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Gelneren [198K]

3 years ago

9

A server fills an ice cream cone with ice cream. The cone has a diameter of 3 inches and is 8 inches tall. Approximately how muc

h ice cream is needed to fill the cone?

1 answer:

irina1246 [14]3 years ago

3 0

Answer:

V≈75.4

Step-by-step explanation:

V=πr^2h/3

v=3.14(3)^2(8)/3=75.4

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dentify all of the following solutions of square root of x minus 8 end root plus 8 equals x. (1 point)

viva [34]

Answer:gh

Step-by-step explanation:

3 0

2 years ago

Find the difference of 1-3/8

Vilka [71]

1-3=-2 so -2/8 is -1/4 so negative one fourth

3 0

2 years ago

<img src="https://tex.z-dn.net/?f=%28x%5E%7B2%7D%20%2Bx-3%29%3A%20%28x%5E%7B2%7D%20-4%29%5Cgeq%201" id="TexFormula1" title="(x^{

Jlenok [28]

Answer:

$x>2$

Step-by-step explanation:

When given the following inequality;

$(x^2+x-3):(x^2-4)\geq1$

Rewrite in a fractional form so that it is easier to work with. Remember, a ratio is another way of expressing a fraction where the first term is the numerator (value over the fraction) and the second is the denominator(value under the fraction);

$\frac{x^2+x-3}{x^2-4}\geq1$

Now bring all of the terms to one side so that the other side is just a zero, use the idea of inverse operations to achieve this:

$\frac{x^2+x-3}{x^2-4}-1\geq0$

Convert the (1) to have the like denominator as the other term on the left side. Keep in mind, any term over itself is equal to (1);

$\frac{x^2+x-3}{x^2-4}-\frac{x^2-4}{x^2-4}\geq0$

Perform the operation on the other side distribute the negative sign and combine like terms;

$\frac{(x^2+x-3)-(x^2-4)}{x^2-4}\geq0\\\\\frac{x^2+x-3-x^2+4}{x^2-4}\geq0\\\\\frac{x+1}{x^2-4}\geq0$

Factor the equation so that one can find the intervales where the inequality is true;

$\frac{x+1}{(x-2)(x+2)}\geq0$

Solve to find the intervales when the equation is true. These intervales are the spaces between the zeros. The zeros of the inequality can be found using the zero product property (which states that any number times zero equals zero), these zeros are as follows;

$-1, 2, -2$

Therefore the intervales are the following, remember, the denominator cannot be zero, therefore some zeros are not included in the domain

$x\leq-2\\-2$

Substitute a value in these intervales to find out if the inequality is positive or negative, if it is positive then the interval is a solution, if it is negative then it is not a solution. This is because the inequality is greater than or equal to zero;

$x\leq-2$ -> negative

$-2$ -> neagtive

$-1\leq x$ -> neagtive

$x>2$ -> positive

Therefore, the solution to the inequality is the following;

$x>2$

6 0

2 years ago

Every prime number greater than 10 has a digit in the ones place that is include in wich set of numbers A- 1,3,7,9 B-1,3,5,9 D-1

marta [7]

That's very interesting. I had never thought about it before.
Let's look through all of the ten possible digits in that place,
and see what we can tell:

-- 0:
   A number greater than 10 with a 0 in the units place is a multiple of
   either 5 or 10, so it's not a prime number.

-- 1:
    A number greater than 10 with a 1 in the units place could be
    a prime (11, 31 etc.) but it doesn't have to be (21, 51).

-- 2:
   A number greater than 10 with a 2 in the units place has 2 as a factor
   (it's an even number), so it's not a prime number.

-- 3:
   A number greater than 10 with a 3 in the units place could be
   a prime (13, 23 etc.) but it doesn't have to be (33, 63) .

-- 4:
   A number greater than 10 with a 4 in the units place is an even
   number, and has 2 as a factor, so it's not a prime number.

-- 5:
   A number greater than 10 with a 5 in the units place is a multiple
   of either 5 or 10, so it's not a prime number.

-- 6:
   A number greater than 10 with a 6 in the units place is an even
   number, and has 2 as a factor, so it's not a prime number.

-- 7:
   A number greater than 10 with a 7 in the units place could be
   a prime (17, 37 etc.) but it doesn't have to be (27, 57) .

-- 8:
   A number greater than 10 with a 8 in the units place is an even
   number, and has 2 as a factor, so it's not a prime number.

-- 9:
   A number greater than 10 with a 9 in the units place could be
   a prime (19, 29 etc.) but it doesn't have to be (39, 69) .

So a number greater than 10 that IS a prime number COULD have
any of the digits 1, 3, 7, or 9 in its units place.

It CAN't have a 0, 2, 4, 5, 6, or 8 .

The only choice that includes all of the possibilities is 'A' .

4 0

2 years ago

The Bernards are trying to determine their babysitter's rates but are to embarrassed to just ask her. They were charged $20 for

Alina [70]

Answer:

A = 5.5t + 9

Step-by-step explanation:

We assume that the relationship between t and A is linear.

We have two points on a line: (2, 20) and (4, 31).

We now find the equation of the line.

A = mt + b

m = (31 - 20)/(4 - 2) = 11/2 = 5.5

20 = 5.5(2) + b

20 = 11 + b

b = 9

A = 5.5t + 9

3 0

3 years ago