Answer:
is linearly dependent set.
Step-by-step explanation:
Given: 
is a linearly dependent set in set of real numbers R
To show: the set is linearly dependent.
Solution:
If 
is a set of linearly dependent vectors then there exists atleast one 
such that 
Consider 
A linear transformation T: U→V satisfies the following properties:
1. 
2. 
Here, 
∈ U
As T is a linear transformation,
As is a linearly dependent set,
for some 
So, for some
Therefore, set is linearly dependent.
70 / 35 = 2
So the 70 pound dog weights double the weight of the 35 pound dog, so he will eat twice as much.
2 x 2.5 = 5 cups
Meaning he will eat 5 cups of dry dog food a day.
By the same logic a 140 pound dog would have to eat twice what a 70 pound dog eats, that is 10 cups of dry dog food a day, so 7 1/2 cups are not enough.
Challenge: per pound a dog eats more than a cat. Since you know a 140 pound dog eats 10 cups, you can dive 140 by 10, to find out that each cup feeds 14 pounds, so half a cup would be the amount of food to feed a 7 pound dog, meaning that a dog weighing 1 pound less than a 8 pound cat would need the same amount of food, so it eats more than a cat.
The distance across the lake, a is 611 yards
<h3>Right angle triangle:</h3>
Right angle triangle have one of its angles as 90 degrees. Therefore, the side a can be found using trigonometric ratios,
Therefore,
tan 47° = opposite / adjacent
tan 47° = a / 570
cross multiply
a = 570 tan 47°
a = 570 × 1.07236871002
a = 611.250164714
a = 611 yards
learn more on right triangle here: brainly.com/question/18450490
62 triangle a and 59 triangle b
Answer:
We can see that this is dependent probability. We can find dependent probability of happening event A then event B by multiplying probability of event A by probability of event B given that event A already happened.
Step-by-step explanation:
In our case event A is pirate hitting captain's ship and event B is captain missing pirate's ship. We have been given that pirate shoots first so pirate's ship can't be hit before pirate shoots his cannons. So probability of hitting captain's ship is 1/3. We have been given that if Captain Ben's ship is already hit then Captain Ben will always miss. So the probability of Captain missing the dread pirate's ship given the pirate Luis hitting the Captain ship is 1. Now to find probability that pirate hits Captain, but Captain misses we will multiply our both probabilities.
