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faltersainse [42]
3 years ago
11

Herbert is decorating the bulletin board in the school's lobby .the bulletin board is a 7 ft by 11ft rectangle,he decides to add

a black boarder around the entire bulletin board . what is the length of boarder that he needs​
Mathematics
1 answer:
Degger [83]3 years ago
8 0

Answer:

36 ft

Step-by-step explanation:

think of it as a rectangle, the shape has 4 sides. two of those sides are 7 ft long and two are 11 ft long so (7*2)+(11*2)

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What is Xavier’s available lot coverage?
slavikrds [6]

Answer:

<h2> <em><u>720</u></em><em><u>0</u></em></h2>

Step-by-step explanation:

<h3>At first:</h3>

<em><u>G</u></em><em><u>i</u></em><em><u>v</u></em><em><u>e</u></em><em><u>n</u></em><em><u>,</u></em>

Back of lot of the trapezoid = 175

Front of lot of the trapezoid = 185

Length of lot of the trapezoid = 100

<em><u>Therefore</u></em><em><u>,</u></em><em><u> </u></em>

Area of the trapezoid

=  \frac{a + b}{2} c

  • <em>[</em><em>On</em><em> </em><em>putting</em><em> </em><em>the</em><em> </em><em>values</em><em>]</em>

=  \frac{175 + 185}{2}  \times 100

  • <em>[</em><em>On</em><em> </em><em>Simplification</em><em>]</em>

= 360 × 50

  • <em>[</em><em>On multiplying</em><em>]</em>

= 18000

<em><u>Hence</u></em><em><u>,</u></em>

We got the <em>T</em><em>otal Area </em><em>as</em><em> </em><em>180</em><em>0</em><em>0</em><em>.</em>

<h3>Now:</h3>

<em><u>As</u></em><em><u> </u></em><em><u>per</u></em><em><u> </u></em><em><u>given</u></em><em><u> </u></em><em><u>equation</u></em><em><u>,</u></em>

  • \frac{40}{100}  =  \frac{x}{total \: area}
  • To find the value of x

<em><u>Solution</u></em><em><u> </u></em><em><u>,</u></em>

=  > \frac{40}{100}  =  \frac{x}{total \: area}

  • <em>[</em><em>On</em><em> </em><em>putting</em><em> </em><em>Total</em><em> </em><em>Area</em><em> </em><em>=</em><em> </em><em>180</em><em>0</em><em>0</em><em>]</em>

=  > \frac{40}{100}  =  \frac{x}{18000}

  • <em>[</em><em>On</em><em> </em><em>cross</em><em> </em><em>multiplication</em><em>]</em>

=> 40 × 18000 = 100 × x

  • <em>[</em><em>O</em><em>n</em><em> </em><em>Simplification</em><em> </em><em>]</em>

=> 720000 = 100x

  • <em>[</em><em>On</em><em> </em><em>dividing</em><em> </em><em>both</em><em> </em><em>sides</em><em> </em><em>with</em><em> </em><em>100</em><em>]</em>

=  >  \frac{720000}{100}  =  \frac{100x}{100}

  • <em>[</em><em>On</em><em> </em><em>Simplification</em><em> </em><em>]</em>

=> x = 7200

<em><u>Hence</u></em><em><u>,</u></em>

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Step-by-step explanation:

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3 years ago
Graph the system of inequalities presented here on your own paper, then use your graph to answer the following questions:
Sav [38]

Answer:

:Part A) The graph in the attached figure (see the explanation)

Part B) The ordered pair is not included in the solution area for the system (see the explanation)

Step-by-step explanation:

Part A) we have

y ----> inequality A

The solution of the inequality A  is the shaded area below the dashed line y=4x-8

The slope of the dashed line is positive

The y-intercept of the dashed line is (0,-8)

The x-intercept of the dashed line is (2,0)

y\geq -\frac{5}{2}x+5 ----> inequality B

The solution of the inequality B  is the shaded area above the solid line y=-\frac{5}{2}x+5

The slope of the solid line is negative

The y-intercept of the solid line is (0,5)

The x-intercept of the solid line is (2,0)

The solution of the system of inequalities is the shaded area between the dashed line and the solid line

see the attached figure

Part B) is the point (5, -8) Included in the solution area for the system?

we know that

If a ordered pair is a solution of the system of inequalities, then the ordered pair must satisfy both inequalities

Substitute the value of x=5, y=-8 in each inequality and then analyze the results

<em>Inequality A</em>

-8

-8 ----> is true

so

the ordered pair satisfy inequality A

<em>Inequality B</em>

-8\geq -\frac{5}{2}(5)+5

-8\geq -7.5 ---> is not true

so

the ordered pair not satisfy the inequality B

therefore

The ordered pair is not included in the solution area for the system

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