<img src="https://tex.z-dn.net/?f=9.10%20%5Ctimes%2015%20%5Ctimes%201.5%20%3D%20" id="TexFormula1" title="9.10 \times 15 \times

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3 years ago

1.5 = " alt="9.10 \times 15 \times 1.5 = " align="absmiddle" class="latex-formula">
show step by step work please I'm doing something wrong but can't figure it out

Mathematics

1 answer:

faust18 [17]3 years ago

5 0

This is my method of doing it, hope it helps.

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Let f(x)=5/x and g(x) =2x^2+5x. What two numbers are not in the domain of F o G?

expeople1 [14]

F(x)=5/x
g(x)=2(x^2)+5x
f(x) has a domain of all real numbers excluding zero
g(x) has a domain of all real numbers
fog(x)=5/(2(x^2)+5x)
fog(x)=5/(x(2x+5))
fog(x) has a domain that excludes both zero and -5/2

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3 years ago

Karen has $1.70 in coins. Karen has 8 coins, all of which are quarters or dimes.

kumpel [21]

x+y=8 and 25x+10y=170 are the linear equations.

x+y≤8 and 25x+10y≤170 are the inequalities.

Step-by-step explanation:

Given,

Worth of coins = $1.70 = 1.70*100 = 170 cents

Number of coins = 8

1 quarter = 25 cents

1 dime = 10 cents

Let,

x represent the number of quarters

y represent the number of dimes

1. Write an equation to represent the amount of coins Karen has.

x+y = 8

2.Write an equation to represent the value of the coins Karen has.

25x+10y=170

x+y=8 and 25x+10y=170 are the linear equations.

For inequalities, the amount cannot increase number of coins and worth but it can be less, therefore,

x+y≤8

25x+10y≤170

x+y≤8 and 25x+10y≤170 are the inequalities.

Keywords: linear equations, addition

Learn more about linear equations at:

#LearnwithBrainly

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4 years ago

A bag contains 4 red marbles and 6 blue marbles. Brandon selects one marble from the bag. Without replacing the marble he then s

IgorC [24]

The probability would be 4/15

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4 years ago

Use the integer tiles to evaluate the following expressions. 6 + 3 = 6 + (–4) = 6 + (–6) =

babunello [35]

Given:

The expressions are:

$6+3$

$6+(-4)$

$6+(-6)$

To find:

The value of given expression by using integer tiles.

Solution:

We have,

$6+3$

Here, both number are positive. When we add 6 and 3 positive integer tiles, we get 9 positive integer tiles as shown in the below figure. So,

$6+3=9$

Similarly,

$6+(-4)$

Here, 6 is positive and -4 is negative. It means we have 6 positive integer tiles and 4 negative integer tiles.

When we cancel the positive and negative integer tiles, we get 2 positive integer tiles as shown in the below figure. So,

$6+(-4)=2$

$6+(-6)$

Here, 6 is positive and -6 is negative. It means we have 6 positive integer tiles and 6 negative integer tiles.

When we cancel the positive and negative integer tiles, we get 0 integer tiles as shown in the below figure. So,

$6+(-6)=0$

Therefore, $6+3=9, 6+(-4)=2,6+(-6)=0$ .

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3 years ago

A particular circle in the standard (x,y) coordinate

matrenka [14]

For any circle with Cartesian equation
$(x-a)^2 + (y-b)^2 = r^2$ ,
we have that the centre of the circle is $(a,b)$ , and the radius of the circle is $r$ .

So in the case that
$(x-5)^2 + y^2 = 38$ ,
we essentially have that
$a = 5, b = 0, r^2 = 38$ .

So the centre of the circle is $(5,0)$ , and the radius is $\sqrt{38}$ .

7 0

4 years ago