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3 years ago

How do you solve these

Mathematics

2 answers:

mestny [16]3 years ago

4 0

For number 1, you solve it like an equation.
My teacher always says this," Undo what has been done."
It means to do the opposite done to the variable.
3/8x>9
3/8 is being multiplied to the variable so if you do the opposite, you divide 3/8 to each side.
x>24

Number 2
21g>105
You divide each side by 21
g>5

MaRussiya [10]3 years ago

3 0

25)divide by 3/8 on both sides

26) divide by 21 both sides

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What is the 4th term in the sequence b(n)=1(-2)^n-1

marishachu [46]

Answer:

<h2>The answer is 8</h2>

Step-by-step explanation:

From the question the formula for the sequence is

$b(n) = 1 ({ - 2})^{n - 1}$

where

n is the nth term in the sequence

Since we are finding the 4th term n = 4

Substitute this value of n into the above formula

That's

$b(4) = 1 ({ - 2})^{4 - 1} \\ = 1 ({ - 2})^{3} \\ = { - 2}^{3}$

We have the final answer as

Hope this helps you

7 0

3 years ago

Pls anybody know this ?

MakcuM [25]

Answer:

Concept: Ordinary Mathematics

Just a general tip: Zoom away further from the camera to capture the question clearer.

Calculate the Total: 31 students
A= 4/31= 13%
(A+B)/31 = 14/31 = 45%
Can't see clearly what it says

8 0

3 years ago

Solve each equation. Round your answers to the nearest ten-thousandth. Please show work. Part 3A

Butoxors [25]

Answer:

19.

log9(5x^2 + 10) - log9(10) = 1

<=> log9((5x^2 + 10)/10) = log9(9)

<=> (5x^2 + 10)/10 = 9

<=> 5x^2 + 10 = 90

<=> 5x^2 = 80

<=> x^2 = 16

<=> x = +/- (4)

20.

log5(2x^2 + 4) + log5(3) = 2

<=> log5((2x^2 + 4) x 3) = log3(9)

<=> 6x^2 + 12 = 9

<=> 6x^2 = -3

=> No real x satisfies. ( x^2 always larger or equal to 0)

21.

log6(8) + log6(7 - 2x^2) = 2

<=> log6(8 x (7 - 2x^2)) = log6(36)

<=> 56 - 16x^2 = 36

<=> 16x^2 = 20

<=> x^2 = 5/4

<=> x = +/- sqrt(5/4)

Hope this helps!

5 0

3 years ago

Solve the system by substitution.<br> y = -4x<br> 23 - 5y = 44

spin [16.1K]

<h3><u>Explanation</u></h3>

Given the system of equations.

$\begin{cases} y = - 4x \\ 23 - 5y = 44 \end{cases}$

Substitute y = -4x in the second equation.

$23 - 5( - 4x) = 44 \\ 23 + 20x = 44 \\ 20x = 44 - 23 \\ 20x = 21 \\ x = \frac{21}{20}$

Substitute the value of x in any given equations. I will substitute the value of x in the first equation.

$y = - 4x \Longrightarrow y = - 4( \frac{21}{20} ) \\ y = \cancel{ - 4}( \frac{21}{ \cancel{20}} ) \Longrightarrow y = - \frac{21}{5} \\ y = - \frac{21}{5}$

Answer Check by substituting both values in two equations.

<u>First</u><u> </u><u>Equation</u>

$y = - 4x \Longrightarrow - \frac{21}{5} = - 4( \frac{21}{20} ) \\ - \frac{21}{5} = \cancel{ - 4}( \frac{21}{ \cancel{20}} ) \\ - \frac{21}{5} = - \frac{21}{5} \: \: \: \checkmark$