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White raven [17]

3 years ago

14

A ship captain is mapping a trip and wants to know the distance the ship will travel over certain time intervals. Assuming that

the ship travels at a constant speed, what is its speed? (Speed uses the per).

1 answer:

olga55 [171]3 years ago

5 0

Answer: 25 miles per hour

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Identify the quadrant where each point is located. (-5,2) (5,-5) (2,-5) (-4,-5)

Tanzania [10]

(-5,2) is located in the II quadrant because the x is negative and the y is positive.

(5,-5) is located in the IV quadrant because the x is positive and the y is negative.

(2,-5) is also located in the IV quadrant because the x is positive and the y is negative.

(-4,-5) is located in the III quadrant because both x and y are negative.

6 0

3 years ago

-1/2x+12=-16 1/2 help<br> Plz I need it quick

Leokris [45]

Subtract 12 from both sides
$- \frac{1}{2} x+12-12=-16 \frac{1}{2} -12$ $- \frac{1}{2}x=-28 \frac{1}{2}$
Divide by - $\frac{1}{2}$ which means multiply by -2
$x=-28 \frac{1}{2} *-2=57$

Hope that helps

7 0

4 years ago

Forty percent of students in a class of 35 students are the girls. How many girls are in the class?

ad-work [718]

40% of 35 would be 14. Therefore 14 girls are in the class. Hope that helps!

5 0

3 years ago

Read 2 more answers

kolezko [41]

Answer:

A (-4,5) B (-3,1) C (-5,2) D (3,1) E (2,5) F(4,4)

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

Find the extreme values of f subject to both constraints. (if an answer does not exist, enter dne.) f(x, y, z) = x + 2y; x + y +

Mariana [72]

Using the method of Lagrange multipliers, we have the Lagrangian

$L(x,y,z,\lambda_1,\lambda_2)=x+2y+\lambda_1(x+y+z-4)+\lambda_2(y^2+z^2-4)$

with partial derivatives (set equal to 0)

$L_x=1+\lambda_1=0\implies\lambda_1=-1$
$L_y=2+\lambda_1+2\lambda_2y=0\implies\lambda_2y=-\dfrac12$
$L_z=\lambda_1+2\lambda_2z=0\implies\lambda_2z=\dfrac12$
$L_{\lambda_1}=x+y+z-4=0$
$L_{\lambda_2}=y^2+z^2-4=0$

From $L_y$ and $L_z$ , we find that $\lambda_2y=-\lambda_2z\implies y=-z$ . Then substituting into $L_{\lambda_2}$ , we find

$y^2+z^2=4\implies2y^2=4\implies y=\pm\sqrt2\implies z=\mp\sqrt2$

and substituting these into $L_{\lambda_1}$ , we get

$x+y+z=4\implies x=4$

So we have two possible critical points, $(4,\pm\sqrt2,\mp\sqrt2)$ , which give extreme values of $4+2\sqrt2$ and $4-2\sqrt2$ , respectively.

5 0

3 years ago