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3 years ago

Please help me solve this problem thankyou Simplify 2/6+1/4

Mathematics

2 answers:

lukranit [14]3 years ago

6 0

Answer:

<h3>The answer is 7/12</h3>

<h2>I HOPE IT HELPS ❤️❤️</h2>

sattari [20]3 years ago

4 0

Answer:

same denominator as 12

2 x 2/ 6 x 2 + 1 x 3/ 4 x 3

4 +3/ 12

= 7/12

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<>) Melinda

Vanyuwa [196]

Answer:

The statement "500 divided by 50 is 10" is a true statement

Step-by-step explanation:

In mathematics, we have a term that we refer to as Place Value.

Place Value in mathematics can be defined as the value that a digit has based on its position or place in a number.

Examples of place value is:

Thousands represented by Th

Hundreds represented by H

Tens represented by T

Units represented by U

In the above question,

500 ÷ 50 gives us 10.

This is true because, 500 as a number, the Digit 5 has a place value of hundreds

50 as a number , the digit 5 has a place values of Tens.

10 as a number, the digit 1 has a place value of Tens.

In mathematics, the multiplication of 2 numbers in a place value of Tens always gives us a place value of hundreds.

For example, 10 × 50 = 500

Likewise, when we divide, a number with a place value of hundreds by a number with a place value of tens, we have a number with a place value of tens.

For example: 500 ÷ 50 = 10.

Therefore, the statement "500 divided by 50 is 10" is a true statement

7 0

3 years ago

What is the quotient when the polynomial 4x2 - 2x - 12 is divided by 2x - 4?

MrRa [10]

Tell me what x equals then I can awnser it do

4 0

3 years ago

Equation that has solutions x = -2 and x = 8

VLD [36.1K]

If you were to solve this equation you would get answers of x=-2 and x=-8

3 0

3 years ago

<img src="https://tex.z-dn.net/?f=%20%5Csqrt%7B%20-%20x%20%5E%7B3%7D%20%7D%20" id="TexFormula1" title=" \sqrt{ - x ^{3} } " alt=

noname [10]

I'm guessing you're given the function $y(x)=2-x^3$ , and you're asked to find the inverse function $y^{-1}(x)$ . To do this, swap $x$ and $y$ , then solve for $y$ :

$x=2-y^3\implies y^3=2-x\implies y=(2-x)^{1/3}=\sqrt[3]{2-x}$

so that the inverse function is

$y^{-1}(x)=\sqrt[3]{2-x}$

Just to verify:

$y(y^{-1}(x))=y(\sqrt[3]{2-x})=2-(\sqrt[3]{2-x})^3=2-(2-x)=x$

$y^{-1}(y(x))=y^{-1}(2-x^3)=\sqrt[3]{2-(2-x^3)}=\sqrt[3]{x^3}=x$

But in case you're actually only interested in computing the square root, first we note that $\sqrt x$ (the real-valued square root) is only defined as long as $x\ge0$ . So $\sqrt{-x^3}$ is defined as long as $-x^3\ge0$ , or $x^3\le0$ , or equivalently $x\le0$ . Under this condition, we could write

$\sqrt{-x^3}=\sqrt{-x\times x^2}=\sqrt{-x}\sqrt{x^2}$

We can simplify this further, but we have to be careful. Suppose $x=-1$ . Then $x^2=(-1)^2=1$ . But we get the same result if $x=1$ , since $x^2=1^2=1$ . There are two possible values of $x$ that given the same value of $x^2$ , so to capture both of them, we take $\sqrt{x^2}=|x|$ , the absolute value of $x$ . Then