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2 years ago

In a group of workers, 16 belonged to a lottery syndicate and 4 did not.

Mathematics

1 answer:

Tatiana [17]2 years ago

8 0

Answer:

<h2>HI THERE</h2>

the answer is number 1 !!

• correct me if I'm wrong

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A group of 3 friends shared 7 bars of clay equally. How much clay did each student get?

Mazyrski [523]

Answer: 2 1/3

Step-by-step explanation:

Hi, I got my answer by making 3 groups and splitting 7 into them. When you get 1 left over, you separate it into 3, so everyone gets 1/3 of a piece. Therefore the answer is 2 1/3. :)

5 0

3 years ago

Find x using the Law of Cosines. Round to the nearest tenth.

Bad White [126]

The Answer for "X" is going to be 25.63201, rounded to the nearest tenth would be 25.6.

I hope this helped :)

5 0

3 years ago

Given equation x = 6 + i, w = -1 + 5i and z = 4 - 8i, determine each of the following in the form of a + bi

Temka [501]

Answer:

i) 28 - 30i

ii) 36 + 28i

Step-by-step explanation:

i) x = 6 + i ⇒2x = 2(6 + i) = 12 + 2i

z = 4 - 8i ⇒ 4z = 4(4 - 8i) = 16 - 32i

2x + 4z = (12 + 2i) + (16 - 32i) = 28 - 30i

ii) w = -1 + 5i and z = 4 - 8i

w × z = (-1 + 5i)(4 - 8i) = -4 + 8i + 20i - 40 $i^{2}$ ⇒collect like terms

w × z = -4 + 28i - 40 $i^{2}$

∵ $i^{2}=-1$

∴w × z = -4 + 28i - 40(-1) = -4 + 28i + 40 = 36 + 28i

3 0

2 years ago

Help !!! with 14 please :))

BartSMP [9]

Since p is an obtuse angle
& that it's between two acute angles

3 0

2 years ago

The slope of the line tangent to the curve y^2 + (xy+1)^3 = 0 at (2, -1) is ...?

Lina20 [59]

$\frac{d}{dx}(y^2 + (xy+1)^3 = 0)
\\
\\\frac{d}{dx}y^2 + \frac{d}{dx}(xy+1)^3 = \frac{d}{dx}0
\\
\\\frac{dy}{dx}2y + \frac{d}{dx}(xy+1)\ *3(xy+1)^2 = 0
\\
\\\frac{dy}{dx}2y + \left(\frac{d}{dx}(xy)+\frac{d}{dx}1\right)\ * 3(xy+1)^2 = 0
\\
\\\frac{dy}{dx}2y + \left(\frac{d}{dx}xy+x\frac{d}{dx}y+\frac{dx}{dx}\right)\ * 3(xy+1)^2 = 0
\\
\\\frac{dy}{dx}2y + \left(\frac{dx}{dx}y+x\frac{dy}{dx}+\frac{dx}{dx}0\right)\ * 3(xy+1)^2 = 0
\\
\\\frac{dy}{dx}2y + \left(y+x\frac{dy}{dx}\right)\ * 3(xy+1)^2 = 0$
$\frac{dy}{dx}2y + \left(y+x\frac{dy}{dx}\right)\ * 3(xy+1)^2 = 0
\\
\\\frac{dy}{dx}2y + \left(3y+3x\frac{dy}{dx}\right) (xy+1)^2 = 0
\\
\\2y\ y' + \left(3y+3x\ y'\right) (xy+1)^2 = 0
\\
\\2y\ y' + \left(3y+3x\ y'\right) ((xy)^2+2xy+1) = 0
\\
\\2y\ y' + \left(3y+3x\ y'\right) ((xy)^2+2xy+1) = 0
\\
\\2y\ y' + 3x^2y^3 +6xy^2+3y+(3x y' +6x^2y y'+3x^2y y') = 0
\\
\\3x^2y^3 +6xy^2+3y+(3x y' +6x^2y y'+3x^2y y'+2yy' ) = 0
\\
\\y'(3x +6x^2y+3x^2y+2y ) = -3x^2y^3 -6xy^2-3y

$
$y' = \frac{-3x^2y^3 -6xy^2-3y}{(3x +6x^2y+3x^2y+2y )};\ x=2,y=-1
\\
\\y' = \frac{-3(2)^2(-1)^3 -6(2)(-1)^2-3(-1)}{(3(2) +6(2)^2(-1)+3(2)^2(-1)+2(-1) )}
\\
\\y' = \frac{-3(4)(-1) -6(2)(1)+3}{(6 +6(4)(-1)+3(4)(-1)-2 )}
\\
\\y' = \frac{12 -12+3}{(6 -24-12-2 )}
\\
\\y' = \frac{3}{( -32 )}$

7 0

3 years ago