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Anastaziya [24]
3 years ago
7

What is 6.974 rounded to the nearest tenth place?

Mathematics
2 answers:
loris [4]3 years ago
7 0

Answer:

7

Step-by-step explanation:

The tenth place is the 9 in 6.974.

You round up because if the place next to it(the 7) is greater than 5 then you round up. If it is less than 5 the 7 becomes a 0.

But this is a special case because 6.9 would not become a 6.10 it would become a 7.

dezoksy [38]3 years ago
3 0

Answer:

7

Step-by-step explanation:

The digits six is in the 10th Pl. even though the other times place so you have to like go to the nine’s and if the ninth grade on the six Danny do you have to round up because the room was fine and hire you round up for or lower your round down to round up to a six stays at 7 because it doesn’t have to make any changes but if you want to go

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Let a = 3 – 5i and b = –1 + 7i. Which of the following are true statements? Check all that apply.
Alona [7]

Answer: 2nd and 6th options are correct

Step-by-step explanation: It showed me after I got em wrong

7 0
3 years ago
An object is launched at 29.4 meters per
Lerok [7]

Answer:

The reasonable domain for the scenario is option 'a';

a) [0, 7]  

Step-by-step explanation:

For the projectile motion of the object, we are given;

The speed at which the object is launched, v = 29.4 meters per second

The height of the platform from which the object is launched, h = 34.3 meter

The equation for the height of the object as a function of time 'x' is given as follows;

f(x) = -4.9·x² + 29.4·x + 34.3

The domain for the scenario, is given by the possible values of 'x' for the function, which is found as follows;

At the height from which the object is launched, x = 0, and f(x) = 34.3

At the ground level to which the object can drop, f(x) = 0

∴ f(x) = -4.9·x² + 29.4·x + 34.3 = 0

-4.9·x² + 29.4·x + 34.3 = 0

By the quadratic formula, we have;

x = (-29.4 ± √(29.4² - 4 × (-4.9) × 34.3))/(2 × (-4.9)

∴ x = -1, or 7

Given that time is a natural number, we have the reasonable domain for the scenario as the start time when the object is launched, t = 0 to the time the object reaches the ground, t = 7

Therefore, the reasonable domain for the scenario is; 0 ≤ x ≤ 7 or [0, 7].

4 0
3 years ago
Plz help i kind of forgot how to do this<br> its just a review should be easy
Jet001 [13]
This is easy just find out what 5^2 equals which will be 5x5=25 (when dealing with a squared number you will always times it by itself. so if you had 10^2 it would be 100.) so now that we know 5^2 is 25 we then can add the 20 which gets 25+20=45! Hope this helps!
7 0
2 years ago
For the following system, if you isolated x in the first equation to use the Substitution Method, what expression would you subs
Arlecino [84]
First, we write the first equation where the left side contains x and y in the right side. We do as follows:

x = 2y + 6

You, then substitute the equation to the second one.

3(2y + 6) + y = 8
6y + 18 + y = 8
7y = -10
y = -10/7
x = 22/7
7 0
2 years ago
What is wrong with the following equation: 8 + (6/2) = 17 - 5
Sidana [21]
8+6/2=17-5

We can start by reducing the fraction using basic division.

Remember, numerator divided by denominator.

6 \ 2 = 3.

Rewrite the equation-

8+3=17-5

It doesn’t matter which side of the equal sign you start with, but in this case I started with the right hand.

8+3= 11

Rewrite-

11=17-5

Now subtract.


17-5=12

Write the equality.

11=12.

This is false.

11≠12.
6 0
3 years ago
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