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3 years ago

How are the GCF and the LCM alike?

Mathematics

1 answer:

Charra [1.4K]3 years ago

3 0

The product of the GCF and the LCM is equal to the product of the original two numbers.

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Naomi and Josiah work at a furniture store. Naomi is paid $190 per week plus 5.5% of her total sales in dollars, x x, which can

Ivanshal [37]

Answer:

x = $11400

Step-by-step explanation:

Naomi is paid $190 per week plus 5.5% of her total sales in dollars, x, which can be represented by g ( x ) = 190 + 0.055 x.

And Josiah is paid $475 per week plus 3% of his total sales in dollars, x, which can be represented by f ( x ) = 475 + 0.03 x

Let the weekly pat for them are equal.

Then, g(x) = f(x)

⇒ 190 + 0.055 x = 475 + 0.03 x

⇒ 0.025 x = 285

⇒ x = $11400

Therefore, for the value of x = $11400, that will make their weekly pay the same. (Answer)

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3 years ago

the table the number of cakes an pies that gretchen sold at a bake sale last weekend she sold each cake for $8 an each pie for $

kaheart [24]

Answer:

Step-by-step explanation:

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3 years ago

Solve: 12^x2+5x-4 =12^2x+6

lyudmila [28]

Answer:

x=2, x=-5

Step-by-step explanation:

to work more comfortably, the first thing we need to do is work the equation linearly, for that we take advantage of the property of logarithms that tells me that $log(a^{b})=b*log(a)$

in this way, the equation remains as:

$log(12^{x^{2} +5x-4 } )=log(12^{5x+6} ) \\ (x^{2} +5x-4 )*log(12)=(5x+6) *log(12)\\x^{2} +5x-4 = 5x+6$

Now we clear the equation so that it is of the form $ax^{2} +bx+c=0$

$x^{2} +5x-2x=6+4\\ x^{2} +3x-10=0$

finally, we apply the equation to solve second degree equations

$x = \frac{-b \pm \sqrt {b^2-4ac}}{2a}$

$x = \frac{-3 \pm \sqrt {3^2-4*1*(-10)}}{2*1}\\ x = \frac{-3 \pm \sqrt {9+40}}{2}\\ x = \frac{-3 \pm 7}{2}$

x=2 and x=-5

Done

5 0

4 years ago

What percentage would number 4 be

wolverine [178]

Answer:

Step-by-step explanation:

You can think of it as 1.75 you divide by 100 and that would be 0.0175 in other words 175%

7 0

3 years ago

Identify the lower class limits, upper class limits, class width, class midpoints, and class boundaries for the given freque

bezimeni [28]

Answer:

See explanation and results below.

Step-by-step explanation:

We have the following data:

Age (yr) when award was won Frequency 25-34 31 35-44 36 45-54 12 55-64 4 65-74 6 75-84 1 85-94 2

If we order the data we got the following result. We assume that the midpoint is given by:

$Midpoint= \frac{L.Limit +U.Limit}{2}$

The $Class.Width = U.Limit-L.Limit$

And the limits are given by the lower and upper limit.

L.Limit U.Limit Class W Midpoint Frequency Limits

25 34 34-25=9 29.5 31 [25-34]

35 44 44-35=9 39.5 36 [35-44]

45 54 54-45=9 49.5 12 [45-54]

55 64 64-55=9 59.5 4 [55-64]

65 74 74-65=9 69.5 6 [65-74]

75 84 84-75=9 79.5 1 [75-84]

85 94 94-85=9 89.5 2 [85-94]

8 0

3 years ago