HELP PLEASE WITH 5 QUESTIONS!!!

Intersection of a square and a regular pentagon can be ?

Natali5045456 [20]

Answer:

A Triangle

Step-by-step explanation:

When you overlap a square in a pentagon you get a triangle

6 0

3 years ago

Mrs.E bought 3 drinks and 5 sandwiches for $25.05 and Mr.E bought 4 drinks and 2 sandwiches for $13.80. How much does each drink

MAVERICK [17]

Answer:

$1.35

Step-by-step explanation:

Set up a system of equations, where d represents the price of one drink and s represents the price of one sandwich:

3d + 5s = 25.05

4d + 2s = 13.80

We can solve this by elimination by multiplying the top equation by -2 and multiplying the bottom equation by 5.

-6d - 10s = -50.1

20d + 10s = 69

Add them together:

14d = 18.9

d = 1.35

7 0

3 years ago

Please help me with this question :) <br> I will mark you as brainliest

wolverine [178]

$\text{Given that,}\\\\s_1 = 5.95,\\\\s_2 = 5.42\\\\\\F= \dfrac{(s_1)^2}{(s_2)^2} = \dfrac{(5.95)^2}{(5.42)^2} = 1.21$

4 0

3 years ago

Jason and Megan are brother and sister. Jason is 4 years older than Megan.If Jason is 16 yearsold write and solve an equation to

Eva8 [605]

16-4=12 is the answer.

8 0

3 years ago

From a thin piece of cardboard 50 in. by 50 in., square corners are cut out so that the sides can be folded up to make a box. Wh

mixer [17]

Answer:

When dimension of box is 33.33 inches × 33.33 inches ×8.33 then its volume is maximum and is 9259.26 cubic inches.

Step-by-step explanation:

Let h be the length (in inches) of the square corners that has been cut out from the cardboard and that would be the height of the cardboard box.

Since the squares have been cut from cardboard, both sides of the cardboard would reduce by 2h.

Thus, The dimension of box is (50 – 2h) × (50 – 2h) × h in dimensions.

The volume V of rectangular box = (Length × Breadth × Height) cubic inches.

$V=(50-2h) \times (50-2h) \times h$

$V=(50-2h)^2 \times h$ ..............(1)

Using $(a-b)^2=a^2+b^2-2ab$

$V=h(2500+4h^2-200h)$

$V=2500h+4h^3-200h^2$

For obtaining a box of maximum volume, maximize V as a function of h.

Differentiate both sides with respect to h,

$\frac{dV}{dh}=2500+12h^2-400h$

$\frac{dV}{dh}=4(625+3h^2-100h)$

Solving quadratic equation, $625+3h^2-100h$

$\frac{dV}{dh}=4(3h^2-25h-75h+625)$

$\frac{dV}{dh}=4(h(3h-25)-25(3h-25))$

$\frac{dV}{dh}=4((h-25)(3h-25))$

For maximum, $\frac{dV}{dh}=0$

thus, $4((h-25)(3h-25))=0$

⇒ $h= 25$ or $h=\frac{25}{3}$

Now check (1) for $h= 25$ and $h=\frac{25}{3}$ .

$h= 25$ is not possible as when h is 25 inches then length and breadth becomes 0.

When $h=\frac{25}{3}$ .

(1) ⇒ $V=(50-2(\frac{25}{3}))^2 \times\frac{25}{3}=9259.2592593$

This is the maximum volume the box can assume.

Thus, when dimension of box is 33.3 inches × 33.3 inches ×8.3 then its volume is maximum and is 9259.26 cubic inches.

6 0

3 years ago