What is the volume of a cylinder with a diameter of 12 in. and a height of 7 in.?

Answer:

D) 40 miles

Step-by-step explanation:

First we must find the distance between the mall and the house.  The figure formed is a right triangle.  The length of the side from the mall to the house forms the hypotenuse of the triangle.  We can use the Pythagorean theorem to find the length:

a² + b² = c²

The two legs of the triangle, a and b, are 15 and 8:

15² + 8² = c²

225 + 64 = c²

289 = c²

Take the square root of each side:

√289 = √(c²)

17 = c

This makes the total distance

15+8+17= 40 miles

Hey! Pretty easy once you get the hang of these problems.

So you start off with the Pythagorean Theorem. Which is : a^2 + b^2 = c^2

Then just fill it in :

15^2 + 8^2 = c^2

225 + 64 = c2

289 = c^2

17 = c

And last :

15 + 8 + 17 = 40 MILES

hoped this helped!! :))

Quadratic relations and comic sections unit test part 1

11. a. (0, +/- 6)

Answer:

  29

Step-by-step explanation:

The perimeter is the sum of the side lengths:

  4 + 12 + 13 = 29

The perimeter of the triangle is 29. (It is not a right triangle.)

___

A right triangle with side lengths 12 and 13 will have a short side of 5. Its perimeter is 30.

Answer:

To find the zeros of a quadratic function, use the quadratic equation, [tex]x=\frac{-b \pm \sqrt{b^2-4ac} }{2a}[/tex]. We find that the eqautions with zeros at 1 and 4 are b) x² -5x + 4 and d) -2x² + 10x - 8.

Step-by-step explanation:

a) x² + x + 4 --

[tex]x = \frac{-1 \pm \sqrt{1^2-4*1*4} }{2*1}\\x=\frac{-1 \pm \sqrt{1-16}}{2}[/tex]

Because the discriminant (the value inside the square root) is negative, this equation does not have real zeros, so it is not the answer.

b) x² - 5x + 4 --

[tex]x = \frac{5 \pm \sqrt{(-5)^2-4*1*4}}{2*1} \\x=\frac{5 \pm \sqrt{25-16}}{2} \\x = \frac{5 \pm 3}{2}[/tex]

Now, we calculate the two zeros by adding and subtracting the 3.

[tex]x = \frac{5+3}{2} \\x= \frac{8}{2} = 4\\\\x= \frac{5-3}{2} \\x= \frac{2}{2}=1[/tex]

The zeros of this function are 1 and 4, so it is included in our answer.

c) x² + 3x - 4 --

[tex]x = \frac{-3 \pm \sqrt{3^2-4*1*-4}}{2*1} \\x = \frac{-3 \pm \sqrt{9+16}}{2} \\x= \frac{-3 \pm 5}{2}\\\\x=\frac{-3+5}{2}=1\\x=\frac{-3-5}{2} = -4[/tex]

The zeros of this function are -4 and 1, so it is not the answer.

d) -2x² + 10x - 8 --

[tex]x = \frac{-10 \pm \sqrt{10^2-4*(-2)*(-8)} }{2*(-2)} \\x=\frac{-10 \pm \sqrt{100-64} }{-4} \\x = \frac{-10 \pm 6}{-4} \\\\x=\frac{-10 + 6}{-4} =1\\x = \frac{-10-6}{-4} =4[/tex]

The zeros of this function are 1 and 4, so it is included in our answer.

Answer:

B,5/8+5r=8

Step-by-step explanation:

1. B. Counterclockwise

2. C. (30,401)

3. A. t=2(x-3)

4. C. She should have taken both the positive and negative square root

5. C. y=x^2+8x-25/8

6. D. Hyperbola

7. A. Graph A

The graph of the function (f(x) = -14x - 2) is attached below and the two points from which line passes are (-0.143,0) and (0,-2).

Given :

Equation  --  f(x) = -14x - 2

The following steps can be used in order to sketch the graph of the given function:

Step 1 - Write the given function.

f(x) = -14x - 2

Step 2 - Now, evaluate the x-intercept of the above function.

0 = -14x - 2

x = -1/7

Step 3 - Now, determine the y-intercept of the given function.

f(x) = -2

Step 4 - Now, graph the equation of a line that passes through the points (-1/7,0) and (0,-2).

The graph of the function is attached below.

For more information, refer to the link given below:

https://brainly.com/question/14375099

Your answer is the last option, -148°F

The formula to convert temperatures from Celcius to Farenheit is:
(C° x 9/5) + 32 or (C° x 1.8) + 32

So to solve, plug in -100°C and follow the order of operations (PEMDAS).

(-100 x 1.8) + 32
(-180) + 32
-148

-100°C is equal to -148°F

The freezing point of the antifreeze solution in Fahrenheit is -148 °F, calculated using the formula for converting Celsius to Fahrenheit.

To convert a temperature from the Celsius scale to the Fahrenheit scale, you can use the following formula: F = (C  imes 9/5) + 32, where F represents the Fahrenheit temperature, and C represents the Celsius temperature.

In this case, we are given the freezing point of an antifreeze solution, which is -100 °C. Applying the formula, we get F = (-100 x 9/5) + 32. This calculation gives us F = (-180) + 32, which simplifies to F = -148 °F.

So, the freezing point of the antifreeze solution on the Fahrenheit scale is -148 °F.

By using the area of the ring we will see that the radius of the inner circle is 3cm.

I assume that we have some kind of ring. To get the area of the ring, we need to take the area of the circle defined by the outer radius of the ring, and subtract the area defined by the circle with the inner radius of the ring.

Remember that the area of a circle of radius R is:

A = pi*R^2

We know that:

The radius of the outer circle is 5cm, so its area is:

A = pi*(5cm)^2 = pi*25cm^2

And the area of the ring is pi*16 cm^2

Then the area of the inner circle should be such that:

pi*25cm^2 - A' = pi*16cm^2

Then, solving for A'

A' = pi*25cm^2 - pi*16cm^2 = pi*9cm^2 = pi*(3cm)^2

So the radius of the inner circle is 3cm.

If you want to learn more about circles, you can read:

https://brainly.com/question/1559324

Answer:

c. I, II, and III

Step-by-step explanation:

The union of two sets includes all elements from one set and all elements from the second set.

For our sets, this means all elements of set A, which includes region I and region II.  It also means all elements of set B, which includes region III.

Thus the answer is regions I, II and III.

Answer:

C

Step-by-step explanation:

Just took the test

The exact value of the arc length of the curve is 149.4 units

How to determine the exact arc length of the curve

From the question, we have the following parameters that can be used in our computation:

[tex]y = \dfrac35x^\frac53 - \dfrac34x^\frac13 + 8[/tex]

Also, we have the interval to be

-1 ≤ x ≤ 27

This means that the x valus are

x = -1 to x = 27

The arc length of the curve can be calculated using

[tex]\text{Length} = \int\limits^a_b {\sqrt{1 + ((dy)/(dx))^2}} \, dx[/tex]

Recall that

[tex]y = \dfrac35x^\frac53 - \dfrac34x^\frac13 + 8[/tex]

So, we have

[tex]\dfrac{dy}{dy} = x^\frac{2}{3}-\dfrac{1}{4x^\frac{2}{3}}[/tex]

This means that

[tex]\text{Length} = \int\limits^{27}_{-1} {\sqrt{1 + (x^\frac{2}{3}-\dfrac{1}{4x^\frac{2}{3}})^2}} \, dx[/tex]

Using a graphing tool, we have the integrand to be

[tex]\text{Length} = \dfrac{12x^\frac{5}{3}+15\sqrt[3]{x}}{20}|\limits^{27}_{-1}[/tex]

Expand and evaluate

[tex]\text{Length} = 149.4[/tex]

Hence, the exact arc length of the curve is 149.4 units

[tex] |\Omega|=2\cdot6=12\\
|A|=1\cdot2=2\\\\
P(A)=\dfrac{2}{12}=\dfrac{1}{6}\approx17% [/tex]

│FV│ = │VD│

Being D the directrix. Given that the focus F is on the y-axis and the directrix is parallel to the x-axis, then the vertex V will also be on this axis, so h = 0.

As │FV│ = │VD│, then:

[tex]k = \frac{6-2}{2}[/tex], that is the middle point of the segment FD, so:

V(0,2)

Now │FV│= │p│= │2-(-2)│=4

Given that the vertex and focus are below the directrix, then the parabola open down, therefore: [tex]p\ \textless \ 0[/tex]

Lastly, the equation is:

[tex]x^{2} = -4(4)(y-2) = -16y+32[/tex]
[tex]y = -\frac{ x^{2} }{16} + 2[/tex]

The equation of a parabola with a focus at (0,-2) and a directrix of y=6 is x^2 = -16(y - 2).

To find the equation of a parabola with a focus at (0,-2) and a directrix of y=6, you need to use the standard form of the equation of a parabola that opens upwards or downwards. The general form of this type of parabola is  (x - h)^2 = 4p(y - k), where (h,k) is the vertex of the parabola, and p is the distance from the vertex to the focus (if the parabola opens upwards or downward) or to the directrix (if the parabola opens sideway).

Given that the focus is at (0,-2) and the directrix is at y=6, the vertex of the parabola will be located midway between them. The distance between the focus and directrix is 8 units, so the vertex will be 4 units from each, which puts the vertex at (0, 2). Therefore, h=0 and k=2.

Since the focus is below the directrix, our parabola opens downward, and the value of p is negative. The distance p is half the distance between the focus and directrix, so p=-4. Plugging these values into the general form, we get (x - 0)^2 = 4(-4)(y - 2), which simplifies to x^2 = -16(y - 2).

This is the equation that represents the desired parabola.

0.8 or 80%.

The question is asking for the probability that a randomly chosen student is either a boy or a girl not wearing blue. There are 25 students in total, with 15 girls and 10 boys. Out of these, 5 girls and 4 boys are wearing blue. Therefore, the number of girls not wearing blue is 15 - 5 = 10 girls. Since all boys are considered in the probability, regardless of what they wear, we have 10 boys. So, we have 10 girls not wearing blue and 10 boys, totalling 20 students that match the criteria out of 25.

The probability can be calculated as follows:

( P(\text{{boy or girl not wearing blue}}) = \frac{{\text{{number of boys and girls not wearing blue}}}}{{\text{{total number of students}}}} = frac{{20}}{{25}} = 0.8 ) or 80%.

Therefore, the probability that a student picked at random is either a boy or a girl who is not wearing blue is 0.8 or 80%.

The correct answer is:

16√13.

Explanation:

The perimeter of a figure is found by adding together the lengths of all of the sides. The side lengths of this garden are: √13, √13, 7√13 and 7√13. This is because opposite sides in a rectangle are congruent.

Adding these together we have:

√13+√13+7√13+7√13 = (1+1+7+7)√13 = 16√13

Answer: just add 8 + 8 and you will get 16

The probability that a strawberry dessert contains whipped topping is approximately 69.2%

We are given that the probability a dessert contains strawberries (P(S)) is 26%, or 0.26, and the probability that a dessert contains both strawberries and whipped topping (P(S and W)) is 18%, or 0.18.

To find the probability that a randomly chosen strawberry dessert contains whipped topping (P(W|S)), we use the conditional probability formula:

[tex]P(W|S) = \frac{P(S and W)}{P(S)}[/tex]

Substituting the given values, we get:

[tex]P(W|S) = \frac{0.18}{0.26} \approx 0.6923[/tex]

Rounding to the nearest tenth of a percent we get the probability that a randomly chosen strawberry dessert contains whipped topping to be 69.2%.

Answer:

The solutions are B. -4 and C. -5

Step-by-step explanation:

For a quadratic equation of the form [tex]ax^2+bx+c=0[/tex] the solutions are

[tex]x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex]

For [tex]\mathrm{}\quad a=3,\:b=27,\:c=60:\quad x_{1,\:2}=\frac{-27\pm \sqrt{27^2-4\cdot \:3\cdot \:60}}{2\cdot \:3}[/tex]

[tex]x_1=\frac{-27+\sqrt{27^2-4\cdot \:3\cdot \:60}}{2\cdot \:3}\\\\x_1=\frac{-27+\sqrt{9}}{2\cdot \:3}\\\\x_1=\frac{-27+3}{2\cdot \:3}\\\\x_1=\frac{-24}{6} = -4[/tex]

[tex]x_2=\frac{-27-\sqrt{27^2-4\cdot \:3\cdot \:60}}{2\cdot \:3}\\\\x_2=\frac{-27-\sqrt{9}}{2\cdot \:3}\\\\x_2=\frac{-27-3}{2\cdot \:3}\\\\x_2=-\frac{30}{6} = -5[/tex]