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yulyashka [42]
2 years ago
12

Solve for x

Mathematics
1 answer:
Sindrei [870]2 years ago
4 0
33=34x+2 = 31/34 - as a decimal is 0.91176
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The volume of the triangular prim is 250 cubic inches. What is the value of x
Vinvika [58]

Answer:

12 in

Step-by-step explanation:

thank me and rate me plz!!!! thank you :)

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3 years ago
Find the missing lengths of the sides
Ket [755]

ANSWER

The correct answer is C

EXPLANATION

The given triangle is a right triangle. Since two angles are equal, it is a right isosceles triangle.

This implies that, x=8 units.

Using Pythagoras Theorem,

{y}^{2}  =  {8}^{2}  +  {8}^{2}

This implies that:

{y}^{2}  = 64  +  64

{y}^{2}  =128

Take positive square root,

{y}  =  \sqrt{128}

{y}  = 8 \sqrt{2}

The correct answer is C

5 0
3 years ago
Read 2 more answers
If someone could help me it would be awesome
alex41 [277]
The kite is 1500 feet above the ground
3 0
3 years ago
Please helpppp I’m not sure how to solve
Nesterboy [21]

Answer:

If a is 1.. we substitute one in every a in g(a)

4 * 1 + 16

4+ 16 = 20

g(a)= 20

f(20)

-16 + 20 = 4

and 4/4 = 1

so f (g (a)) = a

Step-by-step explanation:

6 0
2 years ago
Does anybody know how to do time with exponential decay
My name is Ann [436]
<span>From the message you sent me:

when you breathe normally, about 12 % of the air of your lungs is replaced with each breath. how much of the original 500 ml remains after 50 breaths

If you think of number of breaths that you take as a time measurement, you can model the amount of air from the first breath you take left in your lungs with the recursive function

b_n=0.12\times b_{n-1}

Why does this work? Initially, you start with 500 mL of air that you breathe in, so b_1=500\text{ mL}. After the second breath, you have 12% of the original air left in your lungs, or b_2=0.12\timesb_1=0.12\times500=60\text{ mL}. After the third breath, you have b_3=0.12\timesb_2=0.12\times60=7.2\text{ mL}, and so on.

You can find the amount of original air left in your lungs after n breaths by solving for b_n explicitly. This isn't too hard:

b_n=0.12b_{n-1}=0.12(0.12b_{n-2})=0.12^2b_{n-2}=0.12(0.12b_{n-3})=0.12^3b_{n-3}=\cdots

and so on. The pattern is such that you arrive at

b_n=0.12^{n-1}b_1

and so the amount of air remaining after 50 breaths is

b_{50}=0.12^{50-1}b_1=0.12^{49}\times500\approx3.7918\times10^{-43}

which is a very small number close to zero.</span>
5 0
3 years ago
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