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4 years ago

11

the memory card on melchers digital camera can hold about 430 pictures Melcher used 18% of memory card while taking pictures at

a family reunion about how many pictures in Melcher take at the family reunion? round to the nearest whole number

1 answer:

Talja [164]4 years ago

4 0

About 24 pictures because 18 divided by 430 is 24

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Plz help me with this <br> 1. 4 + 4y <br> 2. 4(y-1)<br> 3. 4y + 1<br> 4.1/2 (4-2x);2-2x

Lubov Fominskaja [6]

1:4+4y
2:4y-4
3:4y+1
4:2-x;2-2x

6 0

3 years ago

A toll bridge charges $1.00 for passenger cars and $2.75 for other vehicles. suppose that during daytime hours, 60% of all vehic

Sidana [21]

20 Cars on the bridge
60% of 20 ＝ 12 Passenger Car
12×1.00＝$12
20-12＝8 Other Vehicles
8×2.75＝$22
12+22＝$34

So expected revenue is $34

6 0

4 years ago

A hiker in Africa discovers a skull that contains 63% of its original amount of C-14.

tester [92]

Answer:

3820 years

Step-by-step explanation:

The age of the skull can be found using carbon-14 by applying the half-life relationship.

Age of skull is given by;

$t=(\frac{In\frac{N}{N0} }{-0.693} )*5730$

where N=63 N₀=100 -0.693=decay factor of carbon and 5730 =half-life of carbon-14, t =age

substitute values in equation

$t=(\frac{In\frac{63}{100} }{-0.693} )*5730\\\\\\t=(\frac{In0.63}{-0.693} )*5730\\\\\\t=(\frac{ -0.4620}{-0.693} )*5730\\\\\\t=(0.6667)*5730\\\\\\t=3820.29\\\\\\t=3820years$

4 0

4 years ago

What is the volume of a cylinder with base radius 333 and height 888?

ozzi

Answer:

V≈3.09×108

Step-by-step explanation:

Shape: Cylinder

Solved for volume

Radius: 333

Height: 888

Formula: Radius, Height

Exact answer: V≈3.09×108

<u><em>Hope this helps.</em></u>

4 0

4 years ago

How am I supposed to figure this out if csc Theta(0) has a radical number on the top?

Harman [31]

$\bf csc(\theta)=\cfrac{hypotenuse}{opposite}\implies csc(\theta)=\cfrac{\sqrt{7}}{2}\cfrac{\leftarrow hypotenuse=c}{\leftarrow opposite=b}\\\\
-----------------------------\\\\
\textit{so hmm using the pythagorean theorem, we can say that}\\\\
c^2=a^2+b^2\implies \pm \sqrt{c^2-b^2}=a\qquad 
\begin{cases}
c=hypotenuse\\
b=opposite\\
a=adjacent
\end{cases}\\\\
\textit{we'll assume is the + of the root, since you're not}\\
\textit{given further info on it, that means the I quadrant}\\\\
$

$\bf -----------------------------\\\\
cos(\theta)=\cfrac{a}{c}\qquad sec(\theta)=\cfrac{c}{a}\qquad tan(\theta)=\cfrac{b}{a}\qquad cot(\theta)\cfrac{a}{b}
$

so just find the "adjacent" side, using the pythagorean theorem, and even though the square root can give you +/-, use the positive one, assuming the angle is in the 1st quadrant

keeping in mind, that the 2nd quadrant is feasible as well, since the opposite side of +2 can also be in the 2nd quadrant

but anyhow, use the I quadrant

7 0

3 years ago