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Mrac [35]
2 years ago
11

Translate to an equation then solve the equation. Seven times the difference of some number and 2 amounts to the quotient of 196

and 14. (Type the question using x as the variable. Do not simplify.) x= ____
Mathematics
2 answers:
inna [77]2 years ago
8 0

Answer:

x = 4

Step-by-step explanation:

<u>7 * ( x-2) = 196/14</u>

7x -14 = 14

7x = 28

x = 4

snow_tiger [21]2 years ago
3 0
Create the equation, then solve for x

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If the graph crosses the x-axis and appears almost linear at the intercept, it is a single zero. If the graph touches the x-axis and bounces off of the axis, it is a zero with even multiplicity. If the graph crosses the x-axis at a zero, it is a zero with odd multiplicity. The sum of the multiplicities is the degree
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100 divided by 32 will give brainliest
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Find the missing lengths and leave answers as radicals in simplest form?
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Answer:

x=6

y=3

Step-by-step explanation:

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3 years ago
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Please help on the area for 7 8 9 please all i need help is
weeeeeb [17]

The area of the equilateral, isosceles and right angled triangle are 12.6mm², 9.61in² and 16.81yds² respectively.

<h3>What is the area of the equilateral, isosceles and right angle triangle?</h3>

Note that:

The area of an Equilateral triangle is expressed as A = ((√3)/4)a²

Where a is the dimension of the side.

The area of an Isosceles triangle is expressed as A = (ah)/2

Where a is the dimension of the base and h is the height.

The area of a Right angled triangle is expressed as A = (ab)/2

Where a and b is the dimension of the two sides other than the hypotenuse.

For the Equilateral triangle.

Given that;

  • a = 5.4mm
  • Area A = ?

A = ((√3)/4)(5.4mm)²

A = ((√3)/4)( 29.16mm² )

A = 12.6mm²

Area of the Equilateral triangle is 12.6mm²

For the Isosceles triangle.

Given that;

  • Base a = 3.4in
  • Slant height b = 5.9in
  • height h = ?
  • Area A = ?

The height h is the imaginary line drawn upward from the center of a.

First, we calculate the height using Pythagorean theorem

x² = y² + z²

Where x = b = 5.9in, y = a/2 = 3.4in/2 = 1.7in, and z = h

(5.9in)² = (1.7in)² + h²

34.81in² = 2.89in² + h²

h² = 34.81in² - 2.89in²

h² = 31.92in²

h = √31.92in²

h = 5.65in

Now, the area will be;

A = (ah)/2

A = (3.4in × 5.65in )/2

A = 19.21in²/2

A = 9.61in²

Area of the Isosceles triangle is 9.61in².

For the Right angled triangle

Given that;

  • a = 8.2yds
  • b = 4.1yds
  • c = 9.17yds
  • Area A = ?

A = (ab)/2

A = ( 8.2yds × 4.1yds)/2

A = ( 33.62yds²)/2

A = 16.81yds²

Area of the Right angled triangle is 16.81yds²

Therefore, the area of the equilateral, isosceles and right angled triangle are 12.6mm², 9.61in² and 16.81yds² respectively.

Learn more about Pythagorean theorem here: brainly.com/question/343682

#SPJ1

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2 years ago
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General form is ax^2 + bx + c = 0

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