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vampirchik [111]
3 years ago
6

What is the y-intercept of the line? y = 5x - 9​

Mathematics
2 answers:
Liula [17]3 years ago
8 0
The y intercept of this line is -9
Luden [163]3 years ago
5 0

Answer:

-9

Step-by-step explanation:

To find the y-intercept, plug in 0 for x and find out what you get.

y=5(0)-9\\y=0-9\\y=-9

The y-intercept is

(0,-9)

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If 6^4x = 82, what is the value of x?<br> Round your answer to three places.
neonofarm [45]

Answer:

x=0.61485921

Step-by-step explanation:

The given equation is in exponential form:

6^{4x} =82

To find x we need to write it in logarithmic form:

b^{a} =n ⇒ log_{b} n=a

6^{4x} =82 ⇒ log_{ 6 }  82=4x

           2.45943684 = 4x

              \frac{2.45943684}{4} = x

           0.61485921 =x

To prove this, we can substitute the value into the given equation

             6^{4x} =82

6^{4(0.61485921)} =82

               82=82

Therefore, the answer is x=0.61485921

6 0
3 years ago
What is the volume (in cubic units) of a cylinder with a radius of 3 units and a height of 12 units?
MatroZZZ [7]
Report this clown who put the first answer he’s trying to get your ip
8 0
2 years ago
Can someone please help me this is due in 10 mins
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7 0
3 years ago
6. If the net investment function is given by
Pachacha [2.7K]

The capital formation of the investment function over a given period is the

accumulated  capital for the period.

  • (a) The capital formation from the end of the second year to the end of the fifth year is approximately <u>298.87</u>.

  • (b) The number of years before the capital stock exceeds $100,000 is approximately <u>46.15 years</u>.

Reasons:

(a) The given investment function is presented as follows;

I(t) = 100 \cdot e^{0.1 \cdot t}

(a) The capital formation is given as follows;

\displaystyle Capital = \int\limits {100 \cdot e^{0.1 \cdot t}} \, dt =1000 \cdot  e^{0.1 \cdot t}} + C

From the end of the second year to the end of the fifth year, we have;

The end of the second year can be taken as the beginning of the third year.

Therefore,  for the three years; Year 3, year 4, and year 5, we have;

\displaystyle Capital = \int\limits^5_3 {100 \cdot e^{0.1 \cdot t}} \, dt \approx 298.87

The capital formation from the end of the second year to the end of the fifth year, C ≈ 298.87

(b) When the capital stock exceeds $100,000, we have;

\displaystyle  \mathbf{\left[1000 \cdot  e^{0.1 \cdot t}} + C \right]^t_0} = 100,000

Which gives;

\displaystyle 1000 \cdot  e^{0.1 \cdot t}} - 1000 = 100,000

\displaystyle \mathbf{1000 \cdot  e^{0.1 \cdot t}}} = 100,000 + 1000 = 101,000

\displaystyle e^{0.1 \cdot t}} = 101

\displaystyle t = \frac{ln(101)}{0.1} \approx 46.15

The number of years before the capital stock exceeds $100,000 ≈ <u>46.15 years</u>.

Learn more investment function here:

brainly.com/question/25300925

6 0
2 years ago
Help please!<br>If you know geometry please help!<br>​
tamaranim1 [39]
50+50=100

180-100=80

X=80
4 0
3 years ago
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