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Blababa [14]
3 years ago
14

Look at the following numbers: -5, -6, 0, 6

Mathematics
2 answers:
Zigmanuir [339]3 years ago
7 0
C) 6 and -6 will equal out to 0
gladu [14]3 years ago
4 0
C is the answer l.

6 + (-6) = 0
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What is the sum of -1/2 and 8/6 <br> a) -5/6<br><br> B) -11/6<br><br> C) 5/6<br><br> D) 11/6
Leto [7]

Answer:

d 11/6

Step-by-step explanation:

find the common denominate for -1/2 which is 3/6 make sure you remember the rule of integers adding if they have diffrent sign there gonna be postive if they have the same sign there negative look up the role it will help you a lot find the common denominator which and then add like normal if it ask for it s lowest term then find the lowest term have a good day if you have question please type in the comments if  you need more help

3 0
3 years ago
in a proportional relationship the ratio y/x is constant. Show that this ratio is not constant for the equation y=a-14
Colt1911 [192]

To show that the ratio \frac{y}{x} is not constant for the function y=a-14 we only need to evaluate this function at 2 points and then show that the ratio is not the same for those two points. Note that here a represents x.  

To find the counter examples here, I will use the values a=20  and a=28. For a = 20 , the equation tells us that the output is

y=20-14=6. For this value of a the point is point is (20,6). The ratio \frac{y}{x} =\frac{6}{20}. For the a=28, y=28-14=14. For this value of a the  point is (28,14) . The ratio \frac{y}{x} =\frac{14}{28} =\frac{1}{2}. Clearly this ratio is not constant for all ordered pairs.

8 0
2 years ago
Please help me in the box below
Fynjy0 [20]
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6 0
3 years ago
Read 2 more answers
Neptune is an average distance of 4.5 × 10^9 km from the Sun. Estimate the length of the Neptunian year using the fact that the
zlopas [31]

Answer:

Answer:

164.32 earth year

Step-by-step explanation:

distance of Neptune, Rn = 4.5 x 10^9 km

distance of earth, Re = 1.5 x 10^8 km

time period of earth, Te = 1 year

let the time period of Neptune is Tn.

According to the Kepler's third law

T² ∝ R³

\left ( \frac{T_{n}}{T_{e}} \right )^{2}=\left ( \frac{R_{n}}{R_{e}} \right )^{3}

\left ( \frac{T_{n}}{1} \right )^{2}=\left ( \frac{4.5\times10^{9}}}{1.5\times10^{8}}} \right )^{3}

Tn = 164.32 earth years

Thus, the neptune year is equal to 164.32 earth year.

Step-by-step explanation:

3 0
3 years ago
(02.05 MC)
Amiraneli [1.4K]

Table for f(x)

\boxed{\begin{array}{c|c}\boxed{\bf x}&\boxed{\sf f(x)}\\ \sf 0 &\sf 1 \\ \sf 2&\sf 9\\ \sf 4 &\sf 17\end{array}}

Table for g(x)

\boxed{\begin{array}{c|c}\boxed{\bf x}&\boxed{\sf g(x)}\\ \sf 0 &\sf 1 \\ \sf 2&\sf 7\\ \sf 4 &\sf 13\end{array}}

Option C is correct

3 0
2 years ago
Read 2 more answers
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