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3 years ago

G(x) = 2x^3 when g(3) ????

2 answers:

7 0

The answer is 54.

You substitute the 3 into the x and get g(3) = 2(3)^3

Solve the exponent first. 3^3 is 27

multiply 27 with 2 and you get 54

Aleks [24]3 years ago

3 0

Answer:18

Step-by-step explanation:3^3 times 2

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I WILL GIVE BRAINLEST!!! (6 points!)

vlada-n [284]

Answer:

OPTION A: 2x + 3y = 5

Step-by-step explanation:

The product of slopes of two perpendicular lines is -1.

We rewrite the given equation as follows:

2y = 3x + 2

⇒ y = $\frac{3}{2}x + 1$

The general equation of the line is: y = mx + c, where 'm' is the slope of the line.

Here, m = $\frac{3}{2}$ .

Therefore, the slope of the line perpendicular to the line given = $\frac{-2}{3}$ because $\frac{3}{2} \times \frac{-2}{3} = -1$ .

To determine the equation of the line passing through the given point and a slope we use the Slope - One - point formula which is:

y - y₁ = m(x - x₁)

The point is: (x₁, y₁) = (-2, 3)

Therefore, the equation is:

y - 3 = $\frac{-2}{3}$ (x + 2) $

⇒ 3y - 9 = -2(x + 2)

⇒ 3y - 9 = -2x - 4

⇒ 2x + 3y = 5 is the required equation.

6 0

3 years ago

Read 2 more answers

The 7th grade has 159 students. There are twice as many boys in the grade as there are girls.

aleksklad [387]

Answer:

106 boys

Step-by-step explanation:

7th grade = 159 students

Boys to girls ratio:

2:1

Find two thirds of 159:

2/3*159 = 106 boys

159-106= 53 girls

Check:

53*2 = 106

3 0

3 years ago

Read 2 more answers

Let <img src="https://tex.z-dn.net/?f=i" id="TexFormula1" title="i" alt="i" align="absmiddle" class="latex-formula"> be the imag

VLD [36.1K]

$Hey~freckledspots!\\----------------------$

$We~will~solve~for~i^{425}!$

$Rule~of~exponent: a^{b + c} = a^ba^c\\Apply:~i^{425}~=~i^{424}i\\ \\Rule~of~exponent: a^{bc} = (a^{b})^c\\Apply: i^{424} = i(i^2)^{212} \\\\Rule~of~imaginary~number: i^2 = -1\\Apply: i(i^2)^{212} = -1^{212}i\\\\Rule~of~exponent~if~n~is~even: -a^n = a^n\\Apply: -1^{212}i = 1^{212}i\\\\Simplify: 1^{212}i = 1i\\Multiply: 1i * 1 = i\\----------------------\\$

$Now~let's~solve~1^{14}!\\\\Rule~of~exponent: a^{b + c} = a^ba^c\\Apply: i^{14} = (i^2)^7\\\\Rule~of~imaginary~number: i^2 = -1\\Apply: (i^2)^7 = -1^7\\\\Rule~of~exponent~if~n~is~odd: (-a)^n = -a^n\\Apply: -1^7 = -1^7\\\\Simplify: -1^7 = -1\\----------------------\\Now,~we~have: i-1+i^{-14}+i^{44}\\----------------------$

$Now~lets~solve~i^{-14}\\\\Rule~of~exponent: a^{-b} = \frac{1}{a^b} \\Apply: i^{-14} = \frac{1}{i^{14}} \\\\Rule~of~exponent: a^{bc} = (a^b)^c\\Apply: \frac{1}{i^{14}} = \frac{1}{(i^2)^7}\\ \\Rule~of~imagianry~number: i^2 = -1\\Apply: \frac{1}{(i^2)^7} = \frac{1}{-1^7} \\\\Simplify: \frac{1}{-1^7} = \frac{1}{-1} \\\\Rule~of~fractions: \frac{a}{-b} = -\frac{a}{b} \\Apply: \frac{1}{-1} = -\frac{1}{1} = -1\\----------------------\\Now,~we~have: i-1-1+i^44\\----------------------$

$Now~let's~solve~i^{44}!\\\\Rule~of~exponent: a^{bc} = (a^b)^c\\Apply: i^{44} = (i^2)^{22}\\\\Rule~of~imaginary~numbers: i^2 = -1\\Apply: (i^2)^{22} = -1^{22}\\\\Rule~of~exponent~if~n~is~even: (-a)^n = a^n\\Apply: -1^{22} = 1^{22}\\\\Simplify: 1^{22} = 1\\----------------------\\Now,~we~have~i-1-1+1\\----------------------$