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ss7ja [257]
3 years ago
9

Divide 4200 by a multiple of 10 so that the quotient is also a multiple of 10

Mathematics
2 answers:
Aleks [24]3 years ago
5 0

Answer:

4200 divided by 10

Step-by-step explanation:

To make 4200 a multiple of ten after dividing it by a multiple of ten, we first find the number of 5's in 4200 to see how many 10's go into 4200. Since 4200 = 6*7*2*2*5*5, we see that there are 2 fives. This means that there are two tens too. Since we have to keep 4200 a multiple of ten after dividing it by another 10. So the only possible way to do that is dividing 4200 by 10.

klasskru [66]3 years ago
3 0

Answer: Well if you divide 4200 by 42, you will get 100, which is a multiple of 10

Step-by-step explanation:

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Write an equation of a line containing (2, -3) and perpendicular to 3x+4y=14 answer
Sergio039 [100]

Answer:

Step-by-step explanation:

From the condition of perpendicularity, if two lines must be perpendicular to each other, the product of their gradient rather the slope must be equaled to ( -1 ).

Equation of a line ; y = mx+ c, where m1 is the gradient of the line and c us the point of intersection if the line on the y- axis.therefore, the equation

3x + 4y = 14, when re arranged, now becomes, 4y = -3x + 14, divide through by 4, gives, y = -3x/4 + 14.

Therefore, m1, = -3/4 and m2, = 4/3 going by the condition. Since the line passes through the coordinate of (2,3), where x = 2, and y = -3, then, substitute for x and y in y = mx+ c, minding the m2, to find C, therefore,

-3 = 4/3x2 + c. -3 = 8/3 +c

Multiply through by 3 to make it a linear equation,

-9 = 8 + 3c; -9 - 8 = 3c, -17= 3c

c = -17/3.Now substitute for c in the equation ,y = mx+ c, the equation now becomes, y = 4x/3 - 17/3. Therefore the new equation is

3y = -4x - 17.

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3 years ago
Use the change of base formula to approximate the solution to log0.515 = 1 – 2x. Round to the nearest hundredth.
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Step-by-step explanation:

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A random variable can be either discrete or continuous. It is discrete it can assume only a finite number of values, or a countable infinity of values at most.

It is continuous if it can assume values in an interval, or in general, an uncountable infinity of values.

That being said, we have:

Option A is a discrete random variable, because the number of heads in 5 throws can be 0, 1, 2, 3, 4 or 5. So, we have finitely many possible values.

Option B is a discrete random variable, because the number you roll on a die is either1, 2, 3, 4, 5 or 6. So, we have finitely many possible values.

Option C is a discrete random variable, because if there are n students in a class, the number of boys is an integer between 0 and n. So, we have finitely many possible values.

Option D is finally a continuous random variable, because the height of a 10-year-old can be any number (in a suitable range of course).

4 0
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