All categories

3 years ago

Find the slope of the line that passes through the points (3, 6) and (5, 3).

Mathematics

1 answer:

DanielleElmas [232]3 years ago

7 0

Hello!

To find the slope of the line that passes through the points (3, 6) and (5, 3), we will need to use the slope formula.

Slope formula is $\frac{y_{2}-y_{1} }{x_{2}-x_{1}}$ .

Now, we should assign the given points as x1, y1 or x2, y2. The point (3, 6) can be assigned to x1, y1, and (5, 3) to x2, y2.

Now, substitute the given points into the slope formula.

$\frac{3-6}{5-3} = -\frac{3}{2}$

Therefore, the slope of the line is choice A, -3/2.

You might be interested in

The area of a rectangular field is 8811 m2. if the width of the field is 89m, what is the length?

xeze [42]

Length = 99 because...
since you multiply to get the area, in order to find the missing term of an area (in this case its length), you do the opposite of what you started with...so if you multiplied to get the area, you need to divide to get the missing side...
so 8811 divided by 89 = 99
99 x 89 = 8811

4 0

3 years ago

Multiply. Write your answer as a fraction in simplest form 9/10 x 2/3

Oduvanchick [21]

Answer:

3/5

Step-by-step explanation:

9/10 and 2/3 can cross cancel

3 goes into 9, 3 times

2 goes into 10, 5 times

they both go into themselves once

our new fractions are 3/5 and 1/1 which equals 3/5

8 0

3 years ago

Read 2 more answers

A certain chalkboard manufacturer determines that their largest blackboard model has a mean length of 5.00 m and a standard devi

Igoryamba

Answer:

23 chalkboards

Step-by-step explanation:

Given:

Mean length = 5 m

Standard deviation = 0.01

Number of units ordered = 1000

Now,

The z factor = $\frac{\textup{x - Mean}}{\textup{standard deviation}}$

The z factor = $\frac{\textup{4.98 - 5}}{\textup{0.01}}$

Z = - 2

Now, the Probability P( length < 4.98 )

Also, From z table the p-value = 0.0228

therefore,

P( length < 4.98 ) = 0.0228

Hence, out of 1000 chalkboard ordered (0.0228 × 1000) = 23 chalkboard are likely to have lengths of under 4.98 m.

7 0

3 years ago

VEEL

Andre45 [30]

Answer:

$a_n=-3(3)^{n-1}$ ; {-3,-9, -27,- 81, -243, ...}

$a_n=-3(-3)^{n-1}$ ; {-3, 9,-27, 81, -243, ...}

$a_n=3(\frac{1}{2})^{n-1}$ ; {3, 1.5, 0.75, 0.375, 0.1875, ...}

$a_n=243(\frac{1}{3})^{n-1}$ ; {243, 81, 27, 9, 3, ...}

Step-by-step explanation:

The first explicit equation is

$a_n=-3(3)^{n-1}$

At n=1,

$a_1=-3(3)^{1-1}=-3$

At n=2,

$a_2=-3(3)^{2-1}=-9$

At n=3,

$a_3=-3(3)^{3-1}=-27$

Therefore, the geometric sequence is {-3,-9, -27,- 81, -243, ...}.

The second explicit equation is

$a_n=-3(-3)^{n-1}$

At n=1,

$a_1=-3(-3)^{1-1}=-3$

At n=2,

$a_2=-3(-3)^{2-1}=9$

At n=3,

$a_3=-3(-3)^{3-1}=-27$

Therefore, the geometric sequence is {-3, 9,-27, 81, -243, ...}.

The third explicit equation is

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

At n=1,

$a_1=3(\frac{1}{2})^{1-1}=3$

At n=2,

$a_2=3(\frac{1}{2})^{2-1}=1.5$

At n=3,

$a_3=3(\frac{1}{2})^{3-1}=0.75$

Therefore, the geometric sequence is {3, 1.5, 0.75, 0.375, 0.1875, ...}.

The fourth explicit equation is

$a_n=243(\frac{1}{3})^{n-1}$

At n=1,

$a_1=243(\frac{1}{3})^{1-1}=243$

At n=2,

$a_2=243(\frac{1}{3})^{2-1}=81$

At n=3,

$a_3=243(\frac{1}{3})^{3-1}=27$

Therefore, the geometric sequence is {243, 81, 27, 9, 3, ...}.

6 0

3 years ago

Can someone explain to me how I am supposed to do this please?

balandron [24]

I think Sam is correct cuz SQ is a bigger rectangle and for OM it is smaller and we only know the width

8 0

3 years ago