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3 years ago

When solving a system of inequalities, which of the following methods are used?

Mathematics

1 answer:

Sedbober [7]3 years ago

8 0

Answer:

Step-by-step explanation:

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Pls answer 2,3,5,6 and thank you for helping and pls work it out

skad [1K]

Answer:

2: add 4

3: 6

5: multiply by 2

6: 7

Step-by-step explanation:

7 0

3 years ago

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Marina CMI [18]

Answer:

D. 3x - 3

Step-by-step explanation:

To find (f-g)(x) you have to subtract each fuction

(f-g)(x) = f(x) - g(x)

(f-g)(x) = 5x -2 - (2x+1)

(f-g)(x) = 5x -2 -2x -1

(f-g)(x) = 3x -3

7 0

3 years ago

Read 2 more answers

What is the slope of a line perpendicular to 3x- 2y= 6

Tema [17]

Answer:

y = (-2/3)x + 5

2y=3x-6 Then divide by 2. y=3/2x-6 Now it is in slope-intercept form. The slope for this particular equation is 3/2 (m). The slope of a line that is perpendicular to this would be -2/3, the negative reciprocal of 3/2.

Step-by-Step:

HOPE THIS HELPED!!!

5 0

3 years ago

Read 2 more answers

Darla is taking a survey at her school. She finds that 20 out of 100 students have fish as pets. This means that her experimenta

Arisa [49]

Answer:

The second option

Step-by-step explanation:

This is because the more samples she has, the closer her experimental probability will be to the real value.

6 0

3 years ago

Help ASAP show work please thanksss!!!!

Llana [10]

Answer:

$\displaystyle log_\frac{1}{2}(64)=-6$

Step-by-step explanation:

<u>Properties of Logarithms</u>

We'll recall below the basic properties of logarithms:

$log_b(1) = 0$

Logarithm of the base:

$log_b(b) = 1$

Product rule:

$log_b(xy) = log_b(x) + log_b(y)$

Division rule:

$\displaystyle log_b(\frac{x}{y}) = log_b(x) - log_b(y)$

Power rule:

$log_b(x^n) = n\cdot log_b(x)$

Change of base:

$\displaystyle log_b(x) = \frac{ log_a(x)}{log_a(b)}$

Simplifying logarithms often requires the application of one or more of the above properties.

Simplify

$\displaystyle log_\frac{1}{2}(64)$

Factoring $64=2^6$ .

$\displaystyle log_\frac{1}{2}(64)=\displaystyle log_\frac{1}{2}(2^6)$

Applying the power rule:

$\displaystyle log_\frac{1}{2}(64)=6\cdot log_\frac{1}{2}(2)$

Since

$\displaystyle 2=(1/2)^{-1}$

$\displaystyle log_\frac{1}{2}(64)=6\cdot log_\frac{1}{2}((1/2)^{-1})$

Applying the power rule:

$\displaystyle log_\frac{1}{2}(64)=-6\cdot log_\frac{1}{2}(\frac{1}{2})$

Applying the logarithm of the base:

$\mathbf{\displaystyle log_\frac{1}{2}(64)=-6}$

5 0

2 years ago