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2 years ago

11

Similar triangles. Solve for x

2 answers:

Vanyuwa [196]2 years ago

8 0

X would equal 3 i am just guessing so its not 100%

levacccp [35]2 years ago

7 0

(11x-4) / 60 = 70/50
550x -200 = 4200
550x = 4400
x = 4400 / 550
x =8

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Use the given circumference to approximate the surface area of the spherical object. Round the answer to the nearest square cent

Kruka [31]

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3 years ago

How do i check an equation for extraneous solutions?

Marina86 [1]

Example:

$\sqrt x(x+1)=0$

This suggests two solutions, $x=0$ and $x=-1$ .

However, upon plugging these solutions back into the equation, you get

$\sqrt0(0+1)=0(1)=0$

which checks out, but

$\sqrt{-1}(1-1)=0$

does not because $\sqrt x$ is defined only for $x\ge0$ (assuming you're looking for real solutions only). So, we call $x=-1$ an extraneous solution, and the complete solution set (over the real numbers) is $x=0$ .

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2 years ago

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Mandarinka [93]

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3 years ago

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The substraction of a matrix B may be considered as the addition of the matrix (-1)B. Dose the commutative law of addition permi

GuDViN [60]

Answer: The corrected statement is A - B = -B + A.

Step-by-step explanation: Given that the subtraction of a matrix B may be considered as the addition of the matrix (-1)B.

We are given to check whether the commutative law of addition permit us to state that A - B = B - A.

If not, We are to correct the statement.

If the subtraction A - B is considered a the addition A + (-B), then the commutative law should be stated as follows :

A + (-B) = (-B) + A.

That is, A - B = -B + A.

Thus, the corrected statement is A - B = -B + A, not B - A.

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2 years ago

Y=-1/3x^2-4x-5 in vertex form

grigory [225]

Answer:

$Y=-\frac{1}{3}(x+6)^2+7$ [Vertex form]

Step-by-step explanation:

Given function:

$Y=-\frac{1}{3}x^2-4x-5$

We need to find the vertex form which is.,

$y=a(x-h)^2+k$

where $(h,k)$ represents the co-ordinates of vertex.

We apply completing square method to do so.

We have

$Y=-\frac{1}{3}x^2-4x-5$

First of all we make sure that the leading co-efficient is =1.

In order to make the leading co-efficient is =1, we multiply each term with -3.

$-3\times Y=-3\times\frac{1}{3}x^2-(-3)\times4x-(-3)\times 5$

$-3Y=x^2+12x+15$

Isolating $x^2$ and $x$ terms on one side.

Subtracting both sides by 15.

$-3Y-15=x^2+12x-15-15$

$-3Y-15=x^2+12x$

In order to make the right side a perfect square trinomial, we will take half of the co-efficient of $x$ term, square it and add it both sides side.

square of half of the co-efficient of $x$ term = $(\frac{1}{2}\times 12)^2=(6)^2=36$

Adding 36 to both sides.

$-3Y-15+36=x^2+12x+36$

$-3Y+21=x^2+12x+36$

Since $x^2+12x+36$ is a perfect square of $(x+6)$ , so, we can write as:

$-3Y+21=(x+6)^2$

Subtracting 21 to both sides:

$-3Y+21-21=(x+6)^2-21$

$-3Y=(x+6)^2-21$

Dividing both sides by -3.

$\frac{-3Y}{-3}=\frac{(x+6)^2}{-3}-\frac{21}{-3}$

$Y=-\frac{1}{3}(x+6)^2+7$ [Vertex form]

8 0

3 years ago