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3 years ago

Use the rule to write the first five terms of the pattern: Rule: add 8, subtract 4 First term: 13

Mathematics

2 answers:

marysya [2.9K]3 years ago

6 0

Answer:

Step-by-step explanation:

13 + 8 - 4

pav-90 [236]3 years ago

6 0

Answer:

{13, 17, 21, 25, 29}

Step-by-step explanation:

Add 8 and subtract 4 is equivalent to Add 4.

If the first term is 13, the next is 17;

the second term is 21; and

the third term is 25.

You might be interested in

Find the distance of (-7,-5) and (-3,4) round to the nearest tenth

Luba_88 [7]

Answer:

7.2 units

Step-by-step explanation:

Distance² = (Change in y)² + (Change in x)²

Distance² = (4 - -5)² + (-3 - -7)²

Distance² = 36 + 16

Distance² = 52

Distance = 7.211102551

Rounding off the answer to nearest tenth gives 7.2 units

7 0

3 years ago

If 2/3 and -3 are the roots of the equation px^2+7x+q=0, find the values of p and q.

weqwewe [10]

Answer:

p = 3

q = -6

Step-by-step explanation:

Work backwards from the roots:

x = $\frac{2}{3}$

3x = 2

3x - 2 = 0

Other root:

x = -3

x + 3 = 0

Multiply them together:

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

3x² - 9x + 2x - 6 = 0

3x² - 7x - 6 = 0

p = 3

q = -6

4 0

3 years ago

Read 2 more answers

Which of the following sentences could not be used to represent the equation x-5 = -10?

e-lub [12.9K]

A number subtracted from five is equal to negative ten

3 0

3 years ago

Read 2 more answers

If a Subway train traveled 12 km in 8 minutes how many km/h was the train going?

Bond [772]

Answer:

1.5 km/h

Step-by-step explanation:

brainly would be epic :)

6 0

3 years ago

Two chemicals A and B are combined to form a chemical C. The rate, or velocity, of the reaction is proportional to the product o

Orlov [11]

Answer:

25.35 grams of C is formed in 14 minutes

after a long time , the limiting amount of C = 60g ,

A = 0 gram

and B = 30 grams; will remain.

Step-by-step explanation:

From the information given;

Let consider x(t) to represent the number of grams of compound C present at time (t)

It is obvious that x(0) = 0 and x(5) = 10 g;

And for x gram of C;

$\dfrac{2}{3}x$ grams of A is used ;

also $\dfrac{1}{3} x$ grams of B is used

Similarly; The amounts of A and B remaining at time (t) are;

$40 - \dfrac{2}{3}x$ and $50 - \dfrac{1}{3}x$

Therefore ; rate of formation of compound C can be said to be illustrated as ;

$\dfrac{dx}{dt }\propto (40 - \dfrac{2}{3}x)(50-\dfrac{1}{3}x)$

= $k \dfrac{2}{3}( 60-x) \dfrac{1}{3}(150-x)$

where;

k = proportionality constant.

= $\dfrac{2}{9}k (60-x)(150-x)$

By applying the separation of variable;

$\dfrac{1}{(60-x)(150-x)}dx= \dfrac{2}{9}k dt \\ \\ \\$

Solving by applying partial fraction method; we have:

$\{ \dfrac{1}{90(60-x)} - \dfrac{1}{90(150-x)} \}dx = \dfrac{2}{9}kdt$

$\dfrac{1}{90}(\dfrac{1}{x-150}-\dfrac{1}{x-60})dx =\dfrac{2}{9}kdt$

Taking the integral of both sides ; we have:

$\dfrac{1}{90}\int\limits(\dfrac{1}{x-150}- \dfrac{1}{x-60})dx= \dfrac{2}{9}\int\limits kdx$

$\dfrac{1}{90}(In(x-150)-In(x-60)) = \dfrac{2}{9}kt+C$

$\dfrac{1}{90}(In(\dfrac{x-150}{x-60})) = \dfrac{2}{9}kt+C$

$In( \dfrac{x-150}{x-60})= 20 kt + C_1 \ \ \ \ \ where \ \ C_1 = 90 C$

$\dfrac{x-150}{x-60}= Pe ^{20 kt} \ \ \ \ \ where \ \ P= e^{C_1}$

Applying the initial condition x(0) =0 to determine the value of P

Replace x= 0 and t =0 in the above equation.

$\dfrac{0-150}{0-60}= Pe ^{0}$

$\dfrac{5}{2}=P$

Thus;

$\dfrac{x-150}{x-60}=Pe^{20kt} \\ \\ \\ \dfrac{x-150}{x-60}=\dfrac{5}{2}e^{20kt} \\ \\ \\ 2x -300 =5e^{20kt}(x-60)$

$2x - 300 = 5xe^{20kt} - 300 e^{20kt} \\ \\ 5xe^{20kt} -2x = 300 e^{20kt} -300 \\ \\ x(5e^{20kt} -2) = 300 e^{20kt} -300 \\ \\ x= \dfrac{300 e^{20kt}-300}{5e^{20kt}-2}$

Thus;

$x(t)= \dfrac{300 e^{20kt}-300}{5e^{20kt}-2}$

Applying the initial condition for x(7) = 15 , to find the value of k

Replace t = 7 into $x(t)= \dfrac{300 e^{20kt}-300}{5e^{20kt}-2}$

$x(7)= \dfrac{300 e^{20k(7)}-300}{5e^{20k(7)}-2}$

$15= \dfrac{300 e^{140k}-300}{5e^{140k}-2}$

$75e^{140k}-30 ={300 e^{140k}-300}$

$225e^{140k}=270$

$e^{140k}=\dfrac{270}{225}$

$e^{140k}=\dfrac{6}{5}$

$140 k = In (\dfrac{6}{5})$

$k = \dfrac{1}{140}In (\dfrac{6}{5})$

k = 0.0013

Thus;

$x(t)= \dfrac{300 e^{20kt}-300}{5e^{20kt}-2}$

$x(t)= \dfrac{300 e^{20(0.0013)t}-300}{5e^{20(0.0013)t}-2}$

$x(t)= \dfrac{300 e^{(0.026)t}-300}{5e^{(0.026)t}-2}$

The amount of C formed in 14 minutes is ;

$x(14)= \dfrac{300 e^{(0.026)14}-300}{5e^{(0.026)14}-2}$

x(14) = 25.35 grams

Thus 25.35 grams of C is formed in 14 minutes

NOW; The limiting amount of C after a long time is:

$\lim_{t \to \infty} = \lim_{t \to \infty} (\dfrac{300 e^{(0.026)t}-300}{5e^{(0.026)t}-2})$

$\lim_{t \to \infty} (\dfrac{300- 300 e^{(0.026)t}}{2-5e^{(0.026)t}})$

As; $\lim_{t \to \infty} e^{-20kt} = 0$

⇒ $\dfrac{300}{5}$

= 60 grams

Therefore as t → $\infty$ ; x = 60

and the amount of A that remain = $40 - \dfrac{2}{3}x$

= $40 - \dfrac{2}{3}(60)$

= 40 -40

=0 grams

The amount of B that remains = $50 - \dfrac{1}{3}x$

= $50 - \dfrac{1}{3}(60)$

= 50 - 20

= 30 grams

Hence; after a long time ; the limiting amount of C = 60g , and 0 g of A , and 30 grams of B will remain.

I Hope That Helps You Alot!.

5 0

4 years ago

Use the rule to write the first five terms of the pattern: Rule: add 8, subtract 4 First term: 13​

Use the rule to write the first five terms of the pattern: Rule: add 8, subtract 4 First term: 13