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2 years ago

Let the given rhombus be abcd in which ab = bc = cd = da = 20 cm and

Mathematics

1 answer:

natka813 [3]2 years ago

6 0

Answer:

384 cm²

Step-by-step explanation:

Property of rhombus :

Diagonals of rhombus bisect each other and are perpendicular to each other

Given :

Rhombus ABCD and its diagonals AC and BD bisect each other at O
AC ⊥ BD
AB = BC = CD = DA = 20 cm
AC = 24 cm

Solving :

Area of the given rhombus is twice the Area of ΔABC formed by the given diagonal and two adjacent sides.

2 × 1/2 × OB × AC
24 × OB
24 × √AB² - AO²
24 × √20² - (AC/2)²
24 × √400 - 144
24 × √256
24 × 16
384 cm²

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1 Determine the value of k if g(x)=4x+k is a tangent to f(x)=-x²+8x+20

tangare [24]

Answer:

k=24

Step-by-step explanation:

The tangent of the function f at x=a, can be found by differentiating f w.r.t. x and then replacing x with a.

f=-x^2+8x+20

Differentiating both sides:

f'=(-x^2+8x+20)'

By sum rule:

f'=(-x^2)'+(8x)'+(20)'

By constant multiple rule:

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The slope of g is 4 if you compare it to slope intercept form y=mx+b.

So we gave -2a+8=4.

Subtracr 8 on both sides: -2a=-4

Divide both sides by -2: a=2

The tangent line to the curve at x=2 is y=4x+k.

To tind y we must first know the y-coordinate of the point of tangency.

If x=2, then

f(2)=-(2)^2+8(2)+20=-4+16+20=12+20=32

So the point is (2,32).

g(x)=4x+k and we know g(2)=32.

This gives us:

32=4(2)+k

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Given the vertex of a quadratic function, find the axis of symmetry.

motikmotik

(i) The equation of the axis of symmetry is x = - 5.

(ii) The coordinates of the vertex of the parabola are (h, k) = (4, - 18). The x-value of the vertex is 4.

(iii) According to the vertex form of the quadratic equation, the parabola opens down due to negative lead coefficient and has a vertex at (2, 4), which is a maximum.

<h3>How to analyze and interpret quadratic functions</h3>

In this question we must find and infer characteristics from three cases of quadratic equations. (i) In this case we must find a formula of a axis of symmetry based on information about the vertex of the parabola. Such axis passes through the vertex. Hence, the equation of the axis of symmetry is x = - 5.

(ii) We need to transform the quadratic equation into its vertex form to determine the coordinates of the vertex by algebraic handling:

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In a nutshell, the coordinates of the vertex of the parabola are (h, k) = (4, - 18). The x-value of the vertex is 4.

(iii) Now here we must apply a procedure similar to what was in used in part (ii):

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