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True [87]
3 years ago
9

ײ × ×⁴=ײ+⁴ find the ×​

Mathematics
1 answer:
Orlov [11]3 years ago
3 0
You have to do X to the power of two times X to the power of four and multiply that by itself and then multiply X to the power of 2+4 by itself to find the X
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POSSIBLE POINTS: 5.88
Bess [88]

The function of the length z in meters of the side parallel to the wall is A(z) = z/2(210 - z)

<h3>How to write a function of the length z in meters of the side parallel to the wall?</h3>

The given parameters are:

Perimeter = 210 meters

Let the length parallel to the wall be represented as z and the width be x

So, the perimeter of the fence is

P = 2x + z

This gives

210 = 2x + z

Make x the subject

x = 1/2(210 - z)

The area of the wall is calculated as

A = xz

So, we have

A = 1/2(210 - z) * z

This gives

A = z/2(210 - z)

Rewrite as

A(z) = z/2(210 - z)

Hence, the function of the length z in meters of the side parallel to the wall is A(z) = z/2(210 - z)

Read more about functions at

brainly.com/question/1415456

#SPJ1

5 0
1 year ago
Assume that x is a positive acute angle.<br> Given: sin x =28/53<br> Find: sin 2x
alexdok [17]

Answer:

its sophisticated but answer is 3

Step-by-step explanation:

you have to dive deep into it and research

4 0
3 years ago
can someone please help me out with this problem i been trying to figure out what the answer is .. ? or at least give me some ti
k0ka [10]

Answer: sin

Step-by-step explanation:

5 0
3 years ago
Gabriel must practice piano at least 15 hours a week. This week he has practiced 6 hours so far. Piano lessons are part of Gabri
suter [353]

Answer:

9

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
The equation of a parabola is y=x^2-10x+27. Write the equation in vertex form
azamat

Answer:

\large \boxed{y = (x - 5)^{2} + 2 }

Step-by-step explanation:

y = x² - 10x + 27

y = ax² + bx + c

This is the general form of the equation for a parabola.

We must convert it to the vertex form

y = (x - h)² + k, where (h,k) are the coordinates of the vertex.

We can do this by completing the square.

\begin{array}{rcll}y & = & x^{2} - 10x + 27 & \\y - 27 & = & x^{2} - 10x & \text{Subtracted 27 from each side}\\y - 27&= & x^{2} - 10x + 25 - 25 & \text{Added and subtracted (b/2)}^{2}\\y - 27&= & (x - 5)^{2} - 25 & \text{Wrote the first three terms as the square of a binomial}\\y& = & \mathbf{(x - 5)^{2} + 2} & \text{Added 27 to each side}\\\end{array}\\\text{The vertex form of the parabola is $\large \boxed{\mathbf{y = (x - 5)^{2} + 2 }}$}The figure below shows that your parabola has its vertex at (5,2).

4 0
3 years ago
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