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3 years ago

7

Find the GCF of each pair 20 and 36

1 answer:

valina [46]3 years ago

8 0

The GCF of 20 & 36 is 4.

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Help. only questions 3 & 4 .

Maksim231197 [3]

I'm doing 3

For 3, using a table is very similar to a double number because the numbers are matching up in both ways.

On the bottom of a double number line we have like, for example,-- the bottom of the double number line would have batches. And its 1, 2, 3 ,4 , etc.

And on a table, it would be the same, the numbers on both diagrams have the same methods, have same way of lining things up but they're just drawn differently.

Hope this helped!

3 0

3 years ago

Which expression is equivalent to (81x4y16)12 for all positive values of x and y?<br> 40.5x4y16

IrinaVladis [17]

Answer:

$(81x^4y^{16})^{\frac{1}{2}} = 9 x^2 y^8$

Step-by-step explanation:

Given

$(81x^4y^{16})^{\frac{1}{2}}$

Required

Determine an equivalent expression

Express 81 as $9^2$

$(9^2x^4y^{16})^{\frac{1}{2}}$

Apply law of indices

$9^2^{\frac{1}{2}} * x^4^{\frac{1}{2}} * y^{16}^{\frac{1}{2}}$

$9^{\frac{2*1}{2}} * x^{\frac{4*1}{2}} * y^{\frac{16*1}{2}}$

$9^{\frac{2}{2}} * x^{\frac{4}{2}} * y^{\frac{16}{2}}$

$9^1 * x^2 * y^8$

$9 * x^2 * y^8$

$9 x^2 y^8$

Hence:

$(81x^4y^{16})^{\frac{1}{2}} = 9 x^2 y^8$

4 0

3 years ago

Which type of equation can be used to model the data in the table below?

leva [86]

X| 2|4|6jsososodidididjfjfifififififofioff

7 0

3 years ago

X/4<20.5<br> Solve for x<br> example of type of answer: 12≥x

Musya8 [376]

Solve for x by simplifying both sides of the inequality, then isolating the variable.
Inequality Form:

x<82

6 0

2 years ago

Read 2 more answers

A raffle offers one $8000.00 prize, one $4000.00 prize, and five $1600.00 prizes. There are 5000 tickets sold at $5 each. Find t

Harman [31]

Answer:

The expectation is $E(1 )= -\$ 1$

Step-by-step explanation:

From the question we are told that

The first offer is $x_1 = \$ 8000$

The second offer is $x_2 = \$ 4000$

The third offer is $\$ 1600$

The number of tickets is $n = 5000$

The price of each ticket is $p= \$ 5$

Generally expectation is mathematically represented as

$E(x)=\sum x * P(X = x )$

$P(X = x_1 ) = \frac{1}{5000}$ given that they just offer one

$P(X = x_1 ) = 0.0002$

Now

$P(X = x_2 ) = \frac{1}{5000}$ given that they just offer one

$P(X = x_2 ) = 0.0002$

Now

$P(X = x_3 ) = \frac{5}{5000}$ given that they offer five

$P(X = x_3 ) = 0.001$

Hence the expectation is evaluated as

$E(x)=8000 * 0.0002 + 4000 * 0.0002 + 1600 * 0.001$

$E(x)=\$ 4$

Now given that the price for a ticket is $\$ 5$

The actual expectation when price of ticket has been removed is

$E(1 )= 4- 5$

$E(1 )= -\$ 1$

4 0

3 years ago