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3 years ago

Liner equations please help with answer!

Mathematics

1 answer:

svetlana [45]3 years ago

6 0

Answer:

n = - $\frac{4}{7}$

Step-by-step explanation:

Given

$\frac{2}{3}$ (1 + n) = - $\frac{1}{2}$ n

Multiply both sides by 6 ( the LCM of 3 and 2 ) to clear the fractions

4(1 + n) = - 3n ← distribute left side

4 + 4n = - 3n ( add 3n to both sides )

4 + 7n = 0 ( subtract 4 from both sides )

7n = - 4 ( divide both sides by 7 )

n = - $\frac{4}{7}$

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You have 42,784 grams of a radioactive kind of curium. If its half-life is 18 years, how much will be left after 72 years?

s344n2d4d5 [400]

Answer:

2,674.14 g

Step-by-step explanation:

Recall that the formula for radioactive decay is

N = N₀ e^(-λt)

where,

N is the amount left at time t

N₀ is the initial amount when t=0, (given as 42,784 g)

λ = coefficient of radioactive decay

= 0.693 ÷ Half Life

= 0.693 ÷ 18

= 0.0385

t = time elapsed (given as 72 years)

e = exponential constant ( approx 2.7183)

If we substitute these into our equation:

N = N₀ e^(-λt)

= (42,787) (2.7183)^[(-0.0385)(72)]

= (42,787) (2.7183)^(-2.7726)

= (42,787) (0.0625)

= 2,674.14 g

6 0

3 years ago

{2, -1, -4, -7, ...}

lesantik [10]

Answer:

-3 every time

Step-by-step explanation:

3 0

3 years ago

Alex scored 70% in a spelling test with 20 quetions

amm1812

Answer:

This means that Alex answered 14 questions correctly

Step-by-step explanation:

Step 1: Find an equation

20 * 0.7 = x

Step 2: Multiply

20 * 0.7 = x

14 = x

Answer: This means that Alex answered 14 questions correctly

5 0

3 years ago

Read 2 more answers

Identify the interval on which the quadratic function is positive.

Alenkasestr [34]

Answer:

$\textsf{1. \quad Solution: $1 < x < 4$,\quad Interval notation: $(1, 4)$}$

$\textsf{2. \quad Solution: $-2 < x < 4$,\quad Interval notation: $(-2, 4)$}$

Step-by-step explanation:

<h3>Question 1</h3>

The intervals on which a quadratic function is positive are those intervals where the function is above the x-axis, i.e. where y > 0.

The zeros of the quadratic function are the points at which the parabola crosses the x-axis.

As the given quadratic function has a negative leading coefficient, the parabola opens downwards. Therefore, the interval on which y > 0 is between the zeros.

To find the zeros of the given quadratic function, substitute y = 0 and factor:

$\begin{aligned}y&= 0\\\implies -7x^2+35x-28& = 0\\-7(x^2-5x+4)& = 0\\x^2-5x+4& = 0\\x^2-x-4x+4& = 0\\x(x-1)-4(x-1)&= 0\\(x-1)(x-4)& = 0\end{aligned}$

Apply the zero-product property:

$\implies x-1=0 \implies x=1$

$\implies x-4=0 \implies x=4$

Therefore, the interval on which the function is positive is:

Solution: 1 < x < 4
Interval notation: (1, 4)

<h3>Question 2</h3>

The intervals on which a quadratic function is negative are those intervals where the function is below the x-axis, i.e. where y < 0.

The zeros of the quadratic function are the points at which the parabola crosses the x-axis.

As the given quadratic function has a positive leading coefficient, the parabola opens upwards. Therefore, the interval on which y < 0 is between the zeros.

To find the zeros of the given quadratic function, substitute y = 0 and factor:

$\begin{aligned}y&= 0\\\implies 2x^2-4x-16& = 0\\2(x^2-2x-8)& = 0\\x^2-2x-8& = 0\\x^2-4x+2-8& = 0\\x(x-4)+2(x-4)&= 0\\(x+2)(x-4)& = 0\end{aligned}$