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timama [110]
3 years ago
5

10x -2y = 10 in slope intercept form

Mathematics
1 answer:
Georgia [21]3 years ago
8 0
Slope-intercept form: y = mx + b   
the point is to get y alone on one side of the equation:

10x - 2y = 10
-10x         -10x
-2y = -10x + 10
-2y/-2 = -10x/-2 + 10/-2
y = 5x - 5
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A rectangular swimming pool has a base of 200 square feet. The pool has a depth of 5 feet. What is the volume of the pool? 25 ft
givi [52]

Answer:

  • D. 1000 ft³

Step-by-step explanation:

The volume is the product of base area and the height:

  • V = Bh

Find the volume:

  • V = 200 *5 = 1000 ft³

Correct choice is D

6 0
2 years ago
What is the solution to this equation 5x -8(x + 5)= 6-3x
Solnce55 [7]

Answer:

no solution

Step-by-step explanation:

5x - 8(x + 5) = 6 - 3x

Remember to follow PEMDAS. Note the equal sign, what you do to one side, you do to the other.

First, distribute -8 to all terms within the parenthesis

-8(x + 5) = (-8)(x) + (-8)(5) = -8x - 40

5x - 8x - 40 = 6 - 3x

Isolate the variable (x). Add 3x & 40 to both sides

5x - 8x (+3x) - 40 (+40) = 6 (+40) - 3x (+3x)

5x - 8x + 3x = 6 + 40

Simplify. Combine like terms

(5x - 8x + 3x) = (6 + 40)

0 = 46 ; False 0 ≠ 46 ∴ no solution is your answer

~

8 0
3 years ago
Which is bigger 6.5l or 645ml
snow_lady [41]

Answer:

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Step-by-step explanation:

3 0
2 years ago
When the sun is at a certain angle in the sky, a 50 foot building will cast a 20 foot shadow. What is the length of the shadow i
Sunny_sXe [5.5K]

Answer:

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Cauliflower= 10 per kg

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3 0
2 years ago
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Write the equation of a circle for which the endpoints of a diameter are (-2,-2) and (4,-10)
Maurinko [17]

Answer:

Therefore the required Equation of Circle

x^{2}+y^{2}-2x+12y+12=0

Step-by-step explanation:

Given:

End point of Diameter be

point A( x₁ , y₁) ≡ ( -2 ,-2 )

point B( x₂ , y₂) ≡ ( 4 , -10 )

To Find:

Equation of a circle =?

Solution:

When end points of the Diameter are A( x₁ , y₁) , B( x₂ , y₂). then the Equation of Circle is given as

(x-x_{1})(x-x_{2})+(y-y_{1})(y-y_{2})=0

Substituting the end point are

(x-(-2))(x-4)+(y-(-2))(y-(-10))=0\\(x+2))(x-4)+(y+2))(y+10))=0\\

Applying Distributive Property we get

x^{2} -2x-8+y^{2}+12y+20 =0\\\\x^{2}+y^{2}-2x+12y+12=0

Therefore the required Equation of Circle

x^{2}+y^{2}-2x+12y+12=0

6 0
2 years ago
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