Please help will mark brainliest .

What is the slope of the line that passes through the point (9,4) and (3,9)?

Fantom [35]

Step-by-step explanation:

m is for slope of the line that passes through

$m = \frac{y_2 - y_1}{x_2 - x_1}$

$m = \frac{9 - 4}{3 - 9}$

$m = \frac{5}{ - 6} = - \frac{5}{6}$

8 0

2 years ago

What is the image of point (8,-4) under the rotation R90° about the origin?

mr Goodwill [35]

Answer:

D). (-4,-8)

Step-by-step explanation:

An image at (8,-4) if rotated at an angle of 90° wipl have another location.

First of all, an image at (8,-4) is in the fourth quadrant, and if it's to rotate clockwise at 90° iths supposed to be in the third quadrant.

And in the third quadrant both x and y is negative.

So the new position is at (-4,-8)

7 0

3 years ago

Given log630 ≈ 1.898 and log62 ≈ 0.387, log615 =

Tom [10]

$\log_630\approx1.898\\\\\log_62\approx0.387\\\\\log_615=\log_6\dfrac{30}{2}=\log_630-\log_62\approx1.898-0.387=1.511\\\\Used:\\\\\log_ab-\log_ac=\log_a\dfrac{b}{c}$

6 0

3 years ago

Read 2 more answers

Question 5 of 35:<br>find the value of 10!/8!

kap26 [50]

9514 1404 393

Answer:

90

Step-by-step explanation:

10! = 10 × 9 × 8!

so 10!/8! = 10×9 = 90

__

The factorial of an integer is the product of that integer and all the positive integers less than it. So, 3! = 3×2×1, for example. 0! is defined as 1.

7 0

3 years ago

In an experiment, a fair coin is tossed 13 times and the face that appears (H for head or T for tail) for each toss is recorded.

zalisa [80]

Answer:

1) 1 element

2) 13 elements

3) 22 elements

4) 40 elements

Step-by-step explanation:

1) Only one element will have no tails: the event that all the coins are heads.

2) 13 elements will have exactly one tile. Basically you have one element in each position that you can put a tail in.

3) There are ${13 \choose 2} = 78$ elements that have exactly 2 tails. From those elements we have to remove the only element that starts and ends with a tail and in the middle it has heads only and the elements that starts and ends with a head and in the 11 remaining coins there are exactly 2 tails. For the last case, there are ${11 \choose 2} = 55$ possibilities, thus, the total amount of elements with one tile in the border and another one in the middle is 78-55-1 = 22

4) We can have:

A pair at the start/end and another tail in the middle (this includes a triple at the start/end)

One tail at the start/end and a pair in the middle (with heads next to the tail at the start/end)

For the first possibility there are 2 * 11 = 22 possibilities (first decide if the pair starts or ends and then select the remaining tail)

For the second possibility, we have 2*9 = 18 possibilities (first, select if there is a tail at the end or at the start, then put a head next to it and on the other extreme, for the remaining 10 coins, there are 9 possibilities to select 2 cosecutive ones to be tails).

This gives us a total of 18+22 = 40 possibilities.

5 0

3 years ago