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GarryVolchara [31]
3 years ago
15

How many zeros will be in the product of 7 x 5000

Mathematics
2 answers:
cluponka [151]3 years ago
6 0

Answer:

3 zeroes.

Step-by-step explanation:

To find: The number of zeroes in the product of 7 and 5000.

Solution:

In order to find the number of zeroes in the product of 7 and 5000, we first need to multiply 7 and 5000.

The product of 7 and 5000 is 7 \times 5000 = 35000.

So, the product of 7 and 5000 is 35000.

Now, in the number 35000, we can clearly see that there are 3 zeroes in the number.

Hence, there are 3 zeroes in the product of 7 and 5000.

Shalnov [3]3 years ago
3 0
3 0s because 7 times 5000 equals 35000
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30*47382 is the correct answer
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If m<9=130°, what is m<4?
nikklg [1K]

From the diagram provided, the two lines that are left and right of each other are parallel and are cut by a transversal. That means,

\begin{gathered} \angle9=\angle5\text{ (corresponding angles are equal)} \\ \text{Similarly,} \\ \angle8=\angle4\text{ (corresponding angles are equal)} \end{gathered}

If angle 9 equals 130, then angle 8 equals,

\begin{gathered} 130+\angle8=180\text{ (angles on a straight line sum up to 180)} \\ \angle8=180-130 \\ \angle8=50 \end{gathered}

If angle 8 equals angle 4, then angle 4 equals 50 degrees

3 0
2 years ago
Help help help help thanks lol
riadik2000 [5.3K]
f(x)=x^2-81\\
y=x^2-81\\
x^2=y+81\\
x=\pm \sqrt{y+81}\\
f^{-1}(x)=\pm \sqrt{x+81}\\

f(x)=2x-6\\
y=2x-6\\
2x=y+6\\
x=\dfrac{1}{2}y+3\\
f^{-1}(x)=\dfrac{1}{2}x+3\\
f^{-1}(2)=\dfrac{1}{2}\cdot2+3\\
f^{-1}(2)=1+3\\
f^{-1}(2)=4
5 0
3 years ago
How many positive integers can be expressed ad a product of two or more of the prime numbers 5,7,11,and 13 if no one product is
sertanlavr [38]

Answer:

11 positive integers can be expressed.

Step-by-step explanation:

Consider the provided information.

The number of possible prime numbers are 5,7,11,and 13.

There are 4 possible prime numbers.

How many positive integers can be expressed as a product of two or more of the prime numbers, that means there can be product of two numbers, three number or four numbers.

The formula to calculate combinations is: ^nC_r=\frac{n!}{r!(n-r)!}

The number of ways are:

^4C_2+^4C_3+^4C_4=\frac{4!}{2!(4-2)!}+\frac{4!}{3!(4-3)!}+\frac{4!}{4!}

^4C_2+^4C_3+^4C_4=\frac{4!}{2!2!}+\frac{4!}{3!}+1

^4C_2+^4C_3+^4C_4=6+4+1

^4C_2+^4C_3+^4C_4=11

Hence, 11 positive integers can be expressed.

8 0
3 years ago
Mathematical model in real life
matrenka [14]
Well, one example would be how much space is in a cardboard box, to ship something to a friend or relative. 
We know the three measurements:
L = Length, W = Width, and H = Height.

And the formula for a volume of a cube is:
V = L*W*H

So say that:
L = 6 , W = 10 Feet, and H = 3 Feet.
By using the formula above we get the equation: V = 6*10*3
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So the volume of the box, or the space you have to send something is 180 square feet. 

Does this help?

The link I used for some of this info, which has other useful math models is: https://www.mathsisfun.com/algebra/mathematical-models.html
6 0
3 years ago
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