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castortr0y [4]
3 years ago
12

Garrett found the slope of the values in the table: A 2-column table with 3 rows. Column 1 is labeled Years: x with entries 4, 8

, 12. Column 2 is labeled Hourly rate: y with entries 12.00, 13.00, 14.00. 1. slope = StartFraction 12 minus 8 Over 14.00 minus 13.00 EndFraction. 2. slope = StartFraction 4 Over 1.00 EndFraction. 3. slope = 4. Is Garrett’s slope correct? If not, identify his error?
Mathematics
2 answers:
kakasveta [241]3 years ago
5 0

Answer:

B

Step-by-step explanation:

irinina [24]3 years ago
5 0

Answer:

B i took the test and got it right

Step-by-step explanation:

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A card was selected at random from a standard deck of cards. The suit of the card was recorded, and then the card was put back i
aleksklad [387]

Answer:

30%

Step-by-step explanation:

6 0
3 years ago
What's the slope-intercept form that passes through the points (0, -1) and (1, 5)?
Rzqust [24]

Answer:

<h2>         y = 6x - 1</h2>

Step-by-step explanation:

\bold{slope\, (m)=\dfrac{change\ in\ Y}{change\ in\ X}=\dfrac{y_2-y_1}{x_2-x_1}}

(0, -1)    ⇒     x₁ = 0,  y₁ = -1

(1, 5)     ⇒     x₂ = 1,  y₂ = 5

So the slope:

                     \bold{m=\dfrac{5+1}{1-0}=\dfrac{6}{1}=6}

The slope-intercept form of the equation of line is y = mx + b, where m is the slope and b is the y-intercept of the line.

(0, -1)    ⇒     x₀ = 0,  y₀ = -1      ⇒   b = -1

Therefore:

                 y = 6x - 1          ←  the slope-intercept form of the equation

5 0
3 years ago
Whats 290 x 4000 pls help
tigry1 [53]

Answer: it’s 1160000

Step-by-step explanation: hope I helped:)

4 0
3 years ago
Read 2 more answers
Can someone help me out !<br><br>got stuck in this question for an hour.<br><br><br>​
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
How do I find the value of a trigonometric ratio
JulsSmile [24]
Check the picture below.

3 0
3 years ago
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