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2 years ago

Calculate the slope of the line passing through (3.8) and (-4,2).

Mathematics

1 answer:

Talja [164]2 years ago

5 0

Answer:

6/7

comment if you need an explanation

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???????????????????whaaaat?

sergejj [24]

Answer:

dependentt

Step-by-step explanation:

8 0

2 years ago

I NEED URGENT ASSISTANCE

Nata [24]

Answer:

8 C sometimes

9 64x^8 y^11

10 c 1.28r^2/ t^9

Step-by-step explanation:

Algebraic Operations

Some basic rules must be fresh in our minds when trying to simplify complex algebraic expressions. For example, the power rule respect to the product or quotient:

$\displaystyle \left(\frac{x^n.y^m}{z^p}\right)^q= \frac{x^{qn}.y^{qm}}{z^{qp}}$

$\displaystyle \frac{x^n.x^q}{x^p}= x^{n+q-p}$

$\displaystyle x^{-n}=\frac{1}{x^n}$

Let's face the questions at hand

8. A number is raised to a negative exponent is negative?

Following the expressions recalled above, let's pick the expression

$\displaystyle 3^{-2}=\frac{1}{3^2}=\frac{1}{9}$

This is a negative power resulting in a positive number

Now we pick

$\displaystyle (-2)^{-3}=\frac{1}{(-2)^3}=-\frac{1}{8}$

This time, the negative power leads to a negative result, so it doesn't matter the sign of exponent to determine the sign of the result

Answer: C sometimes

9 simplify(4xy^2)^3(xy)^5

$(4xy^2)^3(xy)^5=4^3x^3(y^2)^3x^5y^5$

$=64x^3y^6x^5y^5$

$=64x^8y^{11}$

10 simplify(2t^-3)^3(0.4r)^2

$(2t^{-3})^3(0.4r)^2=2^3(t^{-3})^30.4^2r^2$

$=8t^{-9} 0.16 r^2$

$=1.28t^{-9} r^2$

$\displaystyle \frac{1.28 r^2}{t^9}$

7 0

3 years ago

Find (fog)(x) when f(x)=x^2-7x+12 and g(x)=3/x^2-16

Kazeer [188]

Answer:

$f( \frac{3}{ {x}^{2} - 16 } ) = ({\frac{3}{ {x}^{2} - 16 }})^{2} - 7(\frac{3}{ {x}^{2} - 16 }) + 12 \\ = \frac{9}{ {x}^{4} - 32 {x}^{2} + 256 } - \frac{21}{ {x}^{2} - 16} + 12$

4 0

2 years ago

1000 + 6901 - 64 = ?

Radda [10]

Answer:

7837

Step-by-step explanation:

1000 + 6901 - 64 = ?

=7837

8 0

2 years ago

The performance review scores for several employees in a work group are shown in the table. Which of the following measures woul

german

If I understand what you are asking, you can use the median as an average since it is not affected by the mean ( median is a measure of center). Since a mean is affected by outliers, it might lower the score if there are very low outliers. But note that if there are outliers way greater it can increase the average. You can use the Inter Quartile Range if shown on a box plot ( measure of variability).

8 0

3 years ago