Select the equivalent expression.

Kate sold a sailboat at a loss of 20%. If the selling price was $64, what was the original cost?

lapo4ka [179]

So she lost 20% of the original selling price.

$64 is 80% of the original price.

64/80=0.8

0.8·100=80 dollars. HOPE THIS HELPS!!!

8 0

3 years ago

Read 2 more answers

Solve the following differential equations or initial value problems. In part (a), leave your answer in implicit form. For parts

shepuryov [24]

Answer:

(a) (y^5)/5 + y^4 = (t^3)/3 + 7t + C

(b) y = arctan(t(lnt - 1) + C)

(c) y = -1/ln|0.09(t + 1)²/t|

Step-by-step explanation:

(a) dy/dt = (t^2 + 7)/(y^4 - 4y^3)

Separate the variables

(y^4 - 4y^3)dy = (t^2 + 7)dt

Integrate both sides

(y^5)/5 + y^4 = (t^3)/3 + 7t + C

(b) dy/dt = (cos²y)lnt

Separate the variables

dy/cos²y = lnt dt

Integrate both sides

tany = t(lnt - 1) + C

y = arctan(t(lnt - 1) + C)

(c) (t² + t) dy/dt + y² = ty², y(1) = -1

(t² + t) dy/dt = ty² - y²

(t² + t) dy/dt = y²(t - 1)

(t² + t)/(t - 1)dy/dt = y²

Separating the variables

(t - 1)dt/(t² + t) = dy/y²

tdt/(t² + t) - dt/(t² + t) = dy/y²

dt/(t + 1) - dt/(t(t + 1)) = dy/y²

dt/(t + 1) - dt/t + dt/(t + 1) = dy/y²

Integrate both sides

ln(t + 1) - lnt + ln(t + 1) + lnC = -1/y

2ln(t + 1) - lnt + lnC = -1/y

ln|C(t + 1)²/t| = -1/y

y = -1/ln|C(t + 1)²/t|

Apply y(1) = -1

-1 = ln|C(1 + 1)²/1|

-1 = ln(4C)

4C = e^(-1)

C = (1/4)e^(-1) ≈ 0.09

y = -1/ln|0.09(t + 1)²/t|

8 0

3 years ago

I need help with this pls!!

SSSSS [86.1K]

Answer : (12,4)
that is the max point

8 0

3 years ago

Graph the image of the given triangle after the transformation with the rule (x, y)→(x−6, y−4) .

zaharov [31]

Answer: See the figure attached.

Step-by-step explanation:

Observe the figure attached. The vertices of the triangle ABC have the following coordinates:

A(2,5); B(6,2) and C(3,-5)

To graph the image (in this case we can identify it as A'B'C'), you must use the rule $(x,y)$ → $(x-6,y-4)$ .

Then, you have to subtract 6 units from the x-coordinate of each vertex and subtract 4 units from the y-coordinate of each vertex.

Therefore, you get:

Vertex A'→ $(2-6,5-4)=(-4,1)$

Vertex B'→ $(6-6,2-4)=(0,-2)$

Vertex C'→ $(3-6,-5-4)=(-3,-9)$

Having the vertices of the image A'B'C', you can graph it (See the figure attached).

7 0

3 years ago

Find the length of DE.

Amanda [17]

I think same as BC because it seem like its a reflected problem and x would be 23 and if 23-5 that would be 18 I think.

4 0

3 years ago