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3 years ago

What the flip is this neither me nor my parents know what this is can u plz help me?

Mathematics

2 answers:

alexandr1967 [171]3 years ago

5 0

I'm going to write the problem out without the x because that can be confused as the variable x.

2 * m * m * t * t * 2 * 2 * m * 2 * t * t

There are 3 m's being multiplied and 4 t's being multiplied. We can write this as m³ and t⁴. There are four 2's being multiplied and 2⁴ = 16.

We can combine these together to get: 16m³t⁴

Answer: $\boxed {16m^3t^4}$

Vitek1552 [10]3 years ago

3 0

Get all the m’s together all the 1’s and all the 2’s

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Two people start from the same point. One walks east at 4 mi/h and the other walks northeast at 8 mi/h. How fast is the distance

kvasek [131]

Answer:

5.89 mi/h

Step-by-step explanation:

This problem can be solved by using different methods. I will use vectors since it's the simplest way in which we can solve it. This can be solved by using related rates of change though.

First, we start by drawing a diagram with the velocity vectors.

A= velocity of the first person

B= velocity of the second person

C= velocity in which they are moving away from each other.

Since there is no acceleration in the problem, we can suppose we are talking about constant speeds, so the velocity at which they are moving away from each other will always remain constant. (It doesn't matter what time it is, the velocity will always be the same)

Having said this we can solve this problem by using the components, by using law of cosines or graphically. I will use law of cosines. The idea is to find the length of side c.

Law of cosines:

$C^{2}=A^{2}+B^{2}-2ABcos \gamma$

so we can solve the formula for C so we get:

$C=\sqrt{A^{2}+B^{2}-2ABcos \gamma}$

and now we can substitute the values we know:

$C=\sqrt{(4)^{2}+(8)^{2}-2(4)(8)cos 45^{o}}$

$C=\sqrt{16+64-32\sqrt{2}}$

$C=\sqrt{80-32\sqrt{2}} mph$

if we want an exact answer, then that will be the exact answer, which approximates to:

C=5.89 mph

4 0

3 years ago

Which ordered pair represents the y-intercept of the graph of the equation y = -3x - 9?

german

The y-intercept is the point where the graph crosses the y-axis. It is found by setting x=0. For your equation, that makes the x-term vanish and only the constant remains. The ordered pair is ...
(x, y) = (0, -9)

4 0

3 years ago

Actor the following expressions: 10n + 15 GCF: Check: 1

ZanzabumX [31]

Answer: GCF = 5 (5 goes into both 10 and 15)

Step-by-step explanation:

10n/5= 2n

15/5=3

if we factor it out it would be:

5(2n+3)

5 0

3 years ago

Anna played tennis for 2 ¼ hours on Saturday and 1 ⅘ hours on Sunday. How many hours did she play tennis during the weekend?

Aleonysh [2.5K]

Answer:

The answer is $\frac{9}{20}$

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

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

3 years ago