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Black_prince [1.1K]
3 years ago
11

Can someone help me with this please

Mathematics
1 answer:
scZoUnD [109]3 years ago
5 0
I can’t really see the picture . I need a better one
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The domain of a relation is
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The domain is the set of all first elements of ordered pairs (x-coordinates).
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3 years ago
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A bus travelled from location A to location B in 90 minutes. Calculate the speed of the bus in km/h​
DaniilM [7]
Speed is distance over time. There’s no distance in the question and the time is 1 hr 30 mins hence 1.5 hours.
6 0
3 years ago
Someone help me please
Masteriza [31]

Answer:

(5,1)

Step-by-step explanation:

Hi there!

Midpoint = \displaystyle (\frac{x_1+x_2}{2},\frac{y_1+y_2}{2} ) where the two endpoints are (x_1,y_1) and (x_2,y_2)

Plug in the given information:

Midpoint = (5,3), Endpoint = (5,5)

(5,3)=\displaystyle (\frac{5+x_2}{2},\frac{5+y_2}{2} ) where  (x_2,y_2) is the other endpoint

Solve for x_2:

\displaystyle \frac{5+x_2}{2} =5\\\\5+x_2 =10\\x_2 =5

Solve for y_2:

\displaystyle\frac{5+y_2}{2}=3 \\\\{5+y_2}=6\\y_2=1

Therefore, the other endpoint (x_2,y_2) is (5,1).

I hope this helps!

4 0
3 years ago
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0.9 of Karen's Science score is equal to 0.75 of her Math score and 0.8 of her Chemistry score. Given that Karen scored 96 marks
DochEvi [55]

You have to formulate equations for this problem.

Let S = Science score

      M = Math score

       C = Chemistry score

To illustrate the given:

0.9S = 0.75M

0.9S = 0.8C

You are given that Karen’s Math score is 96 marks. You have to substitute the Math score to the first equation.

0.9S = 0.75(96)

0.9S = 72

S = 80

Therefore, Karen’s Science score is 80. Now, you have to substitute the Science score to the second equation.

0.9(80) = 0.8C

0.8C = 72

C = 90

So, Karen’s Chemistry score is 90.

Therefore, the total score of the 3 subjects is 266 (96 + 80 + 90).

5 0
3 years ago
180 product of a prime factor using indices
zvonat [6]
Start with 180. 
<span>Is 180 divisible by 2? Yes, so write "2" as one of the prime factors, and then work with the quotient, 90. </span>

<span>Is 90 divisible by 2? Yes, so write "2" (again) as another prime factor, then work with the quotient, 45. </span>

<span>Is 45 divisible by 2? No, so try a bigger divisor. </span>
<span>Is 45 divisible by 3? Yes, so write "3" as a prime factor, then work with the quotient, 15 </span>

<span>Is 15 divisible by 3? [Note: no need to revert to "2", because we've already divided out all the 2's] Yes, so write "3" (again) as a prime factor, then work with the quotient, 5. </span>

<span>Is 5 divisible by 3? No, so try a bigger divisor. </span>
Is 5 divisible by 4? No, so try a bigger divisor (actually, we know it can't be divisible by 4 becase it's not divisible by 2)
<span>Is 5 divisible by 5? Yes, so write "5" as a prime factor, then work with the quotient, 1 </span>

<span>Once you end up with a quotient of "1" you're done. </span>

<span>In this case, you should have written down, "2 * 2 * 3 * 3 * 5"</span>
5 0
3 years ago
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