The figure is made up of a square and a rectangle. Find the area of the shaded region.
Answers
Answer:
40 square metres
Step-by-step explanation:
The shaded region is of a triangle, whose area is denoted by: A = (1/2) * b * h, where b is the base and h is the height.
Since the left figure is a square with side lengths 10, we know that the height of the triangle is also 10 metres. The right figure is a rectangle with length 4. Since the total base length of the entire figure is 18 and the base of the square is 10, then the width of the rectangle is 18 - 10 = 8 metres.
This width is also the base of the triangle, so b = 8.
Now plug these values into the equation:
A = (1/2) * b * h
A = (1/2) * 8 * 10 = (1/2) * 80 = 40
The area is 40 square metres.
Answer:
40 m²
Step-by-step explanation:
Area of any triangle:
½ × base × height
base = 18 - 10 = 8
height = 10
Area:
½ × 8 × 10
40 m²
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Answer:
Answer: 2x+3y=-14
Y needs to be changed in order to maintain parallelism
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Answers
x^3y^2 + 4x^4y + 10x^9
x^4y^2 + 4x^4y^5 + 10x^2
xy^2 + 4x^3y^8 + 10x^2
x^4y^5 + 4x^3y^2 + 10x^2
Writing the polynomial in standard form means that you have to write the terms by descending degree:
x^4y^5 + 4x^3y^2 + 10x^2
The correct result would be x^4y^5 + 4x^3y^2 + 10x^2.
The polynomial written in standard form is the second one: x⁴y² + 4x⁴y⁵ + 10x².
In standard form, polynomials are arranged in descending order of the exponents of the variables, and the coefficients are written without any exponents. The given polynomial follows this format with the terms arranged in descending order of the exponents of x and y, and the coefficients are written as 1, 4, and 10. The exponents for x are 4, 4, and 2, while the exponents for y are 2, 5, and 0 (since any variable raised to the power of 0 is 1).
In conclusion, the second polynomial x⁴y² + 4x⁴y⁵ + 10x² is written in standard form as it follows the arrangement of terms with descending exponents of variables and coefficients written without exponents.
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f(x) = (x + 3)2 − 6
f(x) = (x + 6)2 + 3
f(x) = (x + 6)2 − 6
Answers
By completing the square, the second orderpolynomial in vertexform f(x) = (x + 3)² - 6 is equivalent to the polynomial in standardform f(x) = x² + 6 · x + 3. (Correct choice: B)
How to find the vertex form of the second order polynomial by algebraic means
In this question we must change the form of the second orderpolynomial from standardform into vertexform. A common method consists in completing the square, that is, to transform part of the polynomial into a perfect squaretrinomial. Now we proceed to find the vertexform of the expression:
1) x² + 6 · x + 3 Given
2) x² + 6 · x + 9 - 6 Modulative property/Existence of additive inverse/Definition of addition
3) (x + 3)² - 6 Associative property/Perfect square trinomial/Result
By completing the square, the second orderpolynomial in vertexform f(x) = (x + 3)² - 6 is equivalent to the polynomial in standardform f(x) = x² + 6 · x + 3. (Correct choice: B)
To learn more on second order polynomials in vertex form: brainly.com/question/20333425
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Answer:
B. f(x) = (x + 3)2 − 6
Step-by-step explanation:
I just did this for "completing the square". Hope this helped!
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Answer:
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Answers
Answer:
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Step-by-step explanation:
D for dog
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