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TiliK225 [7]
3 years ago
6

60 min is 20%of blank min

Mathematics
1 answer:
taurus [48]3 years ago
7 0
The answer is 300 min
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Find the 11th term of the following geometric sequence.<br> 1, 3, 9, 27, ...
Natasha_Volkova [10]

Answer:

the answer is 59048

Step-by-step explanation:

as the common ratio is multiplying by 3 and the first term is 1 so from the rule (Tn=ar(power n-1 ) )

so the 11th term is 1*3(power 11-1 ) equal 3 power 10 equal 59048

4 0
3 years ago
What is the solution to the equation 1/4x+2=-5/8x-5
Fed [463]

multiply everything by 8

2x+16 = -5x-40

-16 on both sides

2x = -5x-56

+5x on both sides

7x = -56

x=-8

so the first option

5 0
3 years ago
Read 2 more answers
Decrease 1280 by 67%
blondinia [14]
67% of 1280 is 857.6 substrack it and it's = 422.4
5 0
4 years ago
What is the roots of the quadratic equation? 6x^2+5x-4=0
Rama09 [41]

x = \frac{1}{2} or x = - \frac{4}{3}

consider the factors of the product 6 × - 4 = - 24 which sum to the coefficient of the x- term ( + 5)

the factors are + 8 and - 3 ( split the middle term using these factors

6x² - 3x + 8x - 4 = 0 ( factor by grouping )

3x(2x - 1) + 4(2x - 1 ) ( take out common factor of (2x - 1) )

= (2x - 1)(3x + 4) = 0

equate each factor to zero and solve for x

2x - 1 = 0 ⇒ x = \frac{1}{2}

3x + 4 = 0 ⇒ x = - \frac{4}{3}


4 0
4 years ago
Read 2 more answers
How do you do this question?
alina1380 [7]

Answer:

(8√2) / 15

Step-by-step explanation:

A curve bounded by the y-axis is represented by in terms of dy;

\int \:x\:dt

When the curve crosses the y-axis, x will be 0. In this case x is the function of t, so we have to solve for x(t) = 0;

0 = t^2 + 2t --- (1)

Solution(s) => t = 0, t = 2

dy = (1/2 * 1/√t)dt --- (2)

Our solutions (0, 2) are our limits. The area of the curve is in the form A\:=\:\int _b^a\:f\left(t\right)g'\left(t\right)dt , so now let's introduce the limits of integration, x(t) and dy/dt. Remember, dy/dt = (1/2 * 1/√t) (second equation). 1/2 * 1/√t can be rewritten as 1/2 * t^(-1/2)....

A\:=\:\int _2^0\:\left(t^2-2t\right)\left(\frac{1}{2}t^{-\frac{1}{2}}\right)dt\\\\= \int _2^0\:\left(\frac{1}{2}t^{\frac{3}{2}}-t^{\frac{1}{2}}\right)dt\\\\= \left[\frac{t^{\frac{5}{2}}}{5}-\frac{2t^{\frac{3}{2}}}{3}\right]_2^0\\\\= 0\:-\:\left(\frac{4\sqrt{2}}{5}-\frac{4\sqrt{2}}{3}\right)\\\\= \frac{8\sqrt{2}}{15}

Your solution is 8√2 / 15

7 0
3 years ago
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