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3 years ago

Eleven more than a number sqaured, decreased by six

Mathematics

1 answer:

Evgesh-ka [11]3 years ago

5 0

Answer:

eleven more than a number squared, decreased

by six;

evaluate when t = 5

Evaluating Expressions

Write the expression as 11 +t2 -6

Write the expression as 11 +2t -6

Write the expression as t underscore 2

The value when t = 5 is 16.

The value when t = 5 is 30.

The value when t = 5 is 7.5.

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Each box has 800 spoons.

Step-by-step explanation:

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Can i get some help with #5 and below thankyou

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Step-by-step explanation:

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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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150 / 1250 = 0.12

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