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2 years ago

Not drawn

Mathematics

1 answer:

Sveta_85 [38]2 years ago

6 0

Answer:

Yes, It is a Right Triangle.

Step-by-step explanation:

The hypotenuse is the longest side of the triangle, so for this problem we set the hypotenuse's value to 18.2

From there it doesn't matter which side is which, as both will undergo the same operations.

If the following equation is true, then it is a right triangle.

7²+16.8²= 18.2²

49 + 282.24 = 331.24

The equation is True.

It is a right-angled Triangle.

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What’s the slope-intercept for this equation???

andrezito [222]

Y= 2/3x + 10 (or 11?) is the equation

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2 years ago

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Adam wants to ride her bicycle more than 80 miles this week, he has already ridden his bicycle 18 miles. Which inequality could

timofeeve [1]

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3 years ago

SHOW YOUR FORMULA AND SOLUTION

spayn [35]

Answer:

2500 kg

Step-by-step explanation:

Given:

• Force produced by the truck = F = 15,000 Newtons Acceleration of the truck = a = 6 m/s²

To find:

Mass of the truck if force is 15000 newtons and acceleration is 6 m/s²

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Force = Mass Acceleration

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11 months ago

If 2452=60q+52r and q and r are positive integers what is one possible value for q+r

Alex73 [517]

The simplest solution is:
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7 0

2 years ago

Find the solution to this system of equations <br> 3x+2y+3z=3 4x-5y+7z=1 2x+3y-2z=6 <br> x=? Y=? Z=?

Ede4ka [16]

Answer:

The solution is $x=2,\ y=0,\ z=-1.$

Step-by-step explanation:

You are given the system of three equations:

$\left\{\begin{array}{l}3x+2y+3z=3\\4x-5y+7z=1\\2x+3y-2z=6\end{array}\right.$

Multiply the first equation by 4, the second equation by 3 and subtract them. Then multiply the third equation by 2 and subtract it from the second equation:

$\left\{\begin{array}{l}3x+2y+3z=3\\4(3x+2y+3z)-3(4x-5y+7z)=4\cdot 3-3\cdot 1\\4x-5y+7z-2(2x+3y-2z)=1-2\cdot 6\end{array}\right.\Rightarrow \\\\\left\{\begin{array}{l}3x+2y+3z=3\\12x+8y+12z-12x+15y-21z=12-3\\4x-5y+7z-4x-6y+4z=1-12\end{array}\right.$

So,

$\left\{\begin{array}{rl}3x+2y+3z=3\\23y-9z=9\\-11y+11z=-11\end{array}\right.\Rightarrow \left\{\begin{array}{l}3x+2y+3z=3\\23y-9z=9\\y-z=1\end{array}\right.$

Multiply the third equation by 23 and subtract it from the second equation:

$\left\{\begin{array}{rl}3x+2y+3z=3\\23y-9z=9\\23y-9z-23(y-z)=9-23\cdot 1\end{array}\right.\Rightarrow \left\{\begin{array}{rl}3x+2y+3z=3\\23y-9z=9\\23y-9z-23y+23z=9-23 \end{array}\right.$

Hence,

$\left\{\begin{array}{rl}3x+2y+3z=3\\23y-9z=9\\14z=-14 \end{array}\right.\Rightarrow z=-1$

Substitute it into the second equation:

$23y-9\cdot (-1)=9\Rightarrow 23y+9=9\\ \\23y=0\\ \\y=0$

Substitute them into the first equation:

$3x+2\cdot 0+3\cdot (-1)=3\Rightarrow 3x-3=3\\ \\3x=6\\ \\x=2$

The solution is $x=2,\ y=0,\ z=-1.$

3 0

3 years ago