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RSB [31]
3 years ago
12

Item 3

Mathematics
2 answers:
Westkost [7]3 years ago
4 0

Answer:

a = -9

Step-by-step explanation:

a+(-4)=-13

add 4 to each side

a-4+4 = -13+4

a = -9

choli [55]3 years ago
4 0

Answer:

a = - 9

Step-by-step explanation:

a  + (-4) = -13

Add  4 to both sides

a + (-4) + 4 = -13 + 4

       a + 0 = -9

        a = -9

When two numbers have different sign, subtract and the result will have the sign of the bigger number

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Step-by-step explanation:

The number system in most common use today is the Arabic system. It was first developed by the Hindus and was used as early as the 3rd century BC. The introduction of the symbol 0, used to indicate the positional value of digits was very important.

6 0
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GIVING BRAINLIEST FOR CORRECT ANSWER!!!!
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Answer:

All systems have 2 real solutions.

Step-by-step explanation:

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The scatterplot below displays a set of bivariate data along with its least-squares regression line. A scatterplot has horizonta
Llana [10]

Answer:

<em><u>All three choices would occur, and are all correct.</u></em>

Step-by-step explanation:

The y-intercept of the least-squares regression line would decrease.

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The correlation coefficient (r) would get closer to 1.

glad to help :)

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3 years ago
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dimulka [17.4K]

sqrt(1610) = 40.12 = -40.2

sqrt(680) = 36.08 = -36.08

sqrt(410) = 20.25 = -20.25

sqrt(27) = 5.20 = -5.20

 

is that supposed to be 025 or negative 25?

 the way the question is written

 the answer is C & D

 


7 0
3 years ago
Use the given transformation x=4u, y=3v to evaluate the integral. ∬r4x2 da, where r is the region bounded by the ellipse x216 y2
exis [7]

The Jacobian for this transformation is

J = \begin{bmatrix} x_u & x_v \\ y_u & y_v \end{bmatrix} = \begin{bmatrix} 4 & 0 \\ 0 & 3 \end{bmatrix}

with determinant |J| = 12, hence the area element becomes

dA = dx\,dy = 12 \, du\,dv

Then the integral becomes

\displaystyle \iint_{R'} 4x^2 \, dA = 768 \iint_R u^2 \, du \, dv

where R' is the unit circle,

\dfrac{x^2}{16} + \dfrac{y^2}9 = \dfrac{(4u^2)}{16} + \dfrac{(3v)^2}9 = u^2 + v^2 = 1

so that

\displaystyle 768 \iint_R u^2 \, du \, dv = 768 \int_{-1}^1 \int_{-\sqrt{1-v^2}}^{\sqrt{1-v^2}} u^2 \, du \, dv

Now you could evaluate the integral as-is, but it's really much easier to do if we convert to polar coordinates.

\begin{cases} u = r\cos(\theta) \\ v = r\sin(\theta) \\ u^2+v^2 = r^2\\ du\,dv = r\,dr\,d\theta\end{cases}

Then

\displaystyle 768 \int_{-1}^1 \int_{-\sqrt{1-v^2}}^{\sqrt{1-v^2}} u^2\,du\,dv = 768 \int_0^{2\pi} \int_0^1 (r\cos(\theta))^2 r\,dr\,d\theta \\\\ ~~~~~~~~~~~~ = 768 \left(\int_0^{2\pi} \cos^2(\theta)\,d\theta\right) \left(\int_0^1 r^3\,dr\right) = \boxed{192\pi}

3 0
2 years ago
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