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3 years ago

Find the first six terms of the sequence.

Mathematics

1 answer:

Yuliya22 [10]3 years ago

8 0

Answer:

6, 12, 24, 48, 96, 192

Step-by-step explanation:

The first term a1 = 6 and the common ratio is 2.

So it is 6, 6*2, 6*2^2 ....

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I need help with this

azamat

Answer:

Let the amount of Danny be y

Therefore, lance will have 1/3y

Total fraction of cup is 1/1

y+y/3=1

4y/3=1

4y=3

y=3/4

therefore, lance had 3/4 of the cup of water

4 0

2 years ago

Please help i am on a timer

meriva

Answer:

x^2 = 36

Step-by-step explanation:

logx ( 36) = 2

Rewrite this as an exponential equation

We know that loga(b) =c as a^b =c

x^2 = 36

7 0

2 years ago

Write (−3) (−3)(−3) (−3) using exponents.

Tanya [424]

-3^4 (Three to the Fourth power)

8 0

2 years ago

Read 2 more answers

The expression (secx + tanx)2 is the same as _____.

trapecia [35]

Answer:

The expression $\bold{(\sec x+\tan x)^{2} \text { is same as } \frac{1+\sin x}{1-\sin x}}$

Solution:

From question, given that $\bold{(\sec x+\tan x)^{2}}$

By using the trigonometric identity $(a + b)^{2} = a^{2} + 2ab + b^{2}$ the above equation becomes,

$(\sec x+\tan x)^{2} = \sec ^{2} x+2 \sec x \tan x+\tan ^{2} x$

We know that $\sec x=\frac{1}{\cos x} ; \tan x=\frac{\sin x}{\cos x}$

$(\sec x+\tan x)^{2}=\frac{1}{\cos ^{2} x}+2 \frac{1}{\cos x} \frac{\sin x}{\cos x}+\frac{\sin ^{2} x}{\cos ^{2} x}$

$=\frac{1}{\cos ^{2} x}+\frac{2 \sin x}{\cos ^{2} x}+\frac{\sin ^{2} x}{\cos ^{2} x}$

On simplication we get

$=\frac{1+2 \sin x+\sin ^{2} x}{\cos ^{2} x}$

By using the trigonometric identity $\cos ^{2} x=1-\sin ^{2} x$ ,the above equation becomes

$=\frac{1+2 \sin x+\sin ^{2} x}{1-\sin ^{2} x}$

By using the trigonometric identity $(a+b)^{2}=a^{2}+2ab+b^{2}$

we get $1+2 \sin x+\sin ^{2} x=(1+\sin x)^{2}$

$=\frac{(1+\sin x)^{2}}{1-\sin ^{2} x}$

$=\frac{(1+\sin x)(1+\sin x)}{1-\sin ^{2} x}$

By using the trigonometric identity $a^{2}-b^{2}=(a+b)(a-b)$ we get $1-\sin ^{2} x=(1+\sin x)(1-\sin x)$

$=\frac{(1+\sin x)(1+\sin x)}{(1+\sin x)(1-\sin x)}$

$= \frac{1+\sin x}{1-\sin x}$

Hence the expression $\bold{(\sec x+\tan x)^{2} \text { is same as } \frac{1+\sin x}{1-\sin x}}$

8 0

2 years ago

Determine whether the graphs of the given equations are parallel, perpendicular, or neither.

Anna11 [10]

When the slopes are equal, the two lines are parallel.
 When the product of two slopes is -1, then the two lines are perpendicular.
So, y - 4 = 3x + 6 ;
Then, y = 3x + 10 ; the slope of the first line is 3 ;
2x + 6y = 10 ;
Then, 6y = - 2x + 10 ;
y = (-2/6)x + 10/6 ;
Finally, y = (-1/3)x + 5/3; the slope if the second line is -1/3;
Because 3*(-1/3) = -1 the two lines are perpendicular.

4 0

3 years ago