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2 years ago

LMN is a straight angle. Find m LMP and m NMP

Mathematics

1 answer:

viva [34]2 years ago

3 0

Answer:

(-16x + 13)° + (-20x + 23)° = 180° ( St. angle)

-16x° + 13° - 20x° + 23° = 180°

-4x° + 36° = 180°

-4x° = 180°-36°

-4x° = 144°

-x° = 144°/4

-x° = -36°

x° = 36°

Minus Minus cut

I hope you can know this answer

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For every 1 litre of water used to make a medicine, 300 ml of sucrose and 750ml of saline solution are used. Express the amount

Neko [114]

Answer: Required ratio will be 20:6:15.

Step-by-step explanation:

Since we have given that

For every 1 liter of water used to make a medicine, 300 ml of sucrose and 750 ml of saline solution are used.

So, We need to express the amount of water, sucrose and saline solution in its simplest form.

So, all amount will be in same units i.e. ml.

As we know that

$1\ liter=1000\ milimeter\\$

So, Ratio will be expressed as

$\text{Amount of water : Sucrose : saline solution}\\\\=1000:300:750\\\\=200:60:150\\\\=40:12:30\\\\=20:6:15$

Hence, required ratio will be 20:6:15.

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2 years ago

The physician orders 1500 flu injections. There were 300 shots given in September, 450 shots given in October, and 665 shots giv

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2 years ago

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The first terms of an infinite geometric sequence, Un are 2, 6, 18, 54... The first terms of a second infinite geometric sequenc

gladu [14]

Answer:

r = 9 and m = 112

Step-by-step explanation:

$\sum_{k=1}^{225}W_{k}=\sum_{k=0}^{m}4r^{k}$

Write W in terms of U and V.

$\sum_{k=1}^{225}(U_{k}+V_{k})=\sum_{k=0}^{m}4r^{k}\\\sum_{k=1}^{225}U_{k}+\sum_{k=1}^{225}V_{k}=\sum_{k=0}^{m}4r^{k}$

Define U and V using geometric series formula.

$\sum_{k=1}^{225}2(3)^{k-1}+\sum_{k=1}^{225}2(-3)^{k-1}=\sum_{k=0}^{m}4r^{k}$

Use sum of geometric series formula.

$2(\frac{1-(3)^{225}}{1-3})+2(\frac{1-(-3)^{225}}{1-(-3)})=4(\frac{1-(r)^{m+1}}{1-r})$

Simplify.

$-1(1-3^{225})+\frac{1+3^{225}}{2}=4(\frac{1-(r)^{m+1}}{1-r})\\-1+3^{225}+\frac{1}{2}+\frac{3^{225}}{2}=4(\frac{1-(r)^{m+1}}{1-r})\\-\frac{1}{2}+\frac{3(3^{225})}{2}=4(\frac{1-(r)^{m+1}}{1-r})\\\frac{-1+3(3^{225})}{2}=4(\frac{1-(r)^{m+1}}{1-r})\\\frac{-1+3^{226}}{2}=4(\frac{1-(r)^{m+1}}{1-r})\\4\frac{-1+3^{226}}{8}=4(\frac{1-(r)^{m+1}}{1-r})\\4\frac{1-3^{226}}{-8}=4(\frac{1-(r)^{m+1}}{1-r})\\4\frac{1-9^{113}}{1-9}=4(\frac{1-(r)^{m+1}}{1-r})$