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algol [13]
3 years ago
5

WILL GIVE BRAINLIEST!!! Which shape is NOT PROPORTIONAL to the other shapes?

Mathematics
1 answer:
Phoenix [80]3 years ago
4 0
The answer is D

Explanation:
If we use A as a reference sides:3&2
To get from A to B you half both sides to get :1.5&1

To get from A to C you multiply each side by 1.5 to get : 4.5 & 3

But to get from A to D the side that was 3 is multiplying by 1 while the side that was 2 is multiplying by 1.5
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Solve (i) 4/15+x=17/30​
Anna007 [38]

Answer:

3/10

Step-by-step explanation:

4/15+x=17/30

x=17/30-4/15

x=17/30-8/30

x=9/30

simplify

x=3/10

3 0
3 years ago
PLEASE HELP! I AM CONFUSED! Explain your answer, thank you so much! (I will give brainlyest)
antiseptic1488 [7]

Answer:

y+3 = 2(x-4)

Step-by-step explanation:

If we have a slope and a point, we can use the point slope form for an equation of a line

y -y1= m(x-x1)

where m is the slope and (x1,y1) is the point

y--3 = 2(x-4)

y+3 = 2(x-4)

8 0
3 years ago
Read 2 more answers
Six less than three times a number is eighteen. What is the number?
Snezhnost [94]

Answer: 6-3x=18

Step-by-step explanation: You just have to read what the question says for example Six less than three times a number is 6-3x=18

7 0
3 years ago
What is the rate of change for f(x) =7 sin x-1 on the interval from x=0 to x=piover 2
Oksi-84 [34.3K]
\bf slope = {{ m}}= \cfrac{rise}{run} \implies 
\cfrac{{{ f(x_2)}}-{{ f(x_1)}}}{{{ x_2}}-{{ x_1}}}\impliedby 
\begin{array}{llll}
average\ rate\\
of\ change
\end{array}\\\\
-------------------------------

\bf f(x)= 7sin(x)-1  \qquad 
\begin{cases}
x_1=0\\
x_2=\frac{\pi }{2}
\end{cases}\implies \cfrac{f\left( \frac{\pi }{2} \right)-f(0)}{\frac{\pi }{2}-0}
\\\\\\
\cfrac{[7\cdot 1-1]~-~[7\cdot 0-1]}{\frac{\pi }{2}}\implies \cfrac{6-(-1)}{\frac{\pi }{2}}\implies \cfrac{6+1}{\frac{\pi }{2}}\implies \cfrac{7}{\frac{\pi }{2}}\implies \cfrac{14}{\pi }
7 0
4 years ago
FOR ALL YOU STAR WARS FANS OUT THERE MY TEACHER GAVE US A STAR WARS MATH QUESTION. CHECK MY ANSWER MYBE? He asked how fast a bla
kvv77 [185]
(Links in comments below)

I'm using the clip from link [1] as a reference, i.e. the shot of the Tantive IV escaping the Devastator towards the camera, which lasts approximately 5 seconds (2:27 - 2:32). I recorded the particular shot, then carried out a cursory frame-by-frame analysis. The shot itself consists of 371 frames.

At around frame 15 - about 0.233s into the clip - one can barely make out a green projectile being launched from one of the Devastator's batteries. It takes the projectile between 19 and 20 frames (I'll round up to 20) - now at frame 35, about 0.333s later - for it to reach the left edge of the camera's field of view.

Let's suppose the Tantive IV was situated in roughly the same physical position in frame 1 as the projectile was in frame 35.

Now, the Tantive IV exits the screen at around frame 182 - about 3.017s into the clip. According to link [2], the standard CR90 "Corellian" Corvette can reach speeds of up to 950km/h. The Rebels are fleeing the Empire, so it's reasonable to assume that they are moving as fast as they can. Let's also assume they're traveling at a constant rate, and that the tractor beam aboard the Devastator hasn't been activated yet.

Note: Link [2] refers to this maximum speed as "atmosphere speed". I can't find any references as to what this means, so I assume it means something along the lines of "same conditions as Earth's atmosphere". The two ships are flying above Tatooine, which is inhabited in part by humans, so it's probably a safe bet that the atmospheres of Tatooine and Earth aren't that dissimilar.

So if it takes the Tantive IV about 3.017s to move (approximately) the same distance as the Devastator's projectile, and if we assume the Tantive IV is moving at maximum speed, then it must have traveled a distance d of (approximately)

d=\dfrac{950\text{ km}}{1\text{ h}}\times(3.017\text{ s})\implies d\approx0.796\text{ km}

However, it's quite clear from the previous shot that the assumption above is not all that accurate. So let's consider "your" assumption, that the distance between the Tantive IV and the Devastator is about 10 times its own length, i.e. d\approx2.296\text{ km}. Now, if the projectile takes about 0.333s to travel the same distance, then it must have moving at a speed r of (approximately)

2.296\text{ km}=r\times(0.333\text{ s})\implies r\approx24,821.6\dfrac{\text{km}}{\text{h}}
3 0
3 years ago
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