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3 years ago

The Distributive Property to express 14 + 63

Mathematics

1 answer:

Ksivusya [100]3 years ago

7 0

7(2+9) would be the answer as 7 is the greatest common factor

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I need help please...

Ostrovityanka [42]

Answer:

Step-by-step explanation:

I think it's hitting a golf ball in the air, as it is a transformation, but the general shape of it is not affected.

Opening a locker, you are still affecting the general shape.

6 0

3 years ago

1) There are 80 grams of salt in a 32% solution. Find the weight of the solution

alukav5142 [94]

Answer:

Weight of the solution = 250 grams

Step-by-step explanation:

1) Consider the weight of solution as x grams.

32 % solution = 80 grams salt

32% of x = 80

(32/100)*x = 80

Solving for x:

x = (80*100) / 32

x = 250 grams

7 0

3 years ago

Which of these expressions is equivalent to -2(x-5)?

Serga [27]

Answer:

The answer is -2x+10

Step-by-step explanation:

Hope this helps

4 0

3 years ago

Does this graph have 10 or 9 squares, 10 or 9 years, I think I might be going insane

Yuliya22 [10]

Answer:

10 years

Step-by-step explanation:

8 0

2 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

$\sqrt{-x^3}=|x|\sqrt{-x}$

We can't simplify the square root term further than this.

3 0

3 years ago