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3 years ago

The sum of the consecutive numbers is 111. What is the smallest of the three numbers?

2 answers:

8 0

Let
x = first consecutive number
x + 1 = second consecutive number
x + 2 = third consecutive number

The sum of these numbers are 111.
x + x + 1 + x + 2 = 111
3x = 111 - 1 - 2
3x = 108
x = 36

Therefore, the smallest number is 36 because other numbers are 37 and 38.

Serga [27]3 years ago

3 0

Answer=36 + 37 + 38 = 111 (36 is the smallest number)

In algebraic form it is x+ x+1 + x+2 = 111 First you minus the 2 and the 1 to give you 108 and then divide that by 3 to give you x (36). That gives you the smallest of the consecutive numbers. Then add on the one on 36 to give 37 and another one to give 38.

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Problem: A scuba diver swam 30 feet below sea level. After a couple of minutes, he went down 10 feet more. Find the total distan

QveST [7]

Answer: -40 feet

Step-by-step explanation: the scuba diver starts at sea level (0) then swims down 30 feet (-30). after a couple minutes, goes down 10 feet more (-40) the scuba diver will now be -40 down from sea level. so it should be -40 because the scuba diver swam that much down.

lmk if this is incorrect, i'm not quit sure if it is or not. if i'm wrong, i'm so sorry!!

3 0

3 years ago

If Vector u=(5,-7) and v=(-11,3), 2v-6u=_____ and ||2v-6u||≈_____

Sauron [17]

<h2>Answer:</h2>

$2\vec{v}-6\vec{u}=(-52,48) \\ \\ ||2\vec{v}-6\vec{u}||=75.28}$

<h2>Step-by-step explanation:</h2>

In this problem we have two vectors:

$\vec{u}=(5,-7) \ and \ \vec{v}=(-11,3)$

So we need to find two things:

$2\vec{v}-6\vec{u}$

and:

$||2\vec{v}-6\vec{u}||$

FIRST:

In this case we have the multiplication of vectors by scalars. A scalar is a simple number, so:

$2\vec{v}-6\vec{u} \\ \\ Replace \ \vec{v} \ and \ \vec{u} \ by \ the \ given \ vectors: \\ \\ 2(-11,3)-6(5,-7) \\ \\ Multiply \ each \ component \ by \ the \ corresponding \ scalar:\\ \\ (2\times (-11),2\times 3)+(-6\times 5,-6\times (-7)) \\ \\ (-22,6)+(-30,42) \\ \\ Sum \ of \ vectors: \\ \\ (-22-30,6+42) \\ \\ \therefore \boxed{(-52,48)}$

SECOND:

If we name:

$\vec{w}=2\vec{v}-6\vec{u}$

Then, $||2\vec{v}-6\vec{u}||$ is the magnitude of the vector $\vec{w}$ . Therefore:

$||\vec{w}||=||2\vec{v}-6\vec{u}|| \\ \\ ||\vec{w}||=||(-52,48)|| \\ \\ ||\vec{w}||=\sqrt{(-58)^2+48^2} \\ \\ ||\vec{w}||=\sqrt{3364+2304} \\ \\ ||\vec{w}||=\sqrt{5668} \\ \\ \boxed{||\vec{w}||=75.28}$

5 0

2 years ago

Read 2 more answers

PLEASE HELP ME ANSWER! im so desperate bro !!

sergey [27]

a)13

b) skewed

c) IQR

d)Sam's observation is correct

e)6.17

Step-by-step explanation:

Step 1 :

From the histogram we can infer that

0 to 2 texts has been sent by 1 student

2 to 4 texts has been sent by 3 student

4 to 6 texts has been sent by 4 student

6 to 8 texts has been sent by 4 student

8 to 10 texts has been sent by 5 student

Step 2:

From step 1 , we can see that there are total of 17 students represented by the histogram

Step 3:

This is a skewed histogram because we have all large values of students on the right of the histogram and smaller values on the left of the histogram.

In symmetrical histogram, there would be large values on the center and the smaller values on both sides.

Step 4:

IQR would be better recommendation for this as this is a skewed histogram

Standard deviation would be recommended for a symmetrical histogram

Step 5:

Total number of students is 17

Number of students who text between 8 to 10 messages is 5

Hence the students who text between 8 and 10 messages is 5/17

which is nearly 1/3 rd of the total students.

Hence Sam's observation is correct

Step 6:

Mean amount of the texts = (1 * 3 + 3* 3 + 5*4 +7*4 + 9* 5) /17 = 6.17

6 0

3 years ago

Explore the properties of angles formed by

deff fn [24]

Answer:

m<AEC = 105

Step-by-step explanation:

(just completed the assignment) they are vertical angles, so they have the same measure

6 0

2 years ago

Read 2 more answers

6. Find the areas of the rectangles with the following side lengths.

Vlad1618 [11]

7/6 in and 6/7 in is the answer

4 0

2 years ago