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3 years ago

Any of these??? the whole page? Please it's due in 2 hours and I would love the help!

Mathematics

1 answer:

Vinvika [58]3 years ago

6 0

9.This is because it goes through the whole circle passing through the centre point. at the full width
The rest are pretty easy. I'm sure you can manage. Good luck!

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What is the invers function of f(x)=(n+1)^3?

zimovet [89]

Answer:

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

Step-by-step explanation:

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

Graphing both inverses, they will be symmetrical about the line $y=x$ .

8 0

3 years ago

Okay these last two are the last ones i need help on!

icang [17]

21) 5 shelves
22)385mm

8 0

3 years ago

How would i solve this?

Flauer [41]

Answer: C

Step-by-step explanation:

The enequality includes x values equal to or greater than -2 and less than(but not equal to) 1.

so the possible integers would be -2 -1 0.

4 0

4 years ago

P(x)= 3x^3-5x^2-14x-4

nexus9112 [7]

$\displaystyle\\
P(x)=3x^3-5x^2-14x-4\\\\
D_{-4}=\{-4;~-2;~\underline{\bf -1};~1;~2;~4\}\\\\
\text{We observe that } \frac{-1}{3} \text{ is a solution of the equation:}\\
3x^3-5x^2-14x-4=0\\\\
$

$\displaystyle\\
\text{Verification}\\\\
3x^3-5x^2-14x-4=\\\\
=3\times\Big(-\frac{1}{3}\Big)^3-5\times\Big(-\frac{1}{3}\Big)^2-14\times\Big(-\frac{1}{3}\Big)-4=\\\\
=-\frac{1}{9}-\frac{5}{9}+\frac{14}{3}-4=\\\\
=-\frac{6}{9}+\frac{14}{3}-4=\\\\
=-\frac{6}{9}+\frac{42}{9}- \frac{4\times 9}{9}=\\\\
 =-\frac{6}{9}+\frac{42}{9}- \frac{36}{9}= \frac{42-6-36}{9}=\frac{42-42}{9}=\frac{0}{9}=0\\\\
\Longrightarrow~~~P(x)~\vdots~\Big(x+ \frac{1}{3}\Big)\\\\
\Longrightarrow~~~P(x)~\vdots~(3x+1)$

$\displaystyle\\
3x^3-5x^2-14x-4=0\\
~~~~~-5x^2 = x^2 - 6x^2\\
~~~~~-14x =-2x-12x \\
3x^3+x^2 - 6x^2-2x-12x-4=0\\
x^2(3x+1)-2x(3x+1) -4(3x+1)=0\\
(3x+1)(x^2-2x -4)=0\\\\
\text{Solve: } x^2-2x -4=0\\\\
x_{12}= \frac{-b\pm \sqrt{b^2-4ac}}{2a}=\\\\=\frac{2\pm \sqrt{4+16}}{2}=\frac{2\pm \sqrt{20}}{2}=\frac{2\pm 2\sqrt{5}}{2}=1\pm\sqrt{5}\\\\
x_1 =1+\sqrt{5}\\
x_2 =1-\sqrt{5}\\
\Longrightarrow P(x)= 3x^3-5x^2-14x-4 =\boxed{(3x+1)(x-1-\sqrt{5})(x-1+\sqrt{5})}$

7 0

3 years ago

H=(-5) answer needed now!!!!

wariber [46]

Answer:

-2

Step-by-step explanation:

h(-5) represents the y value when x is -5.

7 0

4 years ago