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3 years ago

9

HELP ME WALLAHI HELP PLEASE

2 answers:

larisa86 [58]3 years ago

4 0

Answer:

-8,0 I hope it is a right answers

bearhunter [10]3 years ago

3 0

Answer:

anwer is D

x=6 y=0

Step-by-step explanation:

hope this helps

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Which of the following is best used to measure the distance between two large cities

Ymorist [56]

Miles
i just did the assignment

8 0

3 years ago

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Find an equation of the tangent to the curve x =5+lnt, y=t2+5 at the point (5,6) by both eliminating the parameter and without e

svet-max [94.6K]

ANSWER

$y = 2x -4$

EXPLANATION

Part a)

Eliminating the parameter:

The parametric equation is

$x = 5 + ln(t)$

$y = {t}^{2} + 5$

From the first equation we make t the subject to get;

$x - 5 = ln(t)$

$t = {e}^{x - 5}$

We put it into the second equation.

$y = { ({e}^{x - 5}) }^{2} + 5$

$y = { ({e}^{2(x - 5)}) } + 5$

We differentiate to get;

$\frac{dy}{dx} = 2 {e}^{2(x - 5)}$

At x=5,

$\frac{dy}{dx} = 2 {e}^{2(5 - 5)}$

$\frac{dy}{dx} = 2 {e}^{0} = 2$

The slope of the tangent is 2.

The equation of the tangent through

(5,6) is given by

$y-y_1=m(x-x_1)$

$y - 6 = 2(x - 5)$

$y = 2x - 10 + 6$

$y = 2x -4$

Without eliminating the parameter,

$\frac{dy}{dx} = \frac{ \frac{dy}{dt} }{ \frac{dx}{dt} }$

$\frac{dy}{dx} = \frac{ 2t}{ \frac{1}{t} }$

$\frac{dy}{dx} = 2 {t}^{2}$

At x=5,

$5 = 5 + ln(t)$

$ln(t) = 0$

$t = {e}^{0} = 1$

This implies that,

$\frac{dy}{dx} = 2 {(1)}^{2} = 2$

The slope of the tangent is 2.

The equation of the tangent through

(5,6) is given by

$y-y_1=m(x-x_1)$

$y - 6 = 2(x - 5) =$

$y = 2x -4$

5 0

3 years ago

What is the 8th term of the geometric sequence a2 = 20 and r = 5

algol [13]

Answer:

8th term of geometric sequence is 312500

Step-by-step explanation:

Given : $a_2=20$ and common ratio (r) = 5

We have to find the 8th term of the geometric sequence whose $a_2=20$ and common ratio (r) = 5

Geometric sequence is a sequence of numbers in which next term is found by multiplying by a constant called the common ratio (r).

$a_n=ar^{n-1}$ ......(1)

where $a_n$ is nth term and a is first term.

For given sequence

a can be find using $a_2=20$ and r = 5

Substitute in (1) , we get,

$a_2=ar^{2-1}$

$\Rightarrow 20=a(5)$

$\Rightarrow a=4$

Thus, 8th term of the sequence denoted as $a_8$

Substitute n= 8 in (1) , we get,

$a_8=ar^{8-1} \\\\a_8=(4)r_{7} \\\\a_8=4(5)^7=4 \times 78125=312500$

Thus 8th term of geometric sequence is 312500

8 0

3 years ago

Hey can you please help me posted picture of question

GalinKa [24]

The given equation is:
ax2 + bx + c = 0
We have the resolvent is:
x = (- b +/- root (b2 - 4ac)) / (2a)
The discriminant is:
b2 - 4ac = 0
The solution will be:
x = (- b) / (2a)
Thus, the equation has a real solution.
Answer:
option B

5 0

2 years ago

Y = -1/2x +4<br> x + 2y = -8

Flauer [41]

Answer:

Step-by-step explanation:

b

3 0

2 years ago

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