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Tcecarenko [31]
4 years ago
15

How does 31÷5 look using Long division

Mathematics
1 answer:
Pavel [41]4 years ago
3 0
I could only figure out the partial quotient way im sorry but i think this could help

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The terminal side of an angle in standard position passes through P(15, –8). What is the value of sin theta?
MA_775_DIABLO [31]

Check the picture below.

3 0
3 years ago
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Find the area of a circle that has a circumference of 175.84 inches
marissa [1.9K]
2460.51

A = c^2/4pi
175.84^2/4pi

A≈ 2460.51
7 0
4 years ago
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Point R is on line segment QS. Given RS= 5 and QS<br> 5 and QS = 16, determine the length<br> QR.
amid [387]

The value of QR is 11

<h3>How to determine the length of QR?</h3>

The given parameters are:

RS = 5

QS = 16

Because R is on QS, then we have:

QS = QR + RS

This gives

16 = 5 + QR

Evaluate

QR = 11

Hence, the value of QR is 11


Read more about lengths at:

brainly.com/question/17297081

#SPJ1

8 0
2 years ago
Determine all real numbers a$ such that the inequality $ |x^2 2ax 3a|\le2$ has exactly one solution in x$.
vovikov84 [41]

Answer:

x=-1, S=\{x\:\in\mathbb{R}:x=-1\}

Step-by-step explanation:

For the sake of clarity, assuming you meant:

|x^{2}+2x+3|\le2

1) Absolute Value or Modulus functions has one property that assures us that:

If |a|

2) Solving for x

I) x^{2}+2x+3\geqslant -2\Rightarrow x^{2}+2x+5\geqslant 0 \Rightarrow x\geqslant \frac{-2\pm \sqrt{2^{2}-4*1*5}}{2} x'\geqslant \frac{-2+4i }{2} \:and\:x''\geqslant \frac{-2-4i }{2}

Not defined in Real Set of Numbers

Evaluating x^{2}+2x+3\leqslant 2:

x^{2}+2x+3\leqslant 2\\x^{2}+2x+1\leqslant 0\\(x+1)(x+1)\leqslant 0 \Rightarrow S=\{-1\}

3) So, since for the first case the Discriminant Δ <0, then the solutions presented for x^{2}+2x+3\geqslant -2\: \in \mathbb{C}. The only solution in the Real Set for the inequality |x^{2}+2x+3|\le2 is S=\{-1\}, i.e. x=-1.

8 0
3 years ago
Following are measurements of soil concentrations (in mg/kg) of chromium (Cr) and nickel (Ni) at 20 sites in the area of Clevela
WINSTONCH [101]

Answer:

a) see attached

b) see attached

Step-by-step explanation:

There is existence of outlier in the Ni data and there is none is Cr data.

########################################

# You can try this out in R programming

cr = c(34, 1, 511, 2, 574, 496, 322, 424, 269, 140, 244, 252, 76, 108, 24,

38, 18, 34, 30, 191)

Ni = c(23, 22, 55, 39, 283, 34, 159, 37, 61, 34, 163, 140, 32,

23, 54, 837, 64, 354, 376, 471)

par(mfrow=c(1,2))

hist(cr, col='green')

hist(Ni, col='brown')

par(mfrow=c(1,2))

boxplot(cr, main = 'Boxplot of Cr')

boxplot(Ni, main = 'Boxplot of Ni')

boxplot(cr, Ni)

7 0
3 years ago
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