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Evgesh-ka [11]
4 years ago
10

A number is one tenth less than 1.54. Determine the number

Mathematics
2 answers:
liraira [26]4 years ago
8 0
Subtract 1.54 by 0.2

1.54-0.1=1.44
tester [92]4 years ago
3 0

1.54-.10=1.44

Subtract the two to find your answer

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Which expression represents the 4th term in the binomial expansion of (e+2f)^10
butalik [34]
The answer depends on what is meant by the "fourth term". By the binomial theorem, one answer could be

\dbinom{10}3e^3(2f)^{10-3}=120e^3(2f)^7=15360e^3f^7

By symmetry, another could be

\dbinom{10}3(2f)^3e^{10-3}=960e^7f^3
7 0
3 years ago
Find the slope of the line graphed below.​
Finger [1]

to get the equation of any straight line, we simply need two points off of it, let's use the ones provided in the picture below.

(\stackrel{x_1}{1}~,~\stackrel{y_1}{4})\qquad (\stackrel{x_2}{3}~,~\stackrel{y_2}{-3}) ~\hfill \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{-3}-\stackrel{y1}{4}}}{\underset{run} {\underset{x_2}{3}-\underset{x_1}{1}}} \implies \cfrac{-7}{2}

\begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{4}=\stackrel{m}{-\cfrac{7}{2}}(x-\stackrel{x_1}{1}) \\\\\\ y-4=-\cfrac{7}{2}x+\cfrac{7}{2}\implies y=-\cfrac{7}{2}x+\cfrac{7}{2}+4\implies y=-\cfrac{7}{2}x+\cfrac{15}{2}

8 0
2 years ago
If you are spending $1500 per year city driving a Chevrolet at 30mpg. How much would you save driving a Toyota at 40mpg? $300, $
Vikentia [17]

Answer:

1500÷30=50

40× 50=2000

2000-1500=500.

therefore the answer is500

5 0
3 years ago
Compute the square root of√144
zloy xaker [14]
The square root of 144 is 12
4 0
3 years ago
Read 2 more answers
Its a math test again-
Step2247 [10]

Answer:

A. t > 2.5

Step-by-step explanation:

If she wants to achieve <em>more </em>than 590, we already know the sign would be greater than (>).

All we have to do is divide 590/236, which gives us 2.5.

Therefore, she will have to ride more than 2.5 hours to reach her goal, t > 2.5

Hope this helps :)

Good luck on your test.

3 0
3 years ago
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