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harina [27]
3 years ago
14

X+8y/4=9y solve for x

Mathematics
1 answer:
Cerrena [4.2K]3 years ago
4 0

If the 8y/4 is the only fraction, the answer is x=7y

If the whole input of the equation is in a fraction, the answer is x=28y

:)

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a owner increases the rent office house by 5% at end of each year if current its rent is ₹2500pm how much will be the rent after
sergeinik [125]

Answer:

2950$

Step-by-step explanation:

2500×5%=125$

125×2=250$

2500+250=3960$

5 0
3 years ago
Which line is perpendicular to a line that has a slope of 1/2 ?
Lunna [17]

The answer is HJ i took the test and got it right :)

8 0
4 years ago
Read 2 more answers
Ages Number of students 15-18 6 19-22 4 23-26 5 27-30 7 31-34 10 35-38 9 Based on the frequency distribution above, find the rel
dolphi86 [110]

Answer:

9.8%

Step-by-step explanation:

Given the following information

\left|\begin{array}{c|cccccc}Ages &15-18&19-22&23-26&27-30&31-34&35-38\\$Number of students&6&4&5&7&10&9\end{array}\right|

Total Number of Students =6+4+5+7+10+9=41

Number of Students aged 19-22 =4

Therefore:

The relative frequency for the class with a lower class limit of 19

=\dfrac{4}{41}\times 100\\\\ =9.8\% $(to one decimal place)

8 0
3 years ago
When rabbits were introduced to the continent of Australia they quickly multiplied and spread across the continent since there w
Lady bird [3.3K]

Answer:

a. y=6(1.7472)^x

b. y=6e^{0.558t}

c.13.3 months

Step-by-step explanation:

a.-Given the first term  at t_0 is 6 and the second term at t_3 is 32.

-Let's take rabbit population as a function of time to be

y=ab^x

where y is the population at time x and a the initial population at t_0\\

#We substitute our values to calculate the value of the constant b:

y_x=ab^x\\\\y_3=ab^3\\\\32=6b^3\\\\b=1.472

#Replace b in the population function:

y=ab^x, b=1.7472,a=6\\\\\therefore y=6(1.7472)^x

Hence, the regression for the rabbit population as a function of time x is y=6(1.7472)^x

b. The exponential function in terms of base e is usually expressed as:

A=A_0e^{kt}

Where:

A_0-is the initial population at t_o

A-is the population at time t.

k-is the  exponential growth constant.

e- the exponent

Our function in terms of base exponent is rewritten as:

y=A_0e^{kt}

#Substitute with actual figures to solve for t:

y=A_0e^{kt}, y=32, xt=3, A_0=6\\\\32=6e^{3k}\\\\3k=In (32/6)\\\\k=0.5580

Hence, the regression equation in terms of base e is y=6e^{0.558t}

c. We substitute y with any number higher than 10,000 to estimate the time for the rabbits to exceed 10,000.

-We know that y=6e^{0.558t}.

Therefore we calculate t as(take y=10001):

y=6e^{0.558t}, y=10001\\\\10001=6e^{0.558t}\\\\1666.8333=e^{0.558t}\\\\0.558t= In 1666.8333\\\\t=13.2951

Hence, it takes approximately 13.3 months for the population to exceed 10000

3 0
4 years ago
List these numbers from least to greatest to least to greatest 1.04, 1.33, 1.0494, 1.2
fgiga [73]
Least to greatest

1.04, 1.0494, 1.2, 1.33
5 0
3 years ago
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