Which best describes the resulting three-dimensional figure? a cone with a base radius of 26 cm a cone with a base radius of 14 cm a cylinder with a base radius of 26 cm a cylinder with a base radius of 14 cm

Answer:

(B) No

Step-by-step explanation:

If I remember correctly a solution is a point that is on the shaded area or on a solid line, the point 3,-3 is not on the shaded region and not on a solid line. Even if it was on the line in this problem it wouldn't be a solution because it is a dotted line. Hope this helps!

Answer:

3, 7

Step-by-step explanation:

3, 3, 3, 4, 4, 6, 7, 7, 7

3 and 7

Before we start, we gotta acknowledge that one pound is 16 oz.

Okay, so to solve this we need to make a ratio. We know there are 567 calories for every 9 oz. That'll look like 567/9. And to compare this to pounds we have to convert those three pounds to ounces by multiplying: 3*16 = 48. So now we can grab that ratio and use cross multiplication to find the answer:

567/9 = x/48

9x = 567*48

9x = 27216

x = 3024

So there are 3024 calories in three pounds of the ice cream.

There are 3024 calories in a certain type of ice cream.

A unitary method is a mathematical way of obtaining the value of a single unit and then deriving any no. of given units by multiplying it with the single unit.

We know one pound is equal to 16 ounces.

So, 9 ounces is equal to 9/16 pounds.

Given, There are 567 calories in nine ounces of certain ice cream.

So, There are 567 calories in (9/16) pounds of certain ice cream.

Or, (567)/(9/16).

= 567×(16/9).

= 1008 calories in one pound.

Therefore, There is 1008×3 = 3024 calories in 3 pounds of ice cream.

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Answer:

8 x 10^5

Step-by-step explanation:

Answer:

30 cm

Step-by-step explanation:

just times it by three because theres three sides

Answer:

-12x²

Step-by-step explanation:

Given expression:

[tex]-5x^{2} -7x^{2}[/tex]

Since both terms in the expression have common terms, we can subtract the terms by combining like terms.

[tex]\implies x^{2} (-5 -7)[/tex]

[tex]\implies x^{2} (-12)[/tex]

[tex]\implies -12x^{2}[/tex]

Answer:

[tex]2x^{6}y^{2} }[/tex]

Step-by-step explanation:

Just Simplifying the Fraction.

we get,

[tex]=\frac{12x^{9} y^{4} }{6x^{3}y^{2} } \\\\=2x^{9-3} y^{4-2} \\=2x^{6} y^{2}[/tex]

Answer:

4π

Step-by-step explanation:

since the shaded in area is the tile and the radius is 2 and we only need the SHADED IN area 2^2 is 4 therefore 4pi is the correct answer

Answer: 0.9

Step-by-step explanation: All you need to just keep your decimal (0.9) as it is since 4 and 2 are below 5 then you don't need to round it higher. I hope this helped!

0.942

the nearest tenth of 0.942

to round to the nearest tenth, if the number in hundredths place is 5 or more you'll have to round up, if not keep it as it is and remove all the digits in the right side.

=0.9

hence, the nearest tenth of 0.942 is 0.9

Step-by-step explanation:

[tex]5 {x}^{2} - 4 {y}^{2} + 80 = 0[/tex]

[tex]5 {x}^{2} - 4 {y}^{2} = - 80[/tex]

[tex] - \frac{ 5{x}^{2} }{16} + \frac{ {y}^{2} }{20} = 1[/tex]

[tex] \frac{ {y}^{2} }{20} - \frac{5 {x}^{2} }{16} = 1[/tex]

To find foci,

[tex]c = \sqrt{ {a}^{2} + b {}^{2} } [/tex]

so

[tex]c = \sqrt{20 + 16} [/tex]

[tex]c = \sqrt{36} [/tex]

[tex]c = ±6[/tex]

Since the y term has a greater denomiator, our foci is

(0,6) and (0,-6)

We are told to the write the given expression in standard form ;

[tex]{:\implies \quad \sf \dfrac{6.6\times 10^{-2}}{3.3\times 10^{-4}}}[/tex]

Rewrite as ;

[tex]{:\implies \quad \sf \dfrac{66\times 10\times 10^{-2}}{33\times 10\times 10^{-4}}}[/tex]

[tex]{:\implies \quad \sf \dfrac{2\times 10^{-2}}{10^{-4}}}[/tex]

[tex]{:\implies \quad \sf 2\times \dfrac{10^{-2}}{10^{-4}}}[/tex]

[tex]{:\implies \quad \sf 2\times 10^{-2-(4)}\quad \qquad \bigg\{\because \dfrac{a^m}{a^n}=a^{m-n}\bigg\}}[/tex]

[tex]{:\implies \quad \sf 2\times 10^{-2+4}}[/tex]

[tex]{:\implies \quad \boxed{\bf{2\times 10^{2}}}}[/tex]

This is the required answer

Answer:

Step-by-step explanation:

Remark

The purple lines cross but once. A solution to a set of functions counts how many times the functions cross. Since these two lines cross but once, there is only 1 solution.

If the lines are parallel, they never cross. That means there are no solutions.

If you are given 2 equations and one is the same line as the second one, then there are an infinite number of solutions.

Hi!

Solutions to systems of equations are only where the lines intersect at a point.

That means that if the lines cross over at any point, they have a solution. Lines that don't have a solution -- parallel lines -- do not cross over.

In this graph, we can see that it crosses over at only one point.

That means that this system of equations has exactly one solution.

For a similar problem, see:

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Answer:

Interpolation is used to predict values that exist within a data set, and extrapolation is used to predict values that fall outside of a data set and use known values to predict unknown values. Often, interpolation is more reliable than extrapolation, but both types of prediction can be valuable for different purposes

Answer:

It makes a right traiangle.

Step-by-step explanation:

Graph the points and you will see that it makes a right triangle.

The figure formed by the vertices A(3,5), B(3,1), and C(0,1) is a right-angled triangle.

In this triangle, side BC is the shortest side with a length of 3 units, side AB is the longest side with a length of 4 units, and side AC has a length of 5 units.

The right angle is formed between side BC and side AC at point C (0,1).

Here, we have to identify the figure with the given vertices A (3,5), B (3,1), and C (0,1), we can plot these points on a coordinate plane.

Let's plot the points A(3,5), B(3,1), and C(0,1) on a Cartesian coordinate plane.

A(3,5): This point is located at x-coordinate 3 and y-coordinate 5.

B(3,1): This point is located at x-coordinate 3 and y-coordinate 1.

C(0,1): This point is located at x-coordinate 0 and y-coordinate 1.

Since point A and point B share the same x-coordinate (x = 3), they lie on a vertical line. Point C has a different x-coordinate, so we need to connect it with the others to complete the figure.

The figure formed by connecting the points A, B, and C is a right-angled triangle.

To calculate the distances between the vertices, we'll use the distance formula:

Distance between two points [tex](x_1, y_1)[/tex] and [tex](x_2, y_2)[/tex] is given by:

Distance = √[tex]((x_2 - x_1)^2 + (y_2 - y_1)^2)[/tex]

Let's calculate the distances:

Distance between A(3,5) and B(3,1):

Distance_AB = √[tex]((3 - 3)^2 + (1 - 5)^2)[/tex]

Distance_AB = √(0 + 16)

Distance_AB = √16

Distance_AB = 4 units

Distance between A(3,5) and C(0,1):

Distance_AC = √[tex]((0 - 3)^2 + (1 - 5)^2)[/tex]

Distance_AC = √[tex]((-3)^2 + (-4)^2)[/tex]

Distance_AC = √(9 + 16)

Distance_AC = √25

Distance_AC = 5 units

Distance between B(3,1) and C(0,1):

Distance_BC = √[tex]((0 - 3)^2 + (1 - 1)^2)[/tex]

Distance_BC = √[tex]((-3)^2 + 0^2)[/tex]

Distance_BC = √9

Distance_BC = 3 units

As we calculated, the distances between the vertices of the figure are:

AB = 4 units

AC = 5 units

BC = 3 units

Based on the distances, we can determine that the figure formed by the vertices A(3,5), B(3,1), and C(0,1) is a right-angled triangle.

In this triangle, side BC is the shortest side with a length of 3 units, side AB is the longest side with a length of 4 units, and side AC has a length of 5 units.

The right angle is formed between side BC and side AC at point C (0,1).

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Answer:

10 Quarts

Step-by-step explanation:

First, convert 1/2 gallons into quarts. 1/2x4=2. Then convert 1 1/4 to quarts. 1 1/4x4=5. Finally, add it all togher. 2+5+3=10. So, the answer is 10 quarts.

Answer: [tex]x < -\frac{24}{5}[/tex]

Step-by-step explanation:

You have:

[tex]-\frac{5}{6} x-3 > 1[/tex]

Add 3 to both sides to get:

[tex]-\frac{5}{6} x > 4[/tex]

Multiply both sides by 6 to get:

[tex]-{5} x > 24[/tex]

Divide both sides by -5. When you divide an inequality by a negative number, you must switch the signs. Therefore:

[tex]x < -\frac{24}{5}[/tex]

Answer:

The equation is saying multiply or add negative 5/6 to x, then subtract by positive 3 which makes the number go upwards, but not making it positive, and lastly, the equation is saying the result will be greater than 1.

One of the values for x is 5, because...

[tex]-\frac{5}{6} * 5 = -4\frac{1}{6} - 3 = 7\frac{1}{6} > 1[/tex]

With that information, the statement is now complete and true. If you make 5 or anything more than 5 the value for x, the statement will still be true.

Note that I just used 5 as one of the solutions. There are many more than that.

We are given that:

Circumference = 19.4 cm

Formula:

2πr = Circumference

Rewrite the equation:

The "r" variable represents the radius of the circle. Thus, we need to isolate it on one side to determine the radius. This can be done with the four mathematical operations (+  -  ×  ÷). Since the "÷" is included, we can divide both sides by 2π to isolate the "r" variable.

Divide both sides by 2π:

Substitute the radius as 22/7:

[tex]\textit{ellipse, horizontal major axis} \\\\ \cfrac{(x- h)^2}{ a^2}+\cfrac{(y- k)^2}{ b^2}=1 \qquad \begin{cases} center\ ( h, k)\\ vertices\ ( h\pm a, k)\\ c=\textit{distance from}\\ \qquad \textit{center to foci}\\ \qquad \sqrt{ a ^2- b ^2} \end{cases} \\\\[-0.35em] ~\dotfill[/tex]

[tex]\cfrac{(x+4)^2}{400}~~ + ~~\cfrac{(y+3)^2}{144}~~ = ~~1\implies \cfrac{[x-(-4)]^2}{20^2}~~ + ~~\cfrac{[y-(-3)]^2}{12^2}~~ = ~~1 \\\\\\ \begin{cases} h=-4\\ k=-3\\ a=20\\ b=12 \end{cases}\qquad \qquad c=\sqrt{20^2-12^2}\implies c=\sqrt{256}\implies \boxed{c=16} \\\\[-0.35em] ~\dotfill[/tex]

[tex]\stackrel{center}{(-4,-3)}\qquad vertices \begin{cases} (\stackrel{-4-20}{-24}~~,~~-3)\\\\ (\stackrel{-4+20}{16}~~,~~-3) \end{cases}\qquad foci \begin{cases} (\stackrel{-4-16}{-20}~~,~~-3)\\\\ (\stackrel{-4+16}{12}~~,~~-3) \end{cases}[/tex]

Answer:

I think it would be fine if they notified you but only if they used me for a good reason.

On the other hand...

no matter if it's on the agreement if it is inappropriate on your part it is illegal and you can complain.

Answer:

[tex] r = 8.58~cm [/tex]

Step-by-step explanation:

[tex] S = \dfrac{n}{360^\circ}2\pi r [/tex]

[tex] 38.95~cm = \dfrac{260^\circ}{360^\circ}2\pi r [/tex]

[tex] 38.95~cm = 4.5378r [/tex]

[tex] r = 8.58~cm [/tex]

Answer:

No, -14 is not greater than 9

Step-by-step explanation:

because -14 is a negative interger making the postive a higher/greater number

Answer:

10

Step-by-step explanation:

→ Find the difference in x coordinates

8 - 2= 6

→ Find the difference in y coordinates

-1 - 7 = -8

→ Use Pythagoras

√6² + (-8)² = 10

An expression is defined as a set of numbers, variables, and mathematical operations. The expression that can represent u/v is -x³+x²-1.

In mathematics, an expression is defined as a set of numbers, variables, and mathematical operations formed according to rules dependent on the context.

The expression which can represent the u/v can be found in the following manner,

[tex]\dfrac {u}{v} = \dfrac{x^5-x^4+x^2}{-x^2}\\\\\dfrac {u}{v} = \dfrac{x^2(x^3-x^2+1)}{-x^2}\\\\\dfrac {u}{v} = \dfrac{(x^3-x^2+1)}{-1}\\\\\dfrac {u}{v} = -x^3+x^2-1[/tex]

Thus, the expression that can represent u/v is -x³+x²-1.

Learn more about Expression:

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Answer:

C

Step-by-step explanation:

Answer:

He can make eight 3-pancake servings.

Step-by-step explanation:

6 eggs make 12 pancakes.

12 eggs make 24 pancakes.

24 divided by 3 is 8.

Check the picture below.

since the cliff is 48 feet tall, thus that its initial height.

[tex]~~~~~~\textit{initial velocity in feet} \\\\ h(t) = -16t^2+v_ot+h_o \quad \begin{cases} v_o=\textit{initial velocity}&8\\ \qquad \textit{of the object}\\ h_o=\textit{initial height}&48\\ \qquad \textit{of the object}\\ h=\textit{object's height}&\\ \qquad \textit{at "t" seconds} \end{cases} \\\\\\ h(t)=-16t^2+8t+48[/tex]

well, as you can see in the picture, its maximum is at its vertex, so

[tex]\textit{vertex of a vertical parabola, using coefficients} \\\\ y=\stackrel{\stackrel{a}{\downarrow }}{-16}x^2\stackrel{\stackrel{b}{\downarrow }}{+8}x\stackrel{\stackrel{c}{\downarrow }}{+48} \qquad \qquad \left(-\cfrac{ b}{2 a}~~~~ ,~~~~ c-\cfrac{ b^2}{4 a}\right) \\\\\\ \left(-\cfrac{ 8}{2(-16)}~~~~ ,~~~~ 48-\cfrac{ (8)^2}{4(-16)}\right)\implies \left(\cfrac{1}{4}~~,~~48-\cfrac{64}{-64} \right)[/tex]

[tex]\left( \frac{1}{4}~~,~~48+1 \right)\implies \stackrel{\textit{the ball is at highest}}{\left(\frac{1}{4}~~,~~\stackrel{\downarrow }{49} \right)}[/tex]

well, it hits the ground when y = 0.

[tex]\stackrel{h(t)}{0}=-16t^2+8t+48\implies 0=8(-2t^2+t+6)\implies 2t^2-t-6=0 \\\\\\ (2t+3)(t-2)=0\implies t= \begin{cases} -\frac{3}{2}\\\\ 2~~\textit{\large \checkmark} \end{cases}[/tex]

notice, we didn't use the negative value, since "t" must be greater than 0.