All categories

2 years ago

How to find these key features to sketch a rough graph

Mathematics

1 answer:

Ivahew [28]2 years ago

8 0

There’s no question

You might be interested in

IF THREE DIAGONALS ARE DRAWN INSIDE A HEXAGON WITH EACH ONE PASSING THROUGH THE CENTER POINT OF THE HEXAGON, HOW MANY TRIANGLES

frosja888 [35]

Answer: 6 triangles

Step-by-step explanation:

hope this helps :)

3 0

1 year ago

Find two consecutive even integers such that five times the smaller integer is ten more than three times the larger integer.

quester [9]

5n = 3(n + 2) + 10
5n = 3n + 16
12n = 16
n = 8

8 is smaller integer and 10 is the larger.

8 0

2 years ago

Help please :( points !

mars1129 [50]

Answer: 2/5

Explanation: 10+8+12 = 30 12/30 = .4 .4 = 2/5

4 0

2 years ago

Read 2 more answers

The total surface area of a cuboid is 112cm^2. find the value of x. values given: 10cm and 2cm.

Tatiana [17]

Answer:

x= 5.6

Step-by-step explanation:

we will need to put the information we currently have into an equation:

10•2•x= 112

then multiply 10•2 which equals 20...then the equation wil turn out like this:

20•x=112

now you will isolate the x... but how?? dividing 20 by itslef.. cancelling it out, and if you do one thing to the other side of the equation, you have to do it to the other...

20/20•x=112/20

you will end up with x= 122/20

you can use a calculator for this step, but to practice long division... (check the screen shot thingie)

hope this helped !! ^^

6 0

3 years ago

In the problem below, AB, CD, and EF are two-digit numbers, where A, B, C, D, E and F represent distinct digits from 1 to 9. Is

Oksi-84 [34.3K]

Answer:

<em>When both the conditions hold true, F is prime.</em>

Step-by-step explanation:

AB, CD, and EF are two-digit numbers, where A, B, C, D, E and F represent distinct digits from 1 to 9.

+ CD

--------

1st condition, B and D are consecutive.

Adding B and D gives us F.

Possible values can be (F being the unit value after adding not considering the carry over):

B + D = F

1+2=3

2+3=5

3+4=7

4+5=9

5+6=1

6+7=3

7+8=5

8+9=7

Here F is not prime (because 9 is not prime).

Now, let us consider the 2nd condition as well.

i.e. C = 8

For the following

+ CD

--------

C is 8 then A must be 1 because any value other than 1 for A will make the sum of A and C greater than 9 and there will be a carry which is not the case here.

So, E = 8 + 1 = 9

Now, B and D are consecutive and can not be 1, 8 or 9.

So, possible values are:

B + D = F

2 + 3 = 5

3 + 4 = 7

Here F is prime.

So, when both the conditions hold true, F is prime.

3 0

2 years ago