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iren [92.7K]
2 years ago
8

HELP PLS PLS! WILL GIVE POINTS aND BRAINLYEST ​

Mathematics
2 answers:
Vedmedyk [2.9K]2 years ago
6 0

Answer:

See below ~

Step-by-step explanation:

<u>Blue inequality</u>

  • <u>y ≤ -x + 4</u>

<u></u>

<u>Red Inequality</u>

  • <u>y > 1/4x</u>
Nikitich [7]2 years ago
3 0

Answer:

See below

Step-by-step explanation:

Find the equation of the blue line....the blue shaded area is LESS than or EQUAL to this SOLID line

Blue line =    y = -x+4        so  <u> y <= -x+4</u>

Red line  =    y = 1/4 x             <u>  y > 1/4x   </u>   (dotted line says line is not included so the inequality is >   <u> NOT</u> ≥   )

<u></u>

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the area of a rectangle is 90 in2^. the ratio of the length to the width is 5:2. find the length and the width
-BARSIC- [3]

Length and width of rectangle is 15 inches and 6 inches respectively

<h3><u>Solution:</u></h3>

Given that area of a rectangle is 90 square inch

Ratio of length to the width = 5: 2.

Need to determine length and width of rectangle.  

As ratio of length to the width is 5 : 2

Lets assume length of rectangle = 5x inches and width of rectangle = 2x inches.

<em><u>The formula for area of rectangle is given as:</u></em>

\text { Area of rectangle }=\text { length of rectangle } \times \text { width of rectangle}

Substituting the given value of area of rectangle and assumed value of length and width of rectangle we get:

\begin{array}{l}{90=5 x \times 2 x} \\\\ {=>90=10 x^{2}}\end{array}

On solving the above expression for x we get

\begin{array}{l}{=>\frac{90}{10}=x^{2}} \\\\ {=>x^{2}=9} \\\\ {=>x=\sqrt{9}=3}\end{array}

\begin{array}{l}{\text { Length of rectangle }=5 \times x=5 \times 3=15 \text { inches }} \\\\ {\text { Width of rectangle }=2 \times} x=2 \times} 3=6 \text { inches }}\end{array}

Hence length and width of rectangle is 15 inches and 6 inches.

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3 years ago
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