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3 years ago

What’s not a solution of the inequality 9-3x>-36

Mathematics

1 answer:

lorasvet [3.4K]3 years ago

6 0

Answer:

Step-by-step explanation:

when you plug in all the values for x, the only one that does not make the inequality true is 50.

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Nour and Rana are shopping for a Christmas tree. They are deciding between 222 different types of trees (real and fake) and 444

ioda

From what the statement says the question seems to be to represent the sample space of randomly choosing a type of tree and a color for the decorations

In this case, what we must do is identify the events that exist and the options that each event has, like this:

Event 1:

Tree type: real or fake

Event 2:

Colors for the ornaments: White (W), Silver (S), Gold (G) or Purple (P)

Since in an event there are 2 options and the other 4 options, the final number of options would be 2 * 4, that is, 8 options, represented below:

(real, W), (real, S), (real, G), (real, P), (fake, W), (fake, S), (fake, G), (fake, P)

The above would be the sample space of the events.

Now the probability of 1 coming out is 1 over the event total, or 1/8.

5 0

3 years ago

Read 2 more answers

What is the median of the following numbers? 4, 5, 5, 2, 1, 3, 4, 5, 4, 4, 4, 7

Katen [24]

Answer:

Step-by-step explanation:

Rewrite is order going from least to greatest.

1, 2, 3, 4, 4, 4, 4, 4, 5, 5, 5, 7

Find the middle number.

Because there are an even amount of numbers you have to take the two middle numbers which are 4 and 4 add them which gives you 8 and then divide that by 2( because you added two numbers.)

Therefore the answer is 4

3 0

2 years ago

If 20 is added to the number, the absolute value of the result is 6.

Vilka [71]

Number is number is -14 or -26

7 0

3 years ago

Can someone please help me with these 3 answers. aiaf you can not see them make sure to click on the image.

Dafna11 [192]

For the first one I’m not sure, but for 2 it is B and 3 is 5.

7 0

2 years ago

Can someone help me out ! got stuck in this question for an hour.

Reil [10]

Answer:

See below

Step-by-step explanation:

Considering $\vec{u}, \vec{v}, \vec{w} \in V^3 \lambda \in \mathbb{R}$ , then

$\Vert \vec{u} \cdot \vec{v}\Vert \leq \Vert\vec{u}\Vert \Vert\vec{v}\Vert$ we have $(\vec{u} \cdot \vec{v})^2 \leq (\vec{u} \cdot \vec{u})(\vec{v} \cdot \vec{v}) \quad$$

This is the Cauchy–Schwarz Inequality, therefore

$\left(\sum_{i=1}^{n} u_i v_i \right)^2 \leq \left(\sum_{i=1}^{n} u_i \right)^2 \left(\sum_{i=1}^{n} v_i \right)^2$

We have the equation

$\dfrac{\sin ^4 x }{a} + \dfrac{\cos^4 x }{b} = \dfrac{1}{a+b}, a,b\in\mathbb{N}$

We can use the Cauchy–Schwarz Inequality because $a$ and $b$ are greater than 0. In fact, $a>0 \wedge b>0 \implies ab>0$ . Using the Cauchy–Schwarz Inequality, we have

$\dfrac{\sin ^4 x }{a} + \dfrac{\cos^4 x }{b} =\dfrac{(\sin^2 x)^2}{a}+\dfrac{(\cos^2 x)}{b}\geq \dfrac{(\sin^2 x+\cos^2 x)^2}{a+b} = \dfrac{1}{a+b}$

and the equation holds for

$\dfrac{\sin^2{x}}{a}=\dfrac{\cos^2{x}}{b}=\dfrac{1}{a+b}$

$\implies\quad \sin^2 x = \dfrac{a}{a+b} \text{ and }\cos^2 x = \dfrac{b}{a+b}$

Therefore, once we can write

$\sin^2 x = \dfrac{a}{a+b} \implies \sin^{4n}x = \dfrac{a^{2n}}{(a+b)^{2n}} \implies\dfrac{\sin^{4n}x }{a^{2n-1}} = \dfrac{a^{2n}}{(a+b)^{2n}\cdot a^{2n-1}}$

It is the same thing for cosine, thus

$\cos^2 x = \dfrac{b}{a+b} \implies \dfrac{\cos^{4n}x }{b^{2n-1}} = \dfrac{b^{2n}}{(a+b)^{2n}\cdot b^{2n-1}}$

Once

$\dfrac{a^{2n}}{(a+b)^{2n}\cdot a^{2n-1}}+ \dfrac{b^{2n}}{(a+b)^{2n}\cdot b^{2n-1}} =\dfrac{a^{2n}}{(a+b)^{2n} \cdot \dfrac{a^{2n}}{a} } + \dfrac{b^{2n}}{(a+b)^{2n}\cdot \dfrac{b^{2n}}{b} }$

$=\dfrac{1}{(a+b)^{2n} \cdot \dfrac{1}{a} } + \dfrac{1}{(a+b)^{2n}\cdot \dfrac{1}{b} } = \dfrac{a}{(a+b)^{2n} } + \dfrac{b}{(a+b)^{2n} } = \dfrac{a+b}{(a+b)^{2n} }$

dividing both numerator and denominator by $(a+b)$ , we get

$\dfrac{a+b}{(a+b)^{2n} } = \dfrac{1}{(a+b)^{2n-1} }$

Therefore, it is proved that

$\dfrac{\sin ^{4n} x }{a^{2n-1}} + \dfrac{\cos^{4n} x }{b^{2n-1}} = \dfrac{1}{(a+b)^{2n-1}}, a,b\in\mathbb{N}$

4 0

2 years ago