Ask question

Ask question

All categories

3 years ago

12

Jeremy is selling m of his models on the internet for $12.50 each. Which of the following expressions describes the amount he wi

ll be paid?

1 answer:

dem82 [27]3 years ago

6 0

Answer:

I dont have too much info but its 12.50m

Step-by-step explanation:

12.50(1) is 12.50 12.50(2) is 25 buck and so on and so on

You might be interested in

Solve for x<br> −4(x−7)=36

Ipatiy [6.2K]

-4(x-7)=36
-4x+28=36
-4x=8
x=-2

5 0

2 years ago

Read 2 more answers

Find the distance between points S(6,−3)

Gelneren [198K]

Answer:

.....................

4 0

2 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.

AB

+ CD

--------

EF

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

AB

+ CD

--------

EF

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

4) Find the sum of 6\12 and 1\2 .A) 3\12 B) 9\12 C) 12\12 D) 13\12

Darina [25.2K]

Answer:

12/12

Step-by-step explanation:

6/12= 1/2

so 1/2+1/2 = One Whole and that the same thing as 12/12

5 0

3 years ago

Read 2 more answers

A spacecraft is traveling with a velocity of v0x = 5320 m/s along the +x direction. Two engines are turned on for a time of 739

OlgaM077 [116]

Answer:

The velocities after 739 s of firing of each engine would be 6642.81 m/s in the x direction and 5306.02 in the y direction

Step-by-step explanation:

For a constant acceleration: $v_{f}=v_{0}+at$ , where $v_{f}$ is the final velocity in a direction after the acceleration is applied, $v_{0}$ is the initial velocity in that direction before the acceleration is applied, a is the acceleration applied in such direction, and t is the amount of time during where that acceleration was applied.
<em>Then for the x direction</em> it is known that the initial velocity is $v_{0x} =$ 5320 m/s, the acceleration (the applied by the engine) in x direction is $a_{x}$ 1.79 m/s2 and, the time during the acceleration was applied (the time during the engines were fired) of the is 739 s. Then: $v_{fx}=v_{0x}+a_{x}t=5320\frac{m}{s} +1.79\frac{m}{s^{2} }*739s=6642.81\frac{m}{s}$
In the same fashion, <em>for the y direction</em>, the initial velocity is $v_{0y} =$ 0 m/s, the acceleration in y direction is $a_{y}$ 7.18 m/s2, and the time is the same that in the x direction, 739 s, then for the final velocity in the y direction: $v_{fy}=v_{0y}+a_{y}t=0\frac{m}{s} +7.18\frac{m}{s^{2} }*739s=5306.02\frac{m}{s}$

8 0

3 years ago