All categories

2 years ago

4.5-17 - X 8I need help! 10 pts

Mathematics

1 answer:

sesenic [268]2 years ago

7 0

Answer:

x = -1.5625

Step-by-step explanation:

Solve for x:

-8 x - 12.5 = 0

Add 12.5` to both sides:

(12.5 - 12.5) - 8 x = 12.5

12.5 - 12.5 = 0:

-8 x = 12.5

Divide both sides of -8 x = 12.5 by -8:

(-8 x)/(-8) = 12.5/(-8)

(-8)/(-8) = 1:

x = 12.5/(-8)

12.5/(-8) = -1.5625:

Answer: x = -1.5625

You might be interested in

Transform the following linear equation to the "Aha" line. Give a brief explanation as to what kind of solution(s) you expect th

mote1985 [20]

1+2/3 this is the correct answer

3 0

2 years ago

100,203 in expanded form

Paul [167]

Answer:

100,000 + 200 + 3

Step-by-step explanation:

5 0

2 years ago

Simplify the expression (3 - 4i)(1 + 5i) - (2 - i). Show your work please! I have the answer: 21 + `12i but I need to show my wo

masha68 [24]

Answer:

21+12i i hope this will help you

Step-by-step explanation:

(3-4i)(1+5i)-(2-i)

=[3(1+5i)-4i(1+5i)]-(2-i)

=[3+15i-4i-20i²]-(2-i)

=(3+11i-20i²)-(2-i)

opening brackeys

=3+11i-20i²-2+i

=3+11i+1i-20i²-2

as we know that i²=-1

=3-20(-1)-2+12i

=3+20-2+12i

=21+12i

8 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

2 years ago

Which dilation is shown by this graph?

beks73 [17]

Answer:

1/3

Step-by-step explanation:

Take the far side original length and dilated length 3/9 divide it = 1/3 and that’s the answer

5 0

3 years ago

4.5-17 - X 8I need help! 10 pts​

4.5-17 - X 8I need help! 10 pts