What size does the radius of a sphere need to be for its volume to be larger than its surface area? HINT! It is less than 10. HINT! It is NOT a whole number. If you can show me a whole number for a radius where the surface area and volume are equal, then any radius bigger than that will have a larger volume.

To get full points you will need to SHOW me formulas for surface area of a sphere, volume of a sphere, and calculations on how you found your answer.

The slant height is 5 m

The representation of this situation as an inequality is 1.15x ≤ 50. To find the maximum possible cost of Kristin's meal before the tip, the inequality is solved for x, which results in the value of $43.48.

In this scenario, the cost of Kristin's meal before the tip can be represented by the variable x. The tip is 15% of the cost of the meal, which can be represented as 0.15x. The total cost of the meal, including the tip, is therefore x + 0.15x, which is 1.15x. Given that the total cost cannot exceed $50, we can write the following inequality to represent this situation: 1.15x ≤ 50.

To find the most that her meal can cost before a tip, we need to solve this inequality for x. So, we divide both sides by 1.15: x ≤ 50/1.15. After calculation, we find that x (the cost of the meal before the tip) must be ≤ about $43.48.

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Answer:

[tex]\frac{2m^2n^8}{3}[/tex]

Step-by-step explanation:

The given expression is:

[tex](\frac{4m^{-2}n^8}{9m^{-6}n^{-8}  })^{\frac{1}{2}}[/tex]

Upon simplifying the above equation, we get

Move [tex]m^{-6}[/tex] to teh numerator and Using the negative exponent rule, we get

=[tex](\frac{4m^{-2}n^8m^6}{9n^{-8}  })^{\frac{1}{2}}[/tex]

Move [tex]n^{-8}[/tex] to teh numerator and Using the negative exponent rule, we get

=[tex](\frac{4m^{-2}n^8m^6n^8}{9})^{\frac{1}{2}}[/tex]

=[tex](\frac{4m^{4}n^{16}}{9})^{\frac{1}{2}}[/tex]

=[tex](\frac{(2)^2(m^2)^2(n^8)^2}{(3)^2})^{\frac{1}{2}}[/tex]

=[tex]\frac{2m^2n^8}{3}[/tex]

which is the correct simplified form of the given expression.

The best approximation for the volume of the cone in cubic units is 1361.357 units³.

Volume of a three dimensional shape is the space occupied by the shape.

Given that,

Diameter of the cone = 20

Radius of the cone, r = Half of diameter

                                = 20/2 = 10

Height of the cone, h = 13

Volume of the cone = 1/3 π r² h

                                  = 1/3 π (10)² (13)

                                  = 1361.357 units³

Hence the volume is 1361.357 units³.

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The volume of cylinder is 1962.5 [tex]cm^{2}[/tex].

To find the volume of a cylinder.

The part of geometry concerned with ascertaining lengths, areas, and volumes. Mensuration deals with the size, region, and density of different forms both 2D and 3D.

Given that:

Radius= 5 cm

Height= 0.25 m=25 cm

π=3.14

By the formula of volume of cylinder

Volume=π[tex]r^{2} h[/tex]

TO put the value,

Volume=3.14*5*5*25

Volume=1962.5 [tex]cm^{2}[/tex]

So, the volume of cylinder is 1962.5 [tex]cm^{2}[/tex].

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The amount 16 ounce into grams using the conversion factor 1 ounce = 28.35 grams is 454.2 grams.

To convert ounces to grams, you can use the conversion factor:

1 ounce = 28.35 grams

Now, to convert 16 ounces to grams:

= 16 ounces * 28.35 grams/ounce

= 454.16 grams

Rounding to the nearest tenth:

454.16 grams rounded to the nearest tenth is 454.2 grams.

Therefore, the required quantity is 454.2 grams.

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To convert 16 ounces to grams, we multiply 16 by the conversion factor of 1 ounce to grams (28.35). This gives us 453.6 grams, when rounded to the nearest tenth.

To convert ounces to grams, we use the conversion factor 1 oz = 28.35 g. Given that we are converting 16 ounces to grams, we multiply the number of ounces by the conversion factor.

So, 16 oz * 28.35 g/oz = 453.6 g. Rounding to the nearest tenth, the result is 453.6 g.

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Final answer:

To solve the expression 12.5(18/6) + 5³, divide 18 by 6 to get 3. Multiply 12.5 by 3 to get 37.5. Then, calculate 5³, which equals 125. Adding the results, you get 162.5.

Explanation:

To solve the expression 12.5(18/6) + 5³, follow the order of operations (PEMDAS/BODMAS). First, divide 18 by 6, which equals 3. Next, multiply 12.5 by 3, which equals 37.5. Finally, calculate 5³, which equals 125. Adding the results, 37.5 + 125 = 162.5.

We have been given that

2 cups flour + 3 eggs + 1 tbs oil → 4 waffles

In order to make 4 waffles we need 3 eggs.

Thus, to make 1 waffle we need 3/4 eggs.

Now, in 1 dozen there are 12 waffles. Hence, to make 12 waffles we need

[tex]\frac{3}{4} \times 12\\ \\ =3 \times 3\\ =9[/tex]

Therefore, 9 eggs are needed to make 1 dozen waffles.

D is the correct option.

Final answer:

The correct option is (d) 9. Using the given recipe, we can determine that 9 eggs are needed to make 1 dozen waffles by setting up a proportion and solving for the number of eggs.

Explanation:

The original recipe provided states that 2 cups of flour, 3 eggs, and 1 tablespoon of oil will yield 4 waffles. To find out how many eggs are needed to make 1 dozen (12) waffles, we need to set up a proportion based on the given recipe.

Since 4 waffles require 3 eggs, we can set up the following equation: 4 waffles / 3 eggs = 12 waffles / x eggs. Cross-multiplying gives us 4x = 36, and dividing both sides by 4 gives us x = 9. Therefore, 9 eggs are required to make 1 dozen waffles.

Answer:

x = 3/2

Step-by-step explanation:

That is quite long and drawn out, so we have to get it down to a single quadratic equation, set equal to 0.

Begin by expanding the right side to get

[tex]y+3x-6=-3(x^2-4x+4)+4[/tex]

then multiplying in the -3 to get

[tex]y+3x-6=-3x^2+12x-12+4[/tex]

Now we will combine all the like terms and get everything on one side of the equals sign, and set it equal to 0:

[tex]-3x^2+9x-2=0[/tex]

In order to find the equation of the axis of symmetry we have to put it into vertex form, which is accomplished by completing the square on this quadratic.

In order to complete the square, the leading coefficient has to be a +1.  Ours is a -3, so we will factor that out.  First, though, now that it is set to equal 0, we will move the constant, -2, over to the other side, isolating the x terms.

[tex]-3x^2+9x=2[/tex]

Now we can factor out the -3:

[tex]-3(x^2-3x)=2[/tex]

The rules for completing the square are as follows:

Take half the linear term, square it, and add it to both sides.  Our linear term is 3.  Half of 3 is 3/2, and squaring that gives you 9/4.  We add into the parenthesis on the left a 9/4, but don't forget about that -3 hanging around out front that refuses to be ignored.  We didn't add in just a 9/4, we added in (-3)(9/4) = -27/4:

[tex]-3(x^2-3x+\frac{9}{4})=2-\frac{27}{4}[/tex]

In the process of completing the square we created a perfect square binomial on the left.  Stating that binomial and simplifying the addition on the right gives us:

[tex]-3(x-\frac{3}{2})^2=-\frac{19}{4}[/tex]

We can determine the axis of symmetry at this point.  Because this is a positive x-squared polynomial, the axis of symmetry is in the form of (x = ) and what it equals is the h coordinate of the vertex.  Our h coordinate is 3/2; therefore, the axis of symmetry has the equation x = 3/2

Sample Response: Locate two corresponding ordered pairs of the two shapes. Calculate how much the pre-image values were multiplied by to obtain the image values. This is your scale factor. You can calculate this by dividing the coordinates of the mage by the corresponding coordinates of the pre-image.

The weight of the water in the cube-shaped aquarium will be 1674 lb, which is much greater than the table's maximum weight capacity of 200 lb. Therefore, the aquarium should not be placed on the table. The density of the water remains constant at 62 lb/ft³ regardless of how much the aquarium is filled.

To determine if the aquarium can be placed on a table with a weight limit of 200 lb, we must first calculate the weight of the water the aquarium will hold. We calculate the volume of the aquarium, then multiply it by the density of water, as the density is given as weight per unit volume (specifically 62 lb/ft³).

The volume (V) of a cube is given by V = side³, so for our aquarium with 3 ft long edges, V = 3³ = 27 ft³. The weight of the water (W) is therefore W = density x volume = 62 lb/ft³ x 27 ft³ = 1674 lb. Since 1674 lb exceeds the table's maximum weight capacity of 200 lb, the aquarium should not be placed on the table.

As for part (b), the density of the water does not change whether the aquarium is full or half full. Density is an intensive property meaning it's independent of the quantity of the substance present. Therefore, whether the aquarium is filled or only half filled, the density of the water remains at 62 lb/ft³.

Final answer:

A square cannot be a cross section of a traditional circular cylinder when sliced perpendicular to its height. However, circles, rectangles, and triangles can be cross sections depending on the shape of the cylinder and the slicing.

Explanation:

When considering the cross sections of a cylinder, it is important to remember that the cross-sectional shape remains consistent along the height of the cylinder. A cross section made perpendicular to the height of a traditional circular cylinder is always a circle. Therefore, the correct answer to which shape could not be a cross section of a cylinder is a square, option b. Circles, rectangles, and triangles can all be cross sections of a cylinder depending on the shapes of the top and bottom faces of the cylinder and how the cross section is sliced.

If the cylinder is a right circular cylinder, the only possible perpendicular cross-sectional shape is a circle. A square cross section can occur in a cylinder if the top and bottom faces are square and the cylinder is not circular. Similarly, a rectangular cross section can occur if the cylinder has a rectangular type of cross-section to start with. However, for a circular cylinder, a triangle cannot be a cross section because slicing it will always yield a circular shape due to the round nature of its base.

Answer:

C.

0.0316

Step-by-step explanation: