Answer:
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Answer:
![\[\frac{1}{24}\]](https://tex.z-dn.net/?f=%5C%5B%5Cfrac%7B1%7D%7B24%7D%5C%5D)
Step-by-step explanation:
Let the total number of batteries produced daily be represented by x.
7/8 of the batteries are packaged immediately.
So the total number of packaged batteries = ![\[\frac{7x}{8}\]](https://tex.z-dn.net/?f=%5C%5B%5Cfrac%7B7x%7D%7B8%7D%5C%5D)
Remaining batteries = ![\[x-\frac{7x}{8}\]](https://tex.z-dn.net/?f=%5C%5Bx-%5Cfrac%7B7x%7D%7B8%7D%5C%5D)
= ![\[\frac{x}{8}\]](https://tex.z-dn.net/?f=%5C%5B%5Cfrac%7Bx%7D%7B8%7D%5C%5D)
Two third of the remaining batteries are sent to charity.
So the total number of batteries sent to charity= ![\[\frac{2}{3} * \frac{x}{8}\]](https://tex.z-dn.net/?f=%5C%5B%5Cfrac%7B2%7D%7B3%7D%20%2A%20%5Cfrac%7Bx%7D%7B8%7D%5C%5D)
= ![\[\frac{x}{12}\]](https://tex.z-dn.net/?f=%5C%5B%5Cfrac%7Bx%7D%7B12%7D%5C%5D)
Remaining batteries are faulty.
So the total number of faulty batteries = ![\[\frac{x}{8} - \frac{x}{12}\]](https://tex.z-dn.net/?f=%5C%5B%5Cfrac%7Bx%7D%7B8%7D%20-%20%5Cfrac%7Bx%7D%7B12%7D%5C%5D)
= ![\[\frac{3x - 2x}{24}\]](https://tex.z-dn.net/?f=%5C%5B%5Cfrac%7B3x%20-%202x%7D%7B24%7D%5C%5D)
= ![\[\frac{x}{24}\]](https://tex.z-dn.net/?f=%5C%5B%5Cfrac%7Bx%7D%7B24%7D%5C%5D)
So the faction of batteries which are faulty =
= ![\[\frac{x}{24} \div x\]](https://tex.z-dn.net/?f=%5C%5B%5Cfrac%7Bx%7D%7B24%7D%20%5Cdiv%20x%5C%5D)
= ![\[\frac{1}{24}\]](https://tex.z-dn.net/?f=%5C%5B%5Cfrac%7B1%7D%7B24%7D%5C%5D)
Hi how are you hope all is well
Answer:
x = -9
Explanation:
(3x + 9)^5 = 32x^5
(3 (x + 3))^5 = 32x^5
3^5 (x + 3)^5 = 32x^5
243(x + 3)^5 = 32x^5
243(x + 3)^5 - 32x^5 = 0
x = -9
Answer:
UV=29
Step-by-step explanation:
In right triangles AQB and AVB,
∠AQB = ∠AVB ...(i) {Right angles}
∠QBA = ∠VBA ...(ii) {Given that they are equal}
We know that sum of all three angles in a triangle is equal to 180 degree. So wee can write sum equation for each triangle
∠AQB+∠QBA+∠BAQ=180 ...(iii)
∠AVB+∠VBA+∠BAV=180 ...(iv)
using (iii) and (iv)
∠AQB+∠QBA+∠BAQ=∠AVB+∠VBA+∠BAV
∠AVB+∠VBA+∠BAQ=∠AVB+∠VBA+∠BAV (using (i) and (ii))
∠BAQ=∠BAV...(v)
Now consider triangles AQB and AVB;
∠BAQ=∠BAV {from (v)}
∠QBA = ∠VBA {from (ii)}
AB=AB {common side}
So using ASA, triangles AQB and AVB are congruent.
We know that corresponding sides of congruent triangles are equal.
Hence
AQ=AV
5x+9=7x+1
9-1=7x-5x
8=2x
divide both sides by 2
4=x
Now plug value of x=4 into UV=7x+1
UV=7*4+1=28+1=29
<u>Hence UV=29 is final answer.</u>