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velikii [3]
3 years ago
10

Expand and simplify (3x+5)(4x-1)

Mathematics
1 answer:
Whitepunk [10]3 years ago
5 0
Use FOIL and simply, First times First, Outside times Outside, Inside times Inside, Last times Last.

(3x+5)(4x-1)
3x * 4x = 12x²
3x * -1 = -3x
5 * 4x = 20x
5 * -1 = -5

12x² - 3x + 20x - 5

Now simplify by adding -3x to 20x and get:

12x² + 17x - 5
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Lesson 7 module 4 grade 6 please help me
torisob [31]
Length of rectangle

Rectangle 1) 23m
Rectangle 2) 14yd

Width of rectangle

Rectangle 1) 36m
Rectangle 2) 3.5yd

An expressions is an equations but without the equal sign ( = ) and without the solution ( the answer)

The formula To find the area of a rectangle is

A = base * height
or
A = length * width

Rectangle’s area written in as an expression

Rectangle 1) 23 * 36
Rectangle 2) 14 * 3.5

To find the area, you multiply the numbers

23 * 36 = 828
14 * 3.5 = 49

Rectangle’s area written as a number

Rectangle 1) 828m
Rectangle 2) 49yd
6 0
2 years ago
Read 2 more answers
Plz help me with this question
Nezavi [6.7K]

Answer:

first two is 300, then it is Car A's speed is 50mph and b is 40mph and for the other two it is probably faster and steeper

Step-by-step explanation:

you look at the graph. you see 6 miles and look up along the 6 mile line to see where it intersects the red line. once you find that spot you go left along that line to find the answer. same with the other and it both ends up at 300

for the next two find the slope, which is 40 and 50

and for the next two there are multiple answers but i t is probably faster and steeper

4 0
3 years ago
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dmitriy555 [2]

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2 years ago
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anyanavicka [17]
This is the solution I found

6 0
2 years ago
express answer in exact form. Show all work for full credit. A segment of a circle has a 120 arc and a chord of 8 sq root 3. Fin
Anna35 [415]

Let \alpha be the angle of an arc,  \alpha =120^{\circ}, and r be the radius of the circle.

Then by the cosine rule,

\text{chord}^2=r^2+r^2-2\cdot r\cdot r\cdot \cos \alpha,\\ \\(8\sqrt{3})^2=2r^2-2r^2\cdot \cos 120^{\circ}=2r^2-2r^2\cdot \left(-\dfrac{1}{2}\right)=3r^2,\\ \\192=3r^2,\\ \\r^2=64,\\ \\r=8\ un.

1. Find the area of the sector. Since \alpha =120^{\circ}, then

A_{sector}=\dfrac{\pi r^2}{3}=\dfrac{64\pi}{3}\ un^2.

2. Find the area of the triangle formed by two radii and chord:

A_{triangle}=\dfrac{1}{2}\cdot r\cdot r\cdot \sin \alpha=\dfrac{64}{2}\cdot \dfrac{\sqrt{3}}{2}=16\sqrt{3}\ un^2.

3. The area of the segment is

A_{segment}=A_{sector}-A_{triangle}=\dfrac{64\pi}{3}-16\sqrt{3}=\dfrac{64\pi-48\sqrt{3}}{3}\ un^2.

Answer: \dfrac{64\pi-48\sqrt{3}}{3}\ un^2.

6 0
3 years ago
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