If x varies directly as y write a formula connecting x and y y.use k as a constant of variation.

Please help ASAP Thank you

Degger [83]

Answer:

comparison

Step-by-step explanation:

comparison

7 0

2 years ago

What shape best describes the cross section cut at an angle to the base of a right rectangular prism?

Rufina [12.5K]

Answer:

right angle triangle

6 0

2 years ago

A piece of paint cost $400. There is a 25% mark- up. What is the selling price of the painting?

Sliva [168]

Answer:

$500.

Step-by-step explanation:

6 0

2 years ago

Which pairs of points would result in a line with a rate of change of 2:1? select all that apply

Reil [10]

Answer:

A,C,D

Step-by-step explanation:

Complete Question:

"Which pairs of points would result in a line with a rate of change of 2:1? select all that apply"

A.(−4,−11) and (−2−7)

B.(5,6) and (−5, 12)

C.(0,−3)and (−4,−11)

D. (0,−3) and (5, 7)

Rate of change is the slope. 2:1 would mean a slope of simply, 2.

We need to find slope of each of the choices from A - D and see which one has 2. The formula would be:

$m=\frac{y_2-y_1}{x_2-x_1}$

Where m is the slope and

x_1 is first x point

x_2 is second x point

y_1 is first y point

y_2 is second y point

Let's calculate slope of each of the choices.

A.

$m=\frac{y_2-y_1}{x_2-x_1}\\m=\frac{-7+11}{-2+4}\\m=2$

THis one is okay.

B.

$m=\frac{y_2-y_1}{x_2-x_1}\\m=\frac{12-6}{-5-5}\\m=-\frac{3}{5}$

THis one is not 2:1.

C.

$m=\frac{y_2-y_1}{x_2-x_1}\\m=\frac{-11+3}{-4-0}\\m=2$

THis one is okay.

D.

$m=\frac{y_2-y_1}{x_2-x_1}\\m=\frac{7+3}{5-0}\\m=2$

This one is okay.

Hence, A, C, and D all have rate of change of 2:1

4 0

3 years ago

You must design a closed rectangular box of width w, length l and height h, whose volume is 504 cm3. The sides of the box cost 3

Charra [1.4K]

Answer:

Dimensions will be

Length = 7.23 cm

Width = 7.23 cm

Height = 9.64 cm

Step-by-step explanation:

A closed box has length = l cm

width of the box = w cm

height of the box = h cm

Volume of the rectangular box = lwh

504 = lwh

$h=\frac{504}{lw}$

Sides which involve length and width and height, cost = 3 cents per cm²

Top and bottom of the box costs = 4 cents per cm²

Cost of the sides $C_{s}$ = 3[2(l + w)h] = 6(l + w)h

$C_{s}$ = 3[2(l + w)h]

$C_{s}=6(l+w)(\frac{504}{lw} )$

Cost of the top and the bottom $C_{(t,p)}$ = 4(2lw) = 8lw

Total cost of the box C = $3024\frac{(l+w)}{lw}$ + 8lw

= $3024[\frac{1}{l}+\frac{1}{w}]$ + 8lw

To minimize the cost of the sides

$\frac{dC}{dl}=3024(-l^{-2}+0)+8w=0$

$\frac{3024}{l^{2}}=8w$

$\frac{378}{l^{2}}=w$ ---------(1)

$\frac{dC}{dw}=3024(-w^{-2})+8l=0$

$\frac{3024}{w^{2}}=8l$

$\frac{378}{w^{2}}=l$

$w^{2}=\frac{378}{l}$ -------(2)

Now place the value of w from equation (1) to equation (2)

$(\frac{378}{l^{2}})^{2}=\frac{378}{l}$

$\frac{(378)^{2} }{l^{4}}=\frac{378}{l}$

l³ = 378

l = ∛378 = 7.23 cm

From equation (2)

$w^{2}=\frac{378}{7.23}$

$w^{2}=52.28$

w = 7.23 cm

As lwh = 504 cm³

(7.23)²h = 504

$h=\frac{504}{(7.23)^{2}}$

h = 9.64 cm

6 0

3 years ago