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2 years ago

Ballio is cutting construction paper into rectangles for a project. He needs to cut one rectangle

Mathematics

1 answer:

Alexeev081 [22]2 years ago

8 0

Answer:

200 5/8

Step-by-step explanation:

Since you need to find the area, you multiply 6x15 1/2 and add it to 10 1/4x10 1/2.

You might be interested in

Factor 64 – x15. (4 – x3)(16 + 4x3 + x3) (4 – x3)(16 + 4x3 + x6) (4 – x5)(16 + 4x5 + x5) (4 – x5)(16 + 4x5 + x10)

Katen [24]

Answer:

(4 - $x^{5}$ )(16 + 4 $x^{5}$ + $x^{10}$ )

Step-by-step explanation:

64 - $x^{15}$ ← is a difference of cubes and factors in general as

a³ - b³ = (a - b)(a² + ab + b²), thus

64 - $x^{15}$

= 4³ - ( $x^{5}$ )³

= (4 - $x^{5}$ )(4² + 4 $x^{5}$ + ( $x^{5}$ )² )

= (4 - $x^{5}$ )(16 + 4 $x^{5}$ + $x^{10}$ )

7 0

3 years ago

Read 2 more answers

Which is the best approximation of the volume of a cone with the same base and height as the cylinder?

Sever21 [200]

If the volume of a cylinder is x, then the volume of a cone with same height and diameter is x/3

so

if the cone is 72ft³ then the cone with same base and height is 72/3ft³=24ft³

answer is A

3 0

3 years ago

Christy drove 135 miles in 2.5 hours.what was her average speed in miles per hour

garri49 [273]

54 miles per hour................

8 0

3 years ago

Read 2 more answers

<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