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emmainna [20.7K]
1 year ago
6

I need help on a question

Mathematics
1 answer:
NeX [460]1 year ago
7 0

Jb, this is the solution:

If T = (x + 3, y + 1),

T3 = 3(x + 3, y + 1)

T3 = 3x + 9, 3y + 3

The correct answer is C.

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Riley is 3 years older than his sister. Which of the following expressions shows the sum of their ages if s represents Riley's s
erik [133]
If s is the Riley`s sister`s age and Riley is 3 years older so Riley`s age is:
s/3 so the sum of their ages is:
s + s/3 = 4s/3 :)))
i hope this is helpful
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3 0
4 years ago
ANYONE WANT A FREE BRAINLIEST.....
kap26 [50]
3 inches and 2 inches, 6÷3 & 9÷3.

12×3=36 so divide 6 by 3 and 9 by 3
7 0
3 years ago
Read 2 more answers
J=(7x+13) K=(83-2x) and the sum of the measures of the angles is 141. What is the measure of each angle
natta225 [31]

Answer:

j=76 and k=65

Step-by-step explanation:

j + k = 141 \\ 7x + 13 + 83 +  - 2x = 141 \\ 96 + 5x = 141 \\ 5x = 141 - 96 \\ 5x = 45 \\ x =  \frac{45}{5}  \\ x = 9 \\ j = (7 \times 9 + 13) \\ j =76 \\ k = 83 - 2(9) \\  = 83 - 18 \\  = 65

3 0
3 years ago
How to do question 1?Had no idea how to do both parts
givi [52]
y=\dfrac{\ln x}{3x-6}

Differentiate both sides with respect to x:

\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\frac{3x-6}x-3\ln x}{(3x-6)^2}

When x=1, you have

\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\frac{3-6}1-3\ln1}{(3-6)^2}=\dfrac{-3}9=-\dfrac13

For part (b), we now assume that x and y are functions of an independent variable, which we'll call t (for time). Now differentiating both sides with respect to t, we have

\dfrac{\mathrm dy}{\mathrm dt}=\dfrac{\frac{3x-6}x-3\ln x}{(3x-6)^2}\dfrac{\mathrm dx}{\mathrm dt}

where the chain rule is used on the right side. We're told that y is decreasing at a constant rate of 0.1 units/second, which translates to \dfrac{\mathrm dy}{\mathrm dt}=-0.1. So when x=1, you have

-0.1=\dfrac{\frac{3-6}1-3\ln1}{(3-6)^2}\dfrac{\mathrm dx}{\mathrm dt}
-0.1=-\dfrac13\dfrac{\mathrm dx}{\mathrm dt}
\dfrac{\mathrm dx}{\mathrm dt}=0.3

where the unit is again units/second.
6 0
3 years ago
Coach Burns charted the change in his runners’ 5K times from the first race of the year to the second race of the year. Which
Tom [10]

Yasim is Albert who ran the race at a time that was 15.7 seconds shorter, So, Yasim is the student who improved the most by decreasing the time needed to run the race.

<h2>Given that,</h2>

Coach Burns charted the change in his runners' 5K times from the first race of the year to the second race of the year.

<h3>We have to find,</h3>

Which student improved the most by decreasing the time needed to run the race?

<h3>According to the question,</h3>

Coach Burns charted the change in his runners' 5K times from the first race of the year to the second race of the year.

In the given option Yasim has the lowest value of the time which is -37.5 seconds.

Yasim was the most improved student because Yasim ran the second race in a time that was 37.4 seconds shorter than the first race.

Following Yasim is Albert who ran the race at a time that was 15.7 seconds shorter.

At the third place in improvement is Cody who actually was worse off than the first race along with the 4th place Hannah.

Hence, Yasim students improved the most by decreasing the time needed to run the race.

For more details about the Equation refer to the link given below.

brainly.com/question/3842749

7 0
2 years ago
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