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3 years ago

Which graph below shows the solutions for the linear inequality ys x-3?

Mathematics

1 answer:

liraira [26]3 years ago

4 0

Answer:

Graph A.

Step-by-step explanation:

Given the inequality: $y\leq \dfrac{1}{4}x-3$

Since the sign is "less than or equal to", the line cannot be dotted. Therefore, Options C and D are incorrect.

Since the sign is a "less than" sign, the required region must be below the line. Therefore, the graph which shows the given inequality is Graph A.

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PSYCHO15rus [73]

If you used $8, then you must add 8 to 32 which equals $40

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3 years ago

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What is the radius of the circle given the equation x2 - 6x + y2 + 10y + 18 = 0?

arlik [135]

$The\ equation\ of\ a\ circle:(x-a)^2+(y-b)^2=r^2\\\\(a;\ b)-a\ coordinates\ of\ the\ center\\r-a\ radius$

$x^2-6x+y^2+10y+18=0\\\\x^2-2x\cdot3+y^2+2y\cdot5=-18\\\\x^2-2x\cdot3+3^2+y^2+2y\cdot5+5^2=-18+3^2+5^2\ \ \ |use\ (a\pm b)^2=a^2\pm2ab+b^2\\\\(x-3)^2+(y+5)^2=-18+9+25\\\\(x-3)^2+(y+5)^2=16\\\\therefore\\\\Answer:\boxed{r=\sqrt{16}=4}$

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3 years ago

I need Help fast asap thank you

Jobisdone [24]

Answer:

1.25

Step-by-step explanation:

Because they are equal triangles, all of their sides are equal

so, lets take one side of each triangle (4, 5)

the law:

the scale factor = the larger side ÷ the smaller side

x = 5 ÷ 4

x = 1.25

For checking:

the larger side = the smaller side × the scale factor

Y = 4 × 1.25

Y = 5

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3 years ago

For the function given state the period f(t) =6sin(3t-pi/6)-1

lapo4ka [179]

$\bf \qquad \qquad \qquad \qquad \textit{function transformations}
\\ \quad \\
% function transformations for trigonometric functions
\begin{array}{rllll}
% left side templates
f(x)=&{{ A}}sin({{ B}}x+{{ C}})+{{ D}}
\\\\
f(x)=&{{ A}}cos({{ B}}x+{{ C}})+{{ D}}\\\\
f(x)=&{{ A}}tan({{ B}}x+{{ C}})+{{ D}}

\end{array}$

$\bf \begin{array}{llll}
% right side info
\bullet \textit{ stretches or shrinks}\\
\quad \textit{horizontally by amplitude } |{{ A}}|\\\\
\bullet \textit{ horizontal shift by }\frac{{{ C}}}{{{ B}}}\\
\qquad if\ \frac{{{ C}}}{{{ B}}}\textit{ is negative, to the right}\\\\
\qquad if\ \frac{{{ C}}}{{{ B}}}\textit{ is positive, to the left}\\\\
\end{array}$

$\bf \begin{array}{llll}


\bullet \textit{vertical shift by }{{ D}}\\
\qquad if\ {{ D}}\textit{ is negative, downwards}\\\\
\qquad if\ {{ D}}\textit{ is positive, upwards}\\\\
\bullet \textit{function period or frequency}\\
\qquad \frac{2\pi }{{{ B}}}\ for\ cos(\theta),\ sin(\theta),\ sec(\theta),\ csc(\theta)\\\\
\qquad \frac{\pi }{{{ B}}}\ for\ tan(\theta),\ cot(\theta)
\end{array}$

now, with that template in mind, let's take a peek at yours

$\bf \begin{array}{lllcclll}
f(t)=&6sin(&3t&-\frac{\pi }{6})&-1\\
&\uparrow &\uparrow &\uparrow &\uparrow \\
&A&B&C&D
\end{array}\\\\
-----------------------------\\\\
period\qquad \cfrac{2\pi }{B}\iff\cfrac{2\pi }{3}$

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3 years ago

A jar contains 2 green marbles, 4 blue marbles, 3 yellow marbles, and 2 black marbles. A marble is chosen at random from the jar

inysia [295]

For the first marble would be: 2/11

For the second: 3/11

6 0

3 years ago