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kipiarov [429]
2 years ago
12

Gary received a 3.5% raise in salary. If he is now paid $140 374.85, what was his previous salary?

Mathematics
1 answer:
Ostrovityanka [42]2 years ago
3 0

Answer:

$135461.73

Step-by-step explanation:

140374.85 decrease 3.5% =

140374.85 × (1 - 3.5%) = 140374.85 × (1 - 0.035) = 135461.73025

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What’s 2percent of 300
love history [14]

Answer:

6

Step-by-step explanation:

2%=0.02

0.02*300=6

7 0
3 years ago
PLEASE HELP! Neil has been running a tutoring business since 2005. He charges a monthly fee for weekly tutoring sessions and a p
Alina [70]

Answer:

Step-by-step explanation:

A. Rate of change (or slope) is $50

The initial tutoring fee starts out at $1200 in 2005

How do you know? Each year the cost goes up $50

 2005 = $1200

    2006 = $1250

    2007 = $1300

    2008 = $1350

B.  y=50x+b

3 0
2 years ago
Write the equation<br>x - 4y = 16 in<br>slope-intercept form.​
e-lub [12.9K]

<u><em>Good morning</em></u>,

Step-by-step explanation:

Look at the photo below for the Answer.

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3 0
3 years ago
Use synthetic substitution to find f (2) and f (-1) for the function.<br> f (x) = x3 – x2 – 2x + 3
kotegsom [21]

Answer:

f(2)=7

f(-1)=+1

Step-by-step explanation:

f (2)= 2³-2²+3 =8-4+3=7

f (-1)= (-1)³-(-1)²+3= -1- (+1) +3= -1-1+3=+1

3 0
2 years ago
Please help?
Mazyrski [523]

Answer:

23\sqrt{3}\ un^2

Step-by-step explanation:

Connect points I and K, K and M, M and I.

1. Find the area of triangles IJK, KLM and MNI:

A_{\triangle IJK}=\dfrac{1}{2}\cdot IJ\cdot JK\cdot \sin 120^{\circ}=\dfrac{1}{2}\cdot 2\cdot 3\cdot \dfrac{\sqrt{3}}{2}=\dfrac{3\sqrt{3}}{2}\ un^2\\ \\ \\A_{\triangle KLM}=\dfrac{1}{2}\cdot KL\cdot LM\cdot \sin 120^{\circ}=\dfrac{1}{2}\cdot 8\cdot 2\cdot \dfrac{\sqrt{3}}{2}=4\sqrt{3}\ un^2\\ \\ \\A_{\triangle MNI}=\dfrac{1}{2}\cdot MN\cdot NI\cdot \sin 120^{\circ}=\dfrac{1}{2}\cdot 3\cdot 8\cdot \dfrac{\sqrt{3}}{2}=6\sqrt{3}\ un^2\\ \\ \\

2. Note that

A_{\triangle IJK}=A_{\triangle IAK}=\dfrac{3\sqrt{3}}{2}\ un^2 \\ \\ \\A_{\triangle KLM}=A_{\triangle KAM}=4\sqrt{3}\ un^2 \\ \\ \\A_{\triangle MNI}=A_{\triangle MAI}=6\sqrt{3}\ un^2

3. The area of hexagon IJKLMN is the sum of the area of all triangles:

A_{IJKLMN}=2\cdot \left(\dfrac{3\sqrt{3}}{2}+4\sqrt{3}+6\sqrt{3}\right)=23\sqrt{3}\ un^2

Another way to solve is to find the area of triangle KIM be Heorn's fomula, where all sides KI, KM and IM can be calculated using cosine theorem.

7 0
3 years ago
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