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2 years ago

The sum of the digits of a two-digit number is 11.

Mathematics

1 answer:

r-ruslan [8.4K]2 years ago

3 0

Answer:

8 and 3

Step-by-step explanation:

8+3=11

8-3= 5

You might be interested in

There are 42 males and 67 females participating in a marathon. What percent of the participants are male?

Sergio039 [100]

First, you find the total of both sides.

42 + 67 = 109

Next, you divide the Total by either of the sides, but since we're looking for the percentage of males, it'd be better to do this.

42/109 = 0.39

Multiply it by 100 to change it into percentage form.

0.39 x 100 = 39%

There your answer is 39%

4 0

3 years ago

Kevin has $100 in a savings account that earns 10% annually. The interest is not compounded. How much will he have in 2 years?

Kisachek [45]

He will have 220 that is how much he will have

5 0

3 years ago

Ethan and Layla work at a furniture store. Ethan is paid $235 per week plus 9.5% of his total sales in dollars, xx, which can be

kobusy [5.1K]

Answer:

$18,200

Step-by-step explanation:

We want ...

f(x) = g(x)

53 +.105x = 235 +.095x . . . . use given expressions

.01x = 182 . . . . . . . subtract 53+.095x

x = 182/.01 = 18200 . . . . . divide by the coefficient of x

Each must sell $18,200 for their weekly pay to be the same.

7 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