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3 years ago

7. Marco is making beaded bracelets. Each

Mathematics

1 answer:

stiks02 [169]3 years ago

4 0

One way to find the least common multiple of two numbers is to first list the prime factors of each number.

8 = 2 x 2 x 2

Then multiply each factor the greatest number of times it occurs in either number. If the same factor occurs more than once in both numbers, you multiply the factor the greatest number of times it occurs.

2: three occurrences

3: one occurrence

So, our LCM should be

2 x 2 x 2 x 3 = 24.

So, Marco can buy, at the very least, 24 beads of each color to have equal colors of beads.

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Find the x-intercept of the graph of the equation 2x - 5y = 10.

Vanyuwa [196]

In order to find the x-intercept, you must plug in 0 for y.
2x - 5(0) = 10
2x = 10
Divide by 2 on both sides.
x = 5
Thus, the x-intercept is (5,0)

8 0

3 years ago

Write an equation in slope -intercept form for a line that passes through (-2,7) and (14,21)

Georgia [21]

(-2,7) and (14,21)

Find the slope by using the formula $m = \frac{y2 - y1}{x2 - x1}$

Plug in the numbers. $m = \frac{21 - 7}{14 - - 2} 
 \\ m= \frac{14}{16} = \frac{7}{8} 
$

Write it in point slope form first

y - y1 = m(x - x1)
y - 7 = 7/8(x - -2)
y - 7 = 7/8x + 1.75
+ 7 +7

y = 7/8x + 8.75

Our final equation would be: y = 7/8x + 8.75

7 0

3 years ago

Because the two distributions displayed below have different ranges, they have the different standard deviations.

tigry1 [53]

False, the range is just determined by minising the lowest from the highest, but the variables can still get you the same mean, which can lead to the same standard deviation

6 0

3 years ago

Read 2 more answers

Find three consecutive even integers such that the product of the first and second is 8 more than 38 times the third.

BigorU [14]

Answer:

40, 42 and 44

Step-by-step explanation:

Mark these even consecutive integers as (2n-2), 2n and (2n+2)

Note that since they are integers, n > 0 and 2n is always even

Writing the equation

(2n-2) * 2n = 38 * (2n+2) + 8

// divide both sides by 2

(2n-2) * n = 19 * (2n+2) + 4

$2n^{2}$ - 2n = 38n + 38 + 4

$2n^{2}$ - 2n - 38n - 38 - 4 = 0

$2n^{2}$ - 40n - 42 = 0

// divide both sides by 2

$n^{2}$ - 20n - 21 = 0

// rewrite the equation

$n^{2}$ + n - 21n - 21 = 0

n * (n+1) - 21n - 21 = 0

n * (n+1) - 21 * (n+1) = 0

(n+1) * (n-21) = 0

// separate the two possible cases

n + 1 = 0 and n - 21 = 0

n = -1 n = 21

Note that here n has to be an integer, so case n = -1 can't be a solution. Therefore we use n = 21

Calculating the consecutive even integers

2n - 2 = 2 * 21 - 2 = 40

2n = 2 * 21 = 42

2n + 2 = 2 * 21 + 2 = 44

4 0

3 years ago

A house is valued at $118,000.00. The homeowner decides to add on a two-car garage that increased the value of the home by 15%.H

nasty-shy [4]

Answer:

The value of the house after adding the garage is $135,700.

Step-by-step explanation:

Given,

value of house before adding garage = $118,200.00

we need to find the value of house after adding two car garage.

Solution,

Since after adding two car garage the value of the house increased by 15%.

So firstly we will find out the 15% of the value of the house after adding garage.

So we can say that;

15% of the value of the house after adding garage is equal to 15 divided by 100 the multiplied with the value of the house before adding garage.

15% of the value of the house after adding garage = $\frac{15}{100}\times118,000=\$17,700$

Now, The value of the house after adding garage is equal to the sum of value of house before adding garage and 15% of value of house before adding garage.

We can frame it in equation form as;

The value of the house after adding garage = $\$118,000+\$17,000=\$135,700$

Hence The value of the house after adding the garage is $135,700.

3 0

3 years ago