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bulgar [2K]
3 years ago
15

What are the common multiples of 2 3 and 5

Mathematics
1 answer:
LUCKY_DIMON [66]3 years ago
3 0
Los multiplos de 2 es 2,4 ,6,8,10,12,14, etc Del 3: 3,6,9,12,15,18,21,24,27,30,etc Del 5: 5,10,15,20,25,30,35,40,etc
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The volume of a rectangular box is (x3 - 7x2 - 9x + 63) cubic units. Determine the dimensions of the rectangular box by factorin
Shalnov [3]
Hello,
well done.
As x^3-7x²-9x+63=(x-3)(x+3)(x-7) we must have x>3 and x>-3 and x>7 thus x>7

Explanations:
x^3-7x²-9x+63=x^3-9x+(-7x²+63)=x(x²-9)-7(x²-9)=(x²-9)(x-7)=(x+3)(x-3)(x-7)

6 0
4 years ago
Antonia is keeping track of the number of pages she reads. She started the school year having already read 95 pages of her book,
Airida [17]

f(x) represents the total number of pages Antonia has read; x represents the number of days

4 0
3 years ago
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Which of the following figures could have both line symmetry and rotational symmetry?
Lana71 [14]

Answer:

Triangle  and square!

Step-by-step explanation:

Let me know if im wrong and I will fix it.

6 0
2 years ago
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If you run for 0.4 hours at 7 mph, how fast should you walk during the next 0.8 hours to have the average speed of 5 mph?
Fantom [35]

Answer:

4 mph

Step-by-step explanation:

The average speed of an object is given by the total distance covered by the time taken:

v=\frac{d}{t}

where

d is the total distance covered

t is the time taken

in the first part, the person runs for 0.4 hours at a speed of 7 mph, so the distance covered in the 1st part is

d_1 = v_1 t_1 = (7)(0.4)=2.8 mi

Then the distance covered in the second part is d_2, so the total distance is

d=2.8+d_2 (1)

The total time elapsed is 0.4 hours (first part) + 0.8 hours (second part), so

t=0.4+0.8=1.2 h

So we can write the average speed as

v=\frac{2.8+d_2}{1.2} (1)

We want the average speed to be 5 mph,

v = 5 mph

Therefore we can rearrange eq.(1) to find d2:

d_2 = 1.2v-2.8 = (1.2)(5)-2.8=3.2 mi

And therefore, the speed in the second part must be

v_2=\frac{d_2}{t_2}=\frac{3.2}{0.8}=4 mph

4 0
3 years ago
The following statement is either true​ (in all​ cases) or false​ (for at least one​ example). If​ false, construct a specific e
matrenka [14]

Answer:

False, counterexample below

Step-by-step explanation:

Denote the unit vectors of R^3 by e_1=(1,0,0), e_2=(0,1,0), e_3=(0,0,1). Now consider v_1=e_1, v_2=2e_1 and v_3=e_3. We have that v_1, v_2,v_3 \in \mathbb{R}^3. Also, the vector v_3 is not a linear combination of v_1, v_2 because any linear combination of these two vectors will have third coordinate zero, but v_3 has third coordinate 1 so they can't be equal.

However, the set \{v_1, v_2,v_3\} is not linearly independent, because 2v_1-v_2+0v_3=0 is a non-trivial linear combination of these vectors that equals zero.

4 0
3 years ago
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