All categories

3 years ago

On a vacation, DeVante and his family traveled 382 miles at 64 mph. How long did the trip take? (Round to the nearest hundredth)

1 answer:

3 0

Okay. I'm sure you are familiar with the DST triangle, yes? It will help you solve this problem. Distance/speed = time. 382/64 = 5.96875. Round to the nearest hundredth to get 5.97 hours.

You might be interested in

Todd Has 3/4 of a pizza. Mary has 6/10 of a pizza. How much do they have together?

Naily [24]

Answer:

They have 1 7/20 pizza together.

3 0

3 years ago

Read 2 more answers

Twice the difference of a number and 6 is at least 24

gregori [183]

Answer:

what about 4

Step-by-step explanation:

6 0

3 years ago

Help me please I have the faintest clue

djverab [1.8K]

Answer:

chord CD is 11

Step-by-step explanation:

just add 2 and 9

7 0

3 years ago

On Naomi's cell phone plan, the amount she pays each month for international text messages is proportional to the number of inte

egoroff_w [7]

A. Constant of proportionality in this proportional relationship is; k=5

B. Equation to represent this proportional relationship is : c=t/k

Step-by-step explanation:

A.Given that : the amount she pays each month for international text messages is proportional to the number of international texts she sends, then

$3.20 k = 16 ---------where k is the constant of proportionality

k= 16/3.20 =5

k=5

B. Let c be the cost of sending the texts per month and t be the number of texts sent per month , so

c=t/k

c=t/5 ---------- is the proportionality relationship.

For t=16 , c= 16/5 =$3.20

Learn More

Proportionality :brainly.com/question/11490054

Keywords: cell phone plan, month, international texts, proportional,paid

#LearnwithBrainly

4 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