Points

What value of c makes x2 + 6x + c a perfect square trinomial? 3 6 9 12

miskamm [114]

The answer is 3 jehfhdhhedhhshshdhhsgdjshshdhdh

5 0

3 years ago

A music website charges x dollars for individual songs and y dollars for entire albums. Person A pays $25.92 to download 6 indiv

kati45 [8]

Amount charged for downloading individual songs = x dollars
Amount charged for downloading an entire album = y dollars
In respect to person A:
6x + 2y = 25.92
3x + y = 12.96
y = 12.96 - 3x
In respect to person B:
4x + 3y = 33.93
Putting the value of y from the first equation in the second, we get
4x + 3(12.96 - 3x) = 33.93
4x + 38.88 - 9x = 33.93
- 5x = - 4.95
x = 0.99 dollars
Putting the value of x in the first equation, we get
y = 12.96 - 3x
= 12.96 - 2.97
= 9.99 dollars

7 0

3 years ago

Please help!!

garik1379 [7]

Given:

The radius of the cone = 14 ft

The slant height of the cone = 27 ft

To find the lateral surface area and the total surface area of the given cone.

Formula

The lateral surface area of the cone is

$LSA = \pi rl$ and

The total surface area of the cone is

$TSA = \pi r^{2} + \pi r l$

where,

r be the radius and

l be the slant height of the cone.

Now

Taking r = 14 and l = 27 we get

$LSA = \pi (14)(27)$ sq ft

or, $LSA = 1187.5$ sq ft

Again,

$TSA = \pi(14^2)+\pi (14)(27)$ sq ft

or, $TSA = 1803.3$ sq ft

Hence,

The lateral surface area of the cone is 1187.5 sq ft and the total surface area of the cone is 1803.3 sq ft. Option C.

5 0

3 years ago

which of the following is equivalent to 3 sqrt 32x^3y^6 / 3 sqrt 2x^9y^2 where x is greater than or equal to 0 and y is greater

Nutka1998 [239]

Answer:

$\frac{\sqrt[3]{16y^4}}{x^2}$

Step-by-step explanation:

The options are missing; However, I'll simplify the given expression.

Given

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$

Required

Write Equivalent Expression

To solve this expression, we'll make use of laws of indices throughout.

From laws of indices $\sqrt[n]{a} = a^{\frac{1}{n}}$

So,

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$ gives

$\frac{(32x^3y^6)^{\frac{1}{3}}}{(2x^9y^2)^\frac{1}{3}}$

Also from laws of indices

$(ab)^n = a^nb^n$

So, the above expression can be further simplified to

$\frac{(32^\frac{1}{3}x^{3*\frac{1}{3}}y^{6*\frac{1}{3}})}{(2^\frac{1}{3}x^{9*\frac{1}{3}}y^{2*\frac{1}{3}})}$

Multiply the exponents gives

$\frac{(32^\frac{1}{3}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

Substitute $2^5$ for 32

$\frac{(2^{5*\frac{1}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

$\frac{(2^{\frac{5}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

From laws of indices

$\frac{a^m}{a^n} = a^{m-n}$

This law can be applied to the expression above;

$\frac{(2^{\frac{5}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$ becomes

$2^{\frac{5}{3}-\frac{1}{3}}x^{1-3}*y^{2-\frac{2}{3}}$

Solve exponents

$2^{\frac{5-1}{3}}*x^{-2}*y^{\frac{6-2}{3}}$

$2^{\frac{4}{3}}*x^{-2}*y^{\frac{4}{3}}$

From laws of indices,

$a^{-n} = \frac{1}{a^n}$ ; So,

$2^{\frac{4}{3}}*x^{-2}*y^{\frac{4}{3}}$ gives

$\frac{2^{\frac{4}{3}}*y^{\frac{4}{3}}}{x^2}$

The expression at the numerator can be combined to give

$\frac{(2y)^{\frac{4}{3}}}{x^2}$

Lastly, From laws of indices,

$a^{\frac{m}{n} = \sqrt[n]{a^m}$ ; So,

$\frac{(2y)^{\frac{4}{3}}}{x^2}$ becomes

$\frac{\sqrt[3]{(2y)}^{4}}{x^2}$

$\frac{\sqrt[3]{16y^4}}{x^2}$

Hence,

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$ is equivalent to $\frac{\sqrt[3]{16y^4}}{x^2}$

8 0

3 years ago

Suppose y varies directly with x, and y = 8 when x = –6. What direct variation equation relates x and y? What is the value of y

Agata [3.3K]

Given:
y varies directly with x
y = 8 when x = -6

what is the value of y when x = -2?

y = kx

we need to get the constant of variation.
8 = k(-6)
8/-6 = k
4/-3 = k or - 1 1/3

y = -4/3 (-2)
y = (-4*-2) / 3
y = 8 / 3
y = 2 2/3 OR 2.67

The correct answer is Choice B. y = -1.33x ; 2.67

4 0

3 years ago