I hope that we pass this semester god bless :)) calming it

What is the area of the regular hexagon with a side length of 3 cm and an apothem of 2.6 cm? A. 15.6 cm2 B. 23.4 cm,2 C. 46.8 cm

yarga [219]

Regular hexagons have all equal sides and angles
=((3cmx2.6)/2) x 6
=( 7.8/2) x6
= 3.9 x 6 The answer is B. 23.4cm
= 23.4

7 0

3 years ago

Read 2 more answers

Johney bought 3 pounds of hamburger. He wants to divide it evenly into 5 hamburger patties. How much pound(s) is each hamburger

Helga [31]

0.6 pounds because you would divide 3 by 5 and get 0.6

5 0

3 years ago

Read 2 more answers

True or False? A circle could be circumscribed about the quadrilateral below.

BlackZzzverrR [31]

Answer: false

Step-by-step explanation: the answer is false because suplimentary angled dont add up to 180

3 0

2 years ago

Read 2 more answers

Plzzzzzz help meeeee

kipiarov [429]

Answer:

Step-by-step explanation:

For A, 7x=21, it is asking, 7 multiplied by what number is 21? We can try to figure it out. Let's start with 0.

7(0)≠21

Then 1.

7(1)≠21

7(2)≠21

7(3)=21

Therefore, for question a, the answer is 3.

For question b, it's asking 6 minus what is 2.

With this one, we can say it is 4, since 6-4=2.

Now moving on to B.

a) 2x+4=16

We need to isolate x in order to find it's value.

So we can begin with subtracting 4 from both sides, getting us:

2x=12

Then, if we divide both terms by 2, we get x.

x=12/2

x=6.

b)

3 x (z/4)=9

Again, we need to isolate z.

So, we multiply both sides by 4, getting us

3z=36

then we divide by 3

z=36/3

z=12

c) 16-x=2

Here we simply subtract 16 from each side, then flip the signs to get positive x.

So, 16-x=2

-x=-14

x=14

d) 4k=20

This is essentially asking, 4 times what number is 20?

So we can divide both sides by 4 and get

k=20/4

k=5

5 0

3 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

￼I hope that we pass this semester god bless :)) calming it

I hope that we pass this semester god bless :)) calming it