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3 years ago

Is 3.14512 a rational number

Mathematics

2 answers:

Alexxx [7]3 years ago

6 0

It is also not rational because its does not have a pattern to it.

zvonat [6]3 years ago

3 0

No it is not because you can't make it into a fraction.

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You owe your sister $5.35, and she lends you another $4.50, how much do you owe her now

Margarita [4]

You would now have to pay her $9.85

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3 years ago

Q/19 < 4 pls be correct if you type it !!!!

babymother [125]

Answer:

q<76

Step-by-step explanation:

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2 years ago

Your friend attempted to factor an expression as shown. Find the error in your friends work. Then factor the expression correctl

loris [4]

Answer:

Third step is incorrect. The correct factored form is (x-1)(2x-5).

Step-by-step explanation:

The given expression is

$2x^2-7x+5$

We need to find the factored form of this expression.

Step 1: Given

$2x^2-7x+5$

Step 2: Splitting the middle term method, the middle term can be written as (-5x-2x).

$2x^2-5x-2x+5$

$(2x^2-5x)+(-2x+5)$

Step 3: Taking out common factors from each parenthesis.

$x(2x-5)-1(2x-5)$

Step 4: Taking out common factors.

$(x-1)(2x-5)$

Therefore, the third step is incorrect. The correct factored form is (x-1)(2x-5).

3 0

2 years ago

There was 2/3 of pan of lasagna in the refrigerator. Bill and his friends ate half of what was left. Write a number sentence and

Veseljchak [2.6K]

Half of 2/3 is:

$\frac { 1 }{ 2 } \times \frac { 2 }{ 3 } =\frac { 1 }{ 3 }$

This means that 1/3 was left. This also means that they ate 1/3 of the lasagne.

Model diagram can be seen in attachment.

5 0

3 years ago

Can someone please help me on number 16-ABC

melomori [17]

Answer:

Please check the explanation.

Step-by-step explanation:

Given the inequality

-2x < 10

-6 < -2x

Part a) Is x = 0 a solution to both inequalities

FOR -2x < 10

substituting x = 0 in -2x < 10

-2x < 10

-3(0) < 10

0 < 10

TRUE!

Thus, x = 0 satisfies the inequality -2x < 10.

∴ x = 0 is the solution to the inequality -2x < 10.

FOR -6 < -2x

substituting x = 0 in -6 < -2x

-6 < -2x

-6 < -2(0)

-6 < 0

TRUE!

Thus, x = 0 satisfies the inequality -6 < -2x

∴ x = 0 is the solution to the inequality -6 < -2x

Conclusion:

x = 0 is a solution to both inequalites.

Part b) Is x = 4 a solution to both inequalities

FOR -2x < 10

substituting x = 4 in -2x < 10

-2x < 10

-3(4) < 10

-12 < 10

TRUE!

Thus, x = 4 satisfies the inequality -2x < 10.

∴ x = 4 is the solution to the inequality -2x < 10.

FOR -6 < -2x

substituting x = 4 in -6 < -2x

-6 < -2x

-6 < -2(4)

-6 < -8

FALSE!

Thus, x = 4 does not satisfiy the inequality -6 < -2x

∴ x = 4 is the NOT a solution to the inequality -6 < -2x.

Conclusion:

x = 4 is NOT a solution to both inequalites.

Part c) Find another value of x that is a solution to both inequalities.

solving -2x < 10

$-2x\:$

Multiply both sides by -1 (reverses the inequality)

$\left(-2x\right)\left(-1\right)>10\left(-1\right)$

Simplify

$2x>-10$

Divide both sides by 2

$\frac{2x}{2}>\frac{-10}{2}$

$x>-5$

$-2x-5\:\\ \:\mathrm{Interval\:Notation:}&\:\left(-5,\:\infty \:\right)\end{bmatrix}$

solving -6 < -2x

-6 < -2x

switch sides

$-2x>-6$

Multiply both sides by -1 (reverses the inequality)

$\left(-2x\right)\left(-1\right)$

Simplify

$2x$

Divide both sides by 2

$\frac{2x}{2}$

$x$

$-6$

Thus, the two intervals:

$\left(-\infty \:,\:3\right)$

$\left(-5,\:\infty \:\right)$

The intersection of these two intervals would be the solution to both inequalities.

$\left(-\infty \:,\:3\right)$ and $\left(-5,\:\infty \:\right)$

As x = 1 is included in both intervals.

so x = 1 would be another solution common to both inequalities.

<h3>SUBSTITUTING x = 1</h3>

FOR -2x < 10

substituting x = 1 in -2x < 10

-2x < 10

-3(1) < 10

-3 < 10

TRUE!

Thus, x = 1 satisfies the inequality -2x < 10.

∴ x = 1 is the solution to the inequality -2x < 10.

FOR -6 < -2x

substituting x = 1 in -6 < -2x

-6 < -2x

-6 < -2(1)

-6 < -2

TRUE!

Thus, x = 1 satisfies the inequality -6 < -2x

∴ x = 1 is the solution to the inequality -6 < -2x.

Conclusion:

x = 1 is a solution common to both inequalites.

7 0

3 years ago