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2 years ago

3 erasers cost $4.41. Write a proportion that would solve for the cost of 4 erasers.

Mathematics

2 answers:

Bess [88]2 years ago

7 0

Answer:

$5.88

Step-by-step explanation:

3 erasers ⇒ $4,41

4 erasers ⇒ $A

A = 4 erasers * $4.41 / 3 erasers

A = $ 5.88

quester [9]2 years ago

3 0

Answer:

see below

Step-by-step explanation:

Proportion:

4.41 is to 3 erasers as 'x' is to 4 erasers:

4.41 /3 = x / 4 <======this is your proportion

or rearrange

4 * 4.41/3 = x

(this results in x = $ 5.88 for 4 erasers )

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What is the absolute value of the complex number -4-√2

Zepler [3.9K]

|a+bi| = √(a² + b²)

-4-√2 i -> take a = -4 and b = -√2

|-4-√2 i| = √[ (-4)² + (-√2)² ]
 = √[ 16 + 2 ]
 = √[ 18 ] = √[ 9 * 2 ]
 = 3√2
the absolute value is 3√2

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3 years ago

Which of the following is a solution of x2 + 6x = −18? x = 3 − 3i x = −3 + 3i x = −6 + 3i x = 6 − 3i

Marina86 [1]

Step 1 ) Move all terms right side of the equation.

$\displaystyle\ x^{2} + 6x = -18$

$\displaystyle\ x^{2} + 6x + 18 = -18 + 18$

$\displaystyle\ x^{2} +6x + 18 = 0$

Step 2 ) Apply quadratic formula. (Note: There are 2 solutions)

$\displaystyle\ x^{2} +6x + 18 = 0$

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

$\displaystyle\ x = \frac{-6+\sqrt{6^{2}-4 * 18}}{2} , \displaystyle\ x = \frac{-6-\sqrt{6^{2}-4*18}}{2}$

$\displaystyle\ x = \frac{-6+6i}{2}$ ,

$\displaystyle\ x = \frac{-6-6i}{2}$

Step 3 ) Simplify.

$\displaystyle\ x = \frac{-6+6i}{2}$ ,

$\displaystyle\ x = \frac{-6-6i}{2}$

$\displaystyle\ x = -3+ 3i$ ,

$\displaystyle\ x = -3 - 3i$

Since the options only provide one of the answer we found, the answer is...

$\displaystyle\ x = -3 - 3i$

•

- Marlon Nunez

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3 years ago

Read 2 more answers

2 u + 12 < 23 0,1 d + 8 > 0 3 - 4 r > 7 13 < 2 z + 3 6 ≥ 9 - 0, 15 i

BartSMP [9]

Answer:

Part 1) $u< 5.5$

Part 2) $d > -80$

Part 3) $r < -1$

Part 4) $z>5$

Part 5) $i\geq 20$

Step-by-step explanation:

The question is

Solve each inequality for the indicated variable

Part 1) we have

$2u+12$

subtract 12 both sides

$2u< 23-12\\2u$

Divide by 2 both sides

$u< 5.5$

The solution is the interval (-∞,5.5)

In a number line the solution is the shaded area at left of u=5.5 (open circle)

The number 5.5 is not included in the solution

Part 2) we have

$0.1d+8 >0$

subtract 8 both sides

$0.1d > -8$

Divide by 0.1 both sides

$d > -80$

The solution is the interval (-80,∞)

In a number line the solution is the shaded area at right of d=-80 (open circle)

The number -80 is not included in the solution

Part 3) we have

$3-4r> 7$

Subtract 3 both sides

$-4r>7-3\\-4r>4$

Divide by -4 both sides

Remember that, when you multiply or divide both sides of an inequality by a negative number, you must reverse the inequality symbol

$r < -1$

The solution is the interval (-∞,-1)

In a number line the solution is the shaded area at left of r=-1.1 (open circle)

The number -1.1 is not included in the solution

Part 4) we have

$13< 2z+3$

Subtract 3 both sides

$13-3$

Divide by 2 both sides

$5$

Rewrite

$z>5$

The solution is the interval (5,∞)

In a number line the solution is the shaded area at right of z=5 (open circle)

The number 5 is not included in the solution

Part 5) we have

$6\geq 9-0.15i$

Subtract 9 both sides

$6-9\geq -0.15i\\-3\geq -0.15i$

Divide by -0.15 both sides

Remember that, when you multiply or divide both sides of an inequality by a negative number, you must reverse the inequality symbol

$20\leq i$

Rewrite

$i\geq 20$

The solution is the interval [20,∞)

In a number line the solution is the shaded area at right of i=20 (closed circle)

The number 20 is included in the solution

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3 years ago

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ELEN [110]

Step-by-step explanation:

Distribute:

=−x2+−4x+−4+7

Combine Like Terms:

=−x2+−4x+−4+7

=(−x2)+(−4x)+(−4+7)

=−x2+−4x+3

Answer:

=−x2−4x+3

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2 years ago

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SpyIntel [72]

Check the picture below.

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\begin{array}{lccclll}
P(x) = &{{ -0.05}}x^2&{{ +500}}x&{{ -100000}}\\
&\uparrow &\uparrow &\uparrow \\
&a&b&c
\end{array}\qquad 
\left(-\cfrac{{{ b}}}{2{{ a}}}\quad ,\quad {{ c}}-\cfrac{{{ b}}^2}{4{{ a}}}\right)\\\\
-------------------------------\\\\
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