What is the midpoint of a line segment with the endpoints (-6, -3) and (9, -7)?

A scale diagram of a garden shows the length as 14.5 cm. If the scale is 1:150, what is the actual length? The garden is m in le

zvonat [6]

Answer:

21, 75 m

Step-by-step explanation:

The actual length ?

14, 5 cm × 150 = 2 175 cm = 21, 75 m

3 0

2 years ago

1. John and Carolyn are planning a picnic for some of their friends. In all there will be 28 people. They plan to barbecue steak

Sonja [21]

1. 28 x 3/4 = 21 kg

2. 1 hr = 60 mins

=》 1/4 hr = 60/4 = 15mins

4 0

1 year ago

Read 2 more answers

laura picks 7 apples. some apples are red and some are green. fewer than 4 of the apples are red. how many of lauras apples coul

Schach [20]

4 or more apples could be green because if there are 3 or less red then there must be 4-7 green

8 0

1 year ago

Read 2 more answers

Suppose that 5 out of 13 people are to be chosen to go on a mission trip. In how many ways can these 5 be chosen if the order in

algol [13]

5 People can be chosen in 1287 ways if the order in which they are chosen is not important.

Step-by-step explanation:

Given:

Total number of students= 13

Number of Students to be selected= 5

To Find :

The number of ways in which the 5 people can be selected=?

Solution:

Let us use the permutation and combination to solve this problem

$nCr=\frac{(n)!}{(n-r)!(r)!}$

So here , n =13 and r=5 ,

So after putting the value of n and r , the equation will be

$13C_5=\frac{(13)!}{(13-5)!(5)!}$

$13C_5=\frac{(13 \times12 \times11 \times10 \times9 \times8\times7 \times6 \times5 \times4 \times3 \times2 \times1)}{(8 \times7 \times6 \times5 \times4 \times3 \times2 \times1)(5 \times4 \times3 \times2 \times1)}$

$13C_5=\frac{(13 \times12 \times11 \times10 \times9 )}{((5 \times4 \times3 \times2 \times1)}$

$13C_5=\frac{154440}{120}$

$13C_5= 1287$

8 0

2 years ago

mrs.beluga is driving on a snow covered road with a drag factor of 0.2. she brakes suddenly for a deer. The tires leave a yaw ma

Aneli [31]

First, we are going to find the radius of the yaw mark. To do that we are going to use the formula: $r= \frac{c^2}{8m} + \frac{m}{2}$
where
$c$ is the length of the chord
$m$ is the middle ordinate
We know from our problem that the tires leave a yaw mark with a 52 foot chord and a middle ornate of 6 feet, so $c=52$ and $m=6$ . Lets replace those values in our formula:
$r= \frac{52^2}{8(6)} + \frac{6}{2}$
$r= \frac{2704}{48} +3$
$r= \frac{169}{3} +3$
$r= \frac{178}{3}$

Next, to find the minimum speed, we are going to use the formula: $s= \sqrt{15fr}$
where
$f$ is drag factor
 $r$ is the radius
We know form our problem that the drag factor is 0.2, so $f=0.2$ . We also know from our previous calculation that the radius is $\frac{178}{3}$ , so $r= \frac{178}{3}$ . Lets replace those values in our formula:
$s= \sqrt{(15)(0.2)( \frac{178}{3}) }$
$s= \sqrt{178}$
$s=13.34$ mph

We can conclude that Mrs. Beluga's minimum speed before she applied the brakes was 13.34 miles per hour.

8 0

2 years ago