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2 years ago

15

use the order of operations to simplify the left side of the inequality below. What values of x make the inequality a true state

ment? -1/2 (3^(2)+7) x>32

1 answer:

disa [49]2 years ago

7 0

Answer:

x<−0.00325153

x> 0.00325153

Step-by-step explanation:

-1/2 (3^(2)+7) x >32

-1/2 (3^(9) x >32

-1/2 (19683) x >32

-9841.5 x >32

x<−0.00325153

x> 0.00325153

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There are 5 small boxes in rectangle each box has length of 6 cm find the perimeter of all boxes

masha68 [24]

72cm

you multiply the outside length (6cm each) by how many there are

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3 years ago

Five widgets and three gadgets cost $109. 90.

erma4kov [3.2K]

Answer:

one gadget costs $15.80

Step-by-step explanation:

Let w = cost of one widget

Let g = cost of one gadget

Given:

Five widgets and three gadgets cost $109. 90

⇒ 5w + 3g = 109.9

Given:

One widget and four gadgets cost $75. 70

⇒ w + 4g = 75.7

Rewrite w + 4g = 75.7 to make w the subject:

⇒ w = 75.7 - 4g

Substitute into 5w + 3g = 109.9 and solve for g:

⇒ 5(75.7 - 4g) + 3g = 109.9

⇒ 378.5 - 20g + 3g = 109.9

⇒ 378.5 - 109.9 = 20g - 3g

⇒ 268.6 = 17g

⇒ g = 15.8

Therefore, one gadget costs $15.80

To find the cost of one widget, substitute the found value for g into
w = 75.7 - 4g and solve for w:

⇒ w = 75.7 - 4(15.8)

⇒ w = 75.7 - 63.2

⇒ w = 12.5

Therefore, one widget costs $12.50

5 0

2 years ago

Use the quadratic formula to solve the equation. If necessary, round to the nearest hundredth.

Rzqust [24]

Answer:

a = 6 or 9

Step-by-step explanation:

The "a", "b", and "c" of the quadratic formula are the coefficients of a², a, and the constant term in the given equation:

... a = 2, b = -30, c = 108

Then the quadratic formula tells you the solutions are ...

... (-b ± √(b² -4ac))/(2a)

... = (-(-30) ± √((-30)² -4(2)(108)))/(2(2))

... = (30 ± √(900 -864))/4

... = (30 ± √36)/4

... = (30 ± 6)/4 = {24, 36}/4

... = {6, 9}

_____

The <em>a</em> variable should not be confused with the "a" that is used to name the coefficient of the square of the variable in the quadratic formula. If it is too confusing, rewrite one or the other. For example, you could write ...

... The solution to pa² +qa +r = 0 is ... a = (-q ± √(q²-4pr))/(2p)

where p=2, q=-30, r=108 in the given equation.

5 0

3 years ago

Which point is a solution to the system of inequalities below?

vodomira [7]

The ordered pair which is a solution to the given inequality is: C. (2, 1).

<h3>What is an inequality?</h3>

An inequality can be defined as a mathematical relation that compares two (2) or more integers and variables in an equation based on any of the following arguments:

Less than (<).
Greater than (>).
Less than or equal to (≤).
Greater than or equal to (≥).

Next, we would test the ordered pair with the given inequality to determine a solution as follows:

For ordered pair (4, 4), we have:

3x + 2y < 15

3(4) + 2(4) < 15

12 + 8 < 15

20 < 15 (False).

For ordered pair (3, 3), we have:

3x + 2y < 15

3(3) + 2(3) < 15

9 + 6 < 15

15 < 15 (False).

7x - 4y > 9

7(3) - 4(3) > 9

21 - 12 > 9

9 > 9 (False)

For ordered pair (2, 1), we have:

3x + 2y < 15

3(2) + 2(1) < 15

6 + 2 < 15

8 < 15 (True).

7x - 4y > 9

7(2) - 4(1) > 9

14 - 4 > 9

10 > 9 (True)

For ordered pair (1, 0), we have:

3x + 2y < 15

3(1) + 2(0) < 15

3 + 0 < 15

3 < 15 (True).

7x - 4y > 9

7(1) - 4(0) > 9

7 - 4 > 9

3 > 9 (False)

Read more on inequality here: brainly.com/question/27166555

#SPJ1

7 0

1 year ago

Use a proof by contradiction to show that the square root of 3 is national You may use the following fact: For any integer kirke

Ierofanga [76]

Answer:

1. Let us proof that √3 is an irrational number, using <em>reductio ad absurdum</em>. Assume that $\sqrt{3}=\frac{m}{n}$ where $m$ and $n$ are non negative integers, and the fraction $\frac{m}{n}$ is irreducible, i.e., the numbers $m$ and $n$ have no common factors.

Now, squaring the equality at the beginning we get that

$3=\frac{m^2}{n^2}$ (1)

which is equivalent to $3n^2=m^2$ . From this we can deduce that 3 divides the number $m^2$ , and necessarily 3 must divide $m$ . Thus, $m=3p$ , where $p$ is a non negative integer.

Substituting $m=3p$ into (1), we get

$3= \frac{9p^2}{n^2}$

which is equivalent to

$n^2=3p^2$ .

Thus, 3 divides $n^2$ and necessarily 3 must divide $n$ . Hence, $n=3q$ where $q$ is a non negative integer.

Notice that

$\frac{m}{n} = \frac{3p}{3q} = \frac{p}{q}$ .

The above equality means that the fraction $\frac{m}{n}$ is reducible, what contradicts our initial assumption. So, $\sqrt{3}$ is irrational.

2. Let us prove now that the multiplication of an integer and a rational number is a rational number. So, $r\in\mathbb{Q}$ , which is equivalent to say that $r=\frac{m}{n}$ where $m$ and $n$ are non negative integers. Also, assume that $k\in\mathbb{Z}$ . So, we want to prove that $k\cdot r\in\mathbb{Z}$ . Recall that an integer $k$ can be written as

$k=\frac{k}{1}$ .

Then,

$k\cdot r = \frac{k}{1}\frac{m}{n} = \frac{mk}{n}$ .

Notice that the product $mk$ is an integer. Thus, the fraction $\frac{mk}{n}$ is a rational number. Therefore, $k\cdot r\in\mathbb{Q}$ .

3. Let us prove by <em>reductio ad absurdum</em> that the sum of a rational number and an irrational number is an irrational number. So, we have $x$ is irrational and $p\in\mathbb{Q}$ .

Write $q=x+p$ and let us suppose that $q$ is a rational number. So, we get that

$x=q-p$ .

But the subtraction or addition of two rational numbers is rational too. Then, the number $x$ must be rational too, which is a clear contradiction with our hypothesis. Therefore, $x+p$ is irrational.

7 0

3 years ago