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3 years ago

Give 3 jobs for smooth ER

Mathematics

2 answers:

lawyer [7]3 years ago

7 0

I dont understand what you are saying. :!

castortr0y [4]3 years ago

6 0

The smooth ER detoxifies drugs,stores lipids and regulates and releases calcium ions

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I need help with this math problem GP

Valentin [98]

We have

(a + bx) / (a - bx) = (b + cx) / (b - cx)

==> (a + bx) (b - cx) = (a - bx) (b + cx)

==> ab + (b ² - ac) x - bcx ² = ab + (ac - b ²) x - bcx ²

==> (b ² - ac) x = (ac - b ²) x

==> b ² - ac = ac - b ²

==> 2b ² = 2ac

==> b ² = ac … … … [1]

Similarly, you would find

(a + bx) / (a - bx) = (c + dx) / (c - dx)

==> ad = bc … … … [2]

and

(b + cx) / (b - cx) = (c + dx) / (c - dx)

==> c ² = bd … … … [3]

Now:

c ² = bd ==> b = c ² / d

b ² = ac ==> c = b ² / a

ad = bc ==> d = bc / a

and we find

d / c = (bc / a) / (b ² / a) = c / b

and

c / b = (b ² / a) / (c ² / d) = (b ² d) / (a c ²) = b / a

which is to say, the ratio between d and c is equal to the ratio between c and b, and also equal to the ratio between b and a. Therefore (a, b, c, d) are in a geometric progression.

8 0

3 years ago

Please help me with this problem

salantis [7]

136.75 is the answer to this equation

4 0

4 years ago

Read 2 more answers

If AC = 14, what is the measure of AD? Round your answer to the nearest hundredth.

nikklg [1K]

Part A

Answer: Choice D) $\text{AD} = \sqrt{(2n+5)^2+(4n-3)^2}$

-----------------------------------------------

Explanation:

Let E be the intersection point of the diagonals. For any parallelogram, the diagonals bisect each other. This means that AE = EC = 2n+5 and it also means EB = DE = 4n-3

Focus on triangle AED. This triangle is a right triangle with angle AED as a 90 degree angle (any rhombus has its diagonals that are perpendicular to one another). The legs of this triangle are AE = 2n+5 and ED = 4n-3

Apply the pythagorean theorem to find hypotenuse AD

$(\text{leg1})^2+(\text{leg2})^2 = (\text{hypotenuse})^2\\\\(\text{AE})^2+(\text{ED})^2 = (\text{AD})^2\\\\(\text{AD})^2 = (\text{AE})^2+(\text{ED})^2\\\\\text{AD} = \sqrt{(\text{AE})^2+(\text{ED})^2}\\\\\text{AD} = \sqrt{(2n+5)^2+(4n-3)^2}\\\\$

This points to choice D as the final answer.

===========================================================

Part B

<h3>Answer: 7.07</h3>

-----------------------------------------------

Explanation:

If AC = 14, then half of this is AE = 7. The diagonals bisect each other, so they cut each other in half with E at the midpoint.

Since AE = 2n+5, we can solve for n like so

2n+5 = 7

2n = 7-5

2n = 2

n = 2/2

n = 1

Then we can find the length of AD

$\text{AD} = \sqrt{(2n+5)^2+(4n-3)^2}\\\\\text{AD} = \sqrt{(2*1+5)^2+(4*1-3)^2}\\\\\text{AD} = \sqrt{(2+5)^2+(4-3)^2}\\\\\text{AD} = \sqrt{7^2+1^2}\\\\\text{AD} = \sqrt{49+1}\\\\\text{AD} = \sqrt{50}\\\\\text{AD} \approx 7.0710678\\\\\text{AD} \approx 7.07\\\\$

Segment AD is roughly 7.07 units long. The other external sides of the rhombus (AB, BC and CD) are also this length because the four sides of any rhombus are the same length.

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3 years ago

Can someone help me with this please

yuradex [85]

The Answer to this question is negative eight after substituting in the 2 and solving to get to 9, negative 8 solves this problem.

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3 years ago

A certain quadratic function has the factors

olganol [36]

Answer:

Step-by-step explanation:

If those two functions are the factors of a quadratic, then each of them can be equated to 0 to find the roots. Sometimes the roots are called the zeros.

2x - 5 = 0 Add 5 to both sides

2x - 5 + 5 = 0 + 5 Combine

2x = 5 Divide both sides by 2

2x/2 = 5/2

x = 5/2

You don't really have to find the other factor's 0 point. There is only 1 option that makes x = 5/2. On a test, it is important to take short cuts and save time. I'll find the other root anyway because this is not a test.

3x + 4 = 0 Subtract 4 from both sides

3x + 4-4 = 0 - 4 Combine

3x = - 4 Divide both sides by 3

3x/3 = - 4/3

x = - 4/3

7 0

2 years ago