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3 years ago

Solve for x in the diagram

Mathematics

1 answer:

lutik1710 [3]3 years ago

5 0

Answer:

Step-by-step explanation:

This is a Pythagorean theorem problem. You must do a^2+b^2=c^2

9^2+12^2=225

then, we must find the square root of 225 which is 15.

hope this helps -Cam <3

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What is the common ratio Between successive terms in the sequence? 1.5,1.2,0.96,0.768

lidiya [134]

Answer:

0.8

Step-by-step explanation:

0.768/0.96= 0.8

0.96/1.2 =0.8

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3 years ago

For which x is f(x)?=-3

Rzqust [24]

Answer:

the answer would be x=10

Step-by-step explanation:

cuz you do a t-chart the put x/-3 then + 10 then - witch is

x=10

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3 years ago

What is the solution of x=2+ square root x-2?

alex41 [277]

So I'm going to assume that this question is asking for non extraneous solutions, or solutions that are found in the equation and are valid solutions when plugged back into the equation. So firstly, subtract 2 on both sides of the equation:

$x-2=\sqrt{x-2}$

Next, square both sides:

$(x-2)^2=x-2\\(x-2)(x-2)=x-2\\x^2-4x+4=x-2$

Next, subtract x and add 2 to both sides of the equation:

$x^2-5x+6=0$

Now we are going to be factoring by grouping to find the solution(s). Firstly, what two terms have a product of 6x^2 and a sum of -5x? That would be -3x and -2x. Replace -5x with -2x - 3x:

$x^2-2x-3x+6=0$

Next, factor x^2 - 2x and -3x + 6 separately. Make sure that they have the same quantity on the inside of the parentheses:

$x(x-2)-3(x-2)=0$

Now you can rewrite the equation as $(x-3)(x-2)=0$

Now, apply the Zero Product Property and solve for x as such:

$x-3=0\\x=3\\\\x-2=0\\x=2$

Now, it may appear that the answer is C, however we need to plug the numbers back into the original equation to see if they are true as such:

$2=2+\sqrt{2-2}\\2=2+\sqrt{0}\\2=2+0\\2=2\ \textsf{true}\\\\3=2+\sqrt{3-2}\\3=2+\sqrt{1}\\3=2+1\\3=3\ \textsf{true}$

Since both solutions hold true when x = 2 and x = 3, your answer is C. x = 2 or x = 3.

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3 years ago

7-10. Waits - Conditional Statemeru B-Converse Statement C - Inverse Statement

Eva8 [605]

Answer:

Step-by-step explanation:

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2 years ago

Find the area of the largest rectangle (with sides parallel to the coordinate axes) that can be inscribed in the region enclosed

garik1379 [7]

F(x) = 18-x^2 is a parabola having vertex at (0, 18) and opening downwards.
g(x) = 2x^2-9 is a parabola having vertex at (0, -9) and opening upwards.
By symmetry, let the x-coordinates of the vertices of rectangle be x and -x => its width is 2x.
Height of the rectangle is y1 + y2, where y1 is the y-coordinate of the vertex on the parabola f and y2 is that of g.
=> Area, A
= 2x (y1 - y2)
= 2x (18 - x^2 - 2x^2 + 9)
= 2x (27 - 3x^2)
= 54x - 6x^3
For area to be maximum, dA/dx = 0 and d²A/dx² < 0
=> 54 - 18x^2 = 0
=> x = √3 (note: x = - √3 gives the x-coordinate of vertex in second and third quadrants)

d²A/dx² = - 36x < 0 for x = √3
=> maximum area
= 54(√3) - 6(√3)^3
= 54√3 - 18√3
= 36√3.

4 0

2 years ago