When triangle ABC is rotated about side AB, what figure is formed?

cylinder
prism
pyramid
cone

Answer:

2/9

Step-by-step explanation:

To solve the equation, it was first required to get rid of the fraction by multiplying both sides by 7 and then to isolate r, the running miles, you subtract 105 from both sides resulting in Gisele running 35 miles per week.

The question is asking to find the number of miles Gisele runs each week, which is represented by the variable r in the equation. We start with the equation: 1/7(105 + r) = 20. This equation is derived from the fact that Gisele covers a total of 20 miles each day, for 7 days, and that total comprises both her cycling and running mileages. If she cycles 105 miles each week, then the distance she runs is the remaining part of those total 20 miles she covers each day. We multiply both sides of the equation by 7 to get rid of the fraction: 105 + r = 140. Now, if we subtract 105 from both sides of the equation, we get the solution for r as; r = 140 - 105 = 35 miles. Hence, Gisele runs 35 miles each week.

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Answer: 60 cents

Step-by-step explanation:

The square root of the given algebraic expression [tex]64y^{16}[/tex] is [tex]\bold{8y^8}[/tex]

"It is a mathematical statement which consists of numbers, variables and mathematical operations."

For given question,

We have been given an algebraic expression [tex]64y^{16}[/tex]

We need to find the square root of given algebraic expression.

[tex]\Rightarrow \sqrt{ 64y^{16}}\\\\= \sqrt{ 64\times y^{16}} \\\\=\sqrt{64} \times \sqrt{y^{16}} \\\\=8\times (y^{16})^{\frac{1}{2} }\\\\=8\times y^{\frac{16}{2} }\\\\= 8\times y^8[/tex]

Therefore, the square root of the given algebraic expression [tex]64y^{16}[/tex] is [tex]\bold{8y^8}[/tex]

Learn more about the algebraic expression here:

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I'm pretty sure its 1296π m³ Hope this helps!

1. The equation 5x^2+9x=4, can be rewritten as 5x^2+9x-4=0, where a=5 (the coefficient of the quadratic term), b = 9 (the coefficient of the linear term) and c = -4 (the independent term).

2. No attachment is shown.

3. To use the quadratic formula, the coefficients are: a = 2, b = -1 and c = 10. However, the discriminant, i. e., b^2 -4(a)(c) is negative, therefore, there is no real solution.  

4. You have to use the quadratic formula, with a = 4, b = 3 and c = -10, as follows:

[tex]x = \frac{-b \pm \sqrt{b^2 - 4(a)(c)}}{2(a)} [/tex]

[tex]x = \frac{-3 \pm \sqrt{3^2 - 4(4)(-10)}}{2(4)} [/tex]

[tex]x = \frac{-3 \pm \sqrt{169}}{8} [/tex]

[tex]x_1 = \frac{-3 + 13}{8} [/tex]

[tex]x_1 = \frac{5}{4} [/tex]

[tex]x_2 = \frac{-3 - 13}{8} [/tex]

[tex]x_2 = -2 [/tex]

5. Rewriting the formula, so that all the terms are on one side and ordering the polynomial expression, gives: -x2-6x+27=0. Here a = -1, b = -6 and c = 27.  Using the quadratic formula gives

[tex]x = \frac{-b \pm \sqrt{b^2 - 4(a)(c)}}{2(a)} [/tex]

[tex]x = \frac{6 \pm \sqrt{6^2 - 4(-1)(27)}}{2(-1)} [/tex]

[tex]x = \frac{6 \pm \sqrt{144}}{-2} [/tex]

[tex]x_1 = \frac{6 + 12}{-2} [/tex]

[tex]x_1 = -9 [/tex]

[tex]x_2 = \frac{6 - 12}{-2} [/tex]

[tex]x_2 = 3 [/tex]

6. When the second grade polynomial has all of its coefficients (a, b and c) different than zero, like this case, the best method for solving the equation is the quadratic formula  .

7. When the discriminant is greater than zero, there are two different solutions and when it is equal to zero, it has one double solution.

The discriminant for this case is 7^2 - 4(5)(-4) = 129. So, there are two different solutions .

8. Let's call  the sides of the rectangle a and b.

perimeter formula: 2a + 2b = 54 cm (eq. 1)

area formula: (a)(b) = 176 cm2 (eq. 2)

Isolating a in equation 1, gives (units are omitted):

2a + 2b = 54

2a = 54 - 2b

a = 27 - b (eq. 3)

Replacing equation 3 into equation 2, gives (units are omitted):

(a)(b) = 176

(27 - b)(b) = 176

-b^2 + 27b - 176 = 0

Using quadratic formula, we get

[tex]x = \frac{-27 \pm \sqrt{27^2 - 4(-1)(-176)}}{2(-1)} [/tex]

[tex]x = \frac{-27 \pm \sqrt{25}}{-2} [/tex]

[tex]x_1 = \frac{-27 + 5}{-2} [/tex]

[tex]x_1 = 11 [/tex]

[tex]x_2 = \frac{-27 - 5}{-2} [/tex]

[tex]x_2 = 16 [/tex]

Then, there are two possible values for b, 11 cm and 16 cm. Replacing these values into equation 3 gives:

a = 27 - 11 = 16 cm

a = 27 - 16 = 11 cm

Therefore, the two sides are 11 cm and 16 cm long, despite their names.

Answer:

Option a. [tex]\$6,455[/tex]  

Step-by-step explanation:

we know that    

The compound interest formula is equal to  

[tex]A=P(1+\frac{r}{n})^{nt}[/tex]  

where  

A is the Final Investment Value  

P is the Principal amount of money to be invested  

r is the rate of interest  in decimal

t is Number of Time Periods  

n is the number of times interest is compounded per year

in this problem we have      

[tex]t=4\ years\\ A=\$7,160.06\\ r=0.026\\n=4[/tex]  

substitute in the formula above  and solve for P

[tex]7,160.06=P(1+\frac{0.026}{4})^{4*4}[/tex]  

[tex]7,160.06=P(1+\frac{0.026}{4})^{4*4}[/tex]  

[tex]P=7,160.06/1.109227=\$6,455[/tex]  

Answer:

perfect competition

Step-by-step explanation:

because i did the test.

The fraction of the word 'Supercalifragilisticexpialidocious' that has the letter 'i' is 7/34, which is already in its simplest form.

The word 'Supercalifragilisticexpialidocious' contains 34 letters. To find what fraction of this word has the letter 'i' in it, we first need to count how many times the letter 'i' appears. The letter 'i' appears 7 times in the word. Therefore, the fraction of the word that has the letter 'i' in it is 7/34.

It is important to note that this fraction can be simplified. However, since both 7 and 34 are prime numbers with respect to each other (they have no common factors other than 1), the fraction 7/34 is already in its simplest form.