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sergeinik [125]
3 years ago
14

In 2014 the tenants at Jefferson school's Fall festival Was 650. In 2015 the attendance was 575. What was the percent change in

attendance from 2014 to 2015
Mathematics
1 answer:
Over [174]3 years ago
7 0

Answer: the percent change in attendance from 2014 to 2015 is 11.54%

Step-by-step explanation:

In 2014, the total attendance at Jefferson school's Fall festival was 650.

In 2015, the attendance was 575. The change in the number of attendees between 2014 and 2015 is 650 - 575 = 75

There was a decrease in the number of attendees between 2014 and 2015.

The percent change in attendance from 2014 to 2015 would be attendance in 2015 divided by attendance in 2014 and multiplied by 100. It becomes

75/650 × 100 = 11.54%

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Step-by-step explanation:

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If x) = 4x^2 + 1 and g(x) = x^2 - 5, find (f + g)(x).​
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Answer:

C. 5x² - 4

General Formulas and Concepts:

<u>Algebra I</u>

  • Composite Functions
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Step-by-step explanation:

<u>Step 1: Define</u>

f(x) = 4x² + 1

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<u>Step 2: Find (f + g)(x)</u>

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3 years ago
Find the area of triangle ABC with vertices A(2,1), B (12,2), C (12,8). Hence, or otherwise find the perpendicular distance from
Lapatulllka [165]
<h2>The area of a triangle is =54 square units</h2><h2>The perpendicular distance from B to AC is = \frac{108}{\sqrt{149} } units</h2>

Step-by-step explanation:

Given a triangle ABC with vertices A(2,1),B(12,2) and C(12,8)

x_1=2,y_1=1,x_2=12,y_2=2,x_3=12 and      y_3=8

The area of a triangle is= \frac{1}{2} [x_1(y_2-y_3) +x_2 (y_3- y_1)+x_3(y_1-y_2)]

=|\frac{1}{2} [2(2-8+12(8-1)+12(1-2)]|

=|-54| = 54 square units

The length of AC = \sqrt{(x_1-x_3)^{2} +(y_1-y_3)^2}

                          = \sqrt{(2-12)^{2} +(1-8)^2}

                         =\sqrt{149} units

Let the perpendicular distance from B to AC be = x

According To Problem

\frac{1}{2} \times  x \times \sqrt{149} = 54

⇔x =\frac{108}{\sqrt{149} } units

Therefore the perpendicular distance from B to AC is = \frac{108}{\sqrt{149} } units

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Multiply the fraction by 100%.

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