You have a 1 kilogram bag of sugar You use 435 grams to fill the sugar bowls
How much sugar is left in the bag

Answers

Answer 1

Answer: The answer is 565g. Unless if there are more bowls, you just minus 435 from 565.

Answer 2

Answer: There is 1,000 grams in a kilogram. I just took away 435 from 1,000.
1,000 - 435 = 565 grams left in the bag.

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Consider the figure.Find AB if BC = 4, BD = 5, and AD = 5
What’s the volume of the pic
Paul owns a mobile wood-fired pizza oven operation. A couple of his clients complained about his dough at a recent catering, so he changed his dough to a newer product. Using the old dough, there were 6 complaints out of 385 pizzas. With the new dough, there were 16 complaints out of 340 pizzas. Let p 1 be the proportion of customer complaints with the old dough and p 2 be the proportion of customer complaints with the new dough. State the competing hypotheses to determine if the proportion of customer complaints using the old dough is less than the proportion of customer complaints using the new dough g
Two cities are 3450 miles apart. A plane leaves one of them, traveling towards the other at an average speed of 310 miles per hour. At the same time a plane leaves the other, traveling towards the first, at an average speed of 380 miles per hour. How long will it take them to meet?
The local newspaper has letters to the editor from 70 people. If this number represents 4% of all of the newspapers readers, how many readers does the newspaper

9. What is the value
of the expression?
21.3 + (-34.87)

Answers

Answer:

-13.57

Step-by-step explanation:

21.3 + (-34.87) = -13.57

Mark me as brainliest if you want to.

(a)/(5h+q)=t Find a (a)/ means fraction

Answers

Answer:

Step-by-step explanation:

a/(5h+q) = t

multiply both sides by 5h + q

a / (5h + q) * (5h+q) = t * (5h + q)

a = t * (5h + q)

I would say this is your answer.

You could distribute the t

a = 5ht + tq

which could also be an answer.

If a 5 ft tall man cast an 8 ft long shadow at the same time a tree cast a 24 ft long shadow, how tall is the tree?

Answers

Answer:

15 feet

Step-by-step explanation:

We have 2 similar right triangles with legs height and length of shadows.

height of men : length of shadows of the man = height of tree : length of shadows of the tree

5 : 8 = x : 24

8x = 5* 24

x = 5*24/8 = 15 (feet)

Answer:

15ft

Step-by-step explanation:

5 ft is to 8 ft

A ft is to 24 ft

A = 24*5/8

A = 15ft

15ft

Find the radius and height of a cylindrical soda can with a volume of 256cm^3 that minimize the surface area.B: Compare your answer in part A to a real soda can, which has a volume of 256cm^3, a radius of 2.8 cm, and a height of 10.7 cm, to conclude that real soda cans do not seem to have an optimal design. Then use the fact that real soda cans have a double thickness in their top and bottom surfaces to find the radius and height that minimizes the surface area of a real can (the surface area of the top and bottom are now twice their values in part A.

B: New radius=?

New height=?

Answers

Answer:

A) Radius: 3.44 cm.

Height: 6.88 cm.

B) Radius: 2.73 cm.

Height: 10.92 cm.

Step-by-step explanation:

We have to solve a optimization problem with constraints. The surface area has to be minimized, restrained to a fixed volumen.

a) We can express the volume of the soda can as:

$V=\pi r^2h=256$

This is the constraint.

The function we want to minimize is the surface, and it can be expressed as:

$S=2\pi rh+2\pi r^2$

To solve this, we can express h in function of r:

$V=\pi r^2h=256\n\nh=(256)/(\pi r^2)$

And replace it in the surface equation

$S=2\pi rh+2\pi r^2=2\pi r((256)/(\pi r^2))+2\pi r^2=(512)/(r) +2\pi r^2$

To optimize the function, we derive and equal to zero

$(dS)/(dr)=512*(-1)*r^(-2)+4\pi r=0\n\n(-512)/(r^2)+4\pi r=0\n\nr^3=(512)/(4\pi) \n\nr=\sqrt[3]{(512)/(4\pi) } =\sqrt[3]{40.74 }=3.44$

The radius that minimizes the surface is r=3.44 cm.

The height is then

$h=(256)/(\pi r^2)=(256)/(\pi (3.44)^2)=6.88$

The height that minimizes the surface is h=6.88 cm.

b) The new equation for the real surface is:

$S=2\pi rh+2*(2\pi r^2)=2\pi rh+4\pi r^2$

We derive and equal to zero

$(dS)/(dr)=512*(-1)*r^(-2)+8\pi r=0\n\n(-512)/(r^2)+8\pi r=0\n\nr^3=(512)/(8\pi) \n\nr=\sqrt[3]{(512)/(8\pi)}=\sqrt[3]{20.37}=2.73$

The radius that minimizes the real surface is r=2.73 cm.

The height is then

$h=(256)/(\pi r^2)=(256)/(\pi (2.73)^2)=10.92$

The height that minimizes the real surface is h=10.92 cm.

Final answer:

The minimal surface area for a cylindrical can of 256cm^3 is achieved with radius 3.03 cm and height 8.9 cm under uniform thickness, and radius 3.383 cm and height 7.14 cm with double thickness at top and bottom. Real cans deviate slightly from these dimensions possibly due to practicality.

Explanation:

For a cylinder with given volume, the surface area A, radius r, and height h are related by the formula A = 2πrh + 2πr^2 (if the thickness is uniform) or A = 3πrh + 2πr^2 (if the top and bottom are double thickness). By taking the derivative of A w.r.t r and setting it to zero, we can find the optimal values that minimize A.

For a volume of 256 cm^3, this gives us r = 3.03 cm and h = 8.9 cm with uniform thickness, and r = 3.383 cm and h = 7.14 cm with double thickness at the top and bottom. Comparing these optimal dimensions to a real soda can (r = 2.8 cm, h = 10.7 cm), we see that the real can has similar but not exactly optimal dimensions. This may be due to practical considerations like stability and ease of holding the can.

Learn more about Optimal Dimensions here:

brainly.com/question/32818645

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Jack has a summer job caddying at a golf course. Jack earns $120a week. By rounding his weekly income to the nearest hundred
dollars, find a reasonable estimate for his total income during the
12 week summer.

A) $700
B) $800
C) $1200
D) $1700

Answers

1200 as you would round it down to 100 a week so you would do 100x12

More on the Leaning Tower of Pisa. Refer to the previous exercise. (a) In 1918 the lean was 2.9071 meters. (The coded value is 71.) Using the least-squares equation for the years 1975 to 1987, calculate a predicted value for the lean in 1918. (Note that you must use the coded value 18 for year.)