t</em> = -<em>i</em> - <em>j</em>

Then the circulation is

\displaystyle \int_{\partial C}\mathbf F\cdot\mathbf T\,\mathrm ds = \int_0^1 (7t^3\,\mathbf i+5t^4\,\mathbf j)\cdot(\mathbf i+2t\,\mathbf j)\,\mathrm dt + \int_0^1 (7(1-t)^2\,\mathbf i+5(1-t)^2\,\mathbf j)\cdot(-\mathbf i-\mathbf j)\,\mathrm dt \\\\ = \int_0^1 (7t^3+10t^5)\,\mathrm dt - 12 \int_0^1 (1-t)^2\,\mathrm dt =\boxed{-\frac7{12}}

Alternatively, we can use Green's theorem to compute the circulation, as

\displaystyle\int_{\partial C}\mathbf F\cdot\mathbf T\,\mathrm ds = \iint_C\frac{\partial(5y^2)}{\partial x} - \frac{\partial(7xy)}{\partial y}\,\mathrm dx\,\mathrm dy \\\\ = -7\int_0^1\int_{x^2}^x x\,\mathrm dx \\\\ = -7\int_0^1 xy\bigg|_{y=x^2}^{y=x}\,\mathrm dx \\\\ =-7\int_0^1(x^2-x^3)\,\mathrm dx = -\frac7{12}

The flux of <em>F</em> across ∂<em>C</em> is

\displaystyle \int_{\partial C}\mathbf F\cdot\mathbf N \,\mathrm ds

where <em>N</em> is the <em>normal</em> vector to ∂<em>C</em>. While <em>T</em> = <em>x'(t)</em> <em>i</em> + <em>y'(t)</em> <em>j</em>, the normal vector is <em>N</em> = <em>y'(t)</em> <em>i</em> - <em>x'(t)</em> <em>j</em>.

• on the parabolic segment,

<em>N</em> = 2<em>t</em> <em>i</em> - <em>j</em>

• on the line segment,

<em>N</em> = - <em>i</em> + <em>j</em>

So the flux is

\displaystyle \int_{\partial C}\mathbf F\cdot\mathbf N\,\mathrm ds = \int_0^1 (7t^3\,\mathbf i+5t^4\,\mathbf j)\cdot(2t\,\mathbf i-\mathbf j)\,\mathrm dt + \int_0^1 (7(1-t)^2\,\mathbf i+5(1-t)^2\,\mathbf j)\cdot(-\mathbf i+\mathbf j)\,\mathrm dt \\\\ = \int_0^1 (14t^4-5t^4)\,\mathrm dt - 2 \int_0^1 (1-t)^2\,\mathrm dt =\boxed{\frac{17}{15}}

5 0
3 years ago
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