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ycow [4]
3 years ago
8

How do you divide 300 by 150

Mathematics
1 answer:
Alexxx [7]3 years ago
7 0

Answer:

2

Step-by-step explanation:

300/150=2

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6x+12=2(3x+6) someone please help me solve this
Archy [21]

Answer:

x = all real numbers

infinite solutions

Step-by-step explanation:

6x+12=2(3x+6)

Distribute

6x+12 = 6x+12

Subtract 6x from each side

12 =12

This is always true

This means there are infinite solutions for x

x is all real numbers

7 0
4 years ago
Does cell division always mean an organism is growing?
umka2103 [35]
No cell division dosen't always means is growing, it also means diffrent thing for example in unicellular organisms cell division is the mean of reproduction, and in multicellular organism it is the mean of tissue growth and maintenance. So yeah it means to diffrent words cell division, and growing it dosent means that same thing always I hope you understand it
8 0
4 years ago
0.35/((2/3)x) = 0.45/(x−10)
elena-14-01-66 [18.8K]
X=70 because you have to convert the decimals to fractions so that the .35 equals 7/20 and .45 equals 9/20 then simplify the complex fractions on each side so 7/20 over 2X/3 equals 21/40X and the 9/20 over the X-10 equals 9/20(X-10), then cross multiply to get 420 (X-10)= 320X, next you would distribute the 420 through the parenthesis to get 420X-4200=360X, then add 4200 to both sides so 420X on its own then add 360X on both sides so both X's are on the same side then combine like terms and get 60X on the left, then divide both sides by 60 so X is on its own so X=70
7 0
4 years ago
Read 2 more answers
Find the equation of the parallel to the line 3x-2y=5 and passing through the midpoint of the line segment joining the points (-
levacccp [35]

Given:

The equation of parallel line is:

3x-2y=5

The required line passing through the midpoint of the line segment joining the points (-4,2) and (2,4).​

To find:

The equation of required line.

Solution:

Midpoint of  line segment joining the points (-4,2) and (2,4) is:

Midpoint=\left(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\right)

Midpoint=\left(\dfrac{-4+2}{2},\dfrac{2+4}{2}\right)

Midpoint=\left(\dfrac{-2}{2},\dfrac{6}{2}\right)

Midpoint=\left(-1,3\right)

It means the required line passes through the point (-1,3).

The slope of line Ax+By=C is:

m=\dfrac{-A}{B}

The given equation is:

3x-2y=5

Here, A=3 and B=-2. So, the slope of the line is:

m=\dfrac{-3}{-2}

m=1.5

Slope of parallel lines are same. So, the slope of the required line is 1.5.

The required line passes through the point (-1,3) with slope 1.5. So, the equation of the line is:

y-y_1=m(x-x_1)

y-3=1.5(x-(-1))

y-3=1.5(x+1)

y-3=1.5x+1.5

Adding 3 on both sides, we get

y=1.5x+1.5+3

y=1.5x+4.5

Therefore, the equation of the required line is y=1.5x+4.5.

8 0
3 years ago
Which two angles in the figure are obtuse angles?
ValentinkaMS [17]

Answer: C. ∠SQP and ∠TQR

Step-by-step explanation:

Since, by the given diagram,

m∠TQR  = m∠SQP (vertical angles)

And, m∠TQP < m∠TQR

By adding ∠TQR on both sides,

m∠TQP + m∠TQR < m∠TQR+m∠TQR

⇒m∠TQP + m∠TQR < 2 ∠TQR

⇒ 180° < 2 m∠TQR ( Because ∠TQP and ∠TQR are linear angles)

⇒ 90° < m∠TQR

Hence ∠TQR is a obtuse angle,

⇒  ∠SQP also is a obtuse angle.

4 0
3 years ago
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