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labwork [276]
3 years ago
7

Kyle wants to compare 0.25 to 1/2 using fractions. Which fraction can he use for 0.25? A) 14 B) 25 C) 34 D) 45​

Mathematics
2 answers:
Andreas93 [3]3 years ago
8 0

Answer:

1/4

Step-by-step explanation:

0.25 is 1/4 of a whole use quarters for example

tatyana61 [14]3 years ago
6 0

Answer:

(B) 25

Step-by-step explanation:

0.25

after the point there are two there

means 25/100

25/100=1/4

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2 lines intersect. A line with points T, R, W intersects a line with point S, R, V at point R. 4 angles are created. Labeled clo
Agata [3.3K]

Answer:

45

Step-by-step explanation:

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5 0
3 years ago
Read 2 more answers
What is the slope of the following 12x - y = 30<br> -12
antiseptic1488 [7]

Answer:

12

Step-by-step explanation:

Rearrange into the form y = mx + c, where m is the slope.

12x - y = 30

y = 12x - 30

m = 12

4 0
3 years ago
R0-180(x, y) maps to the same point as 80,180(x, y).<br> True<br> False
mariarad [96]
The anser is false that what i think ):):(:(:(:(:(:
6 0
3 years ago
What is the mean deviation of 4, 3, 4, 5, 6, 6, 8, 4?
MrRa [10]
Calculate mean:
4+3+4+5+6+6+8+4=40
40/8=5

Calculate MAD:
1+2+1+0+1+1+3+1=10
10/8= 1.25

Final answer: 1.25
5 0
4 years ago
Read 2 more answers
For what values of x is log base 0.8 (x+4)&gt;log base 0.4 (x+4)<br>Thank you!
Radda [10]
We are seeking the solution of the inequality:

\displaystyle{ \log_{0.8}(x+4)\ \textgreater \ \log_{0.4}(x+4).


We recall that a log function f(x)=\log_b(x) is either increasing or decreasing:

i) it is increasing if b>1, 

ii) it is decreasing if 0<b<1.

Consider the functions \displaystyle{ \log_{0.8}(x) and \displaystyle{ \log_{0.4}(x).

The graphs of these functions both meet at x=1 (clearly), and after 1 they are both negative. So from 0 to 1 one of them is larger for all x, and from 1 to infinity the other is larger. (Being strictly decreasing, their graphs can only intersect once.)


We can check for a certain convenient point, for example x=0.8:

\displaystyle{ \log_{0.8}(0.8)=1 and

\displaystyle{ \log_{0.4}(0.8)=\log_{0.4}(0.4\cdot 2)=\log_{0.4}(0.4)+\log_{0.4}(\cdot 2)=1+\log_{0.4}(2).

Now, \displaystyle{ \log_{0.4}(2) is negative since we already explained that for x>1 both functions were negative. This means that 

\displaystyle{ \log_{0.8}(0.8)\ \textgreater \ \log_{0.4}(0.8), and since 0.8\in (0, 1), then this is the interval where \displaystyle{ \log_{0.8}(x)\ \textgreater \ \log_{0.4}(x).


So, now considering the functions \displaystyle{ \log_{0.8}(x+4) and \displaystyle{ \log_{0.4}(x+4), we see that 

x+4 must be in the interval (0,1), so we solve:

0<x+4<1, which yields -4<x<-3 after we subtract by 4.


Answer: (-4, -3). Attached is the graph generated using Desmos.

7 0
3 years ago
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