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3 years ago

Which of the inequalities describes the graph below?

Mathematics

1 answer:

Serga [27]3 years ago

7 0

The square in the graph represents one unit. In the diagram, the line has coordinates (0,4) and (4,0). Use the cpordinates, find the gradient and the equation using y = mx + c. So the equation of this line is x + y = 4. Now having obtained the equation of yhe line, the shaded part is below the line, therefore the inequality is : x + y =< 4

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PLS HELP ASAP!!!!<br> Determine whether each relation is a function. Explain your reasoning.

V125BC [204]

Each domain needs to have one range every x can only have 1 y they can be the same y value so...

Yes it is a function

6 0

3 years ago

7.45+(-24.45)+74.17+(-76.52)

Blizzard [7]

Answer:

-19.35

Step-by-step explanation:

Steps

7.45−24.45+74.17−76.52

−17+74.17−76.52

57.17−76.52

-19.35

3 0

3 years ago

HELP ME PLS!!!!

Marina CMI [18]

Answer:

10°

Step-by-step explanation:

360/36 = 10

3 0

3 years ago

A group of hikers walked 2 1/4 km in 3/4 hour.The distance here varies directly with time. What is the constant of variation?

telo118 [61]

The distance depends on the time.

So now we need to find the constant of variation or the constant of proportionality.

To do so we must find y(the dependent variable) and x(the independent variable).

To find the constant(k) we must find y/x

Since y/x=k

Then 2.25/.75=k

k=3

3 0

3 years ago

Read 2 more answers

Describe the translation of the function from the parent function f(x) = x².

kkurt [141]

$\bf \qquad \qquad \qquad \qquad \textit{function transformations}
\\ \quad \\\\

\begin{array}{rllll} 
% left side templates
f(x)=&{{ A}}({{ B}}x+{{ C}})+{{ D}}
\\ \quad \\
y=&{{ A}}({{ B}}x+{{ C}})+{{ D}}
\\ \quad \\
f(x)=&{{ A}}\sqrt{{{ B}}x+{{ C}}}+{{ D}}
\\ \quad \\
f(x)=&{{ A}}(\mathbb{R})^{{{ B}}x+{{ C}}}+{{ D}}
\\ \quad \\
f(x)=&{{ A}} sin\left({{ B }}x+{{ C}} \right)+{{ D}}
\end{array}$

$\bf \begin{array}{llll}
% right side info
\bullet \textit{ stretches or shrinks horizontally by } {{ A}}\cdot {{ B}}\\\\
\bullet \textit{ flips it upside-down if }{{ A}}\textit{ is negative}
\\\\

\end{array}\\$

$\bf \begin{array}{llll}
\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}\\\\
\bullet \textit{ vertical shift by }{{ D}}\\
\qquad if\ {{ D}}\textit{ is negative, downwards}\\\\
\qquad if\ {{ D}}\textit{ is positive, upwards}\\\\
\bullet \textit{ period of }\frac{2\pi }{{{ B}}}
\end{array}
$

now, notice the template above... now let's see your function $\bf y=x^2+4\implies 
\begin{array}{llllll}
y=&1(&1x&+&0)^2+&4\\
&A&B&&C&D
\end{array}$

so.. what do you think was the shift then?

8 0

3 years ago