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3 years ago

9

What is the value of z for the equation

1 answer:

vitfil [10]3 years ago

5 0

Z= -7
So your answer is number 2, -7

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In a geometric sequence, A1 = 0.3 and r=3.Find A12, to the nearest integer

soldi70 [24.7K]

Step-by-step explanation:

We have,

First terms of geometric sequence, a = 0.3

Common ratio, r = 3

It is required to find the 12th term of a GP. The formula of the nth term is given by :

$T_n=ar^{n-1}$

Here, n =12

So,

$T_{12}=0.3\times 3^{12-1}\\\\T_{12}=0.3\times 3^{11}\\\\T_{12}=53144.1$

or

$T_{12}=53144$

So, the 12th term of the GP is 53144.

8 0

4 years ago

Asa has 4 colored pencils. His friend purchases three times as many. If Asa’s friend buys two more 4-packs, which expression sho

daser333 [38]

24

Three times as many as four - 3(4) / 4*3/ 4+4+4

4 x 3 = 12

Then for 2 more 4 packs, it'd be 8 since it's either 2*4 or 4+4 (both equal eight)

And then you'd add them both together, 12+8=20

But since it's asking for both Asa and his friend's total amount, you'd add four more to it since 20 is just his friend's amount.

5 0

3 years ago

The population of a town gradually decreases by about 30

topjm [15]

It would be 3,600 - 30 = 3570

3 0

2 years ago

I need help with this question ??

aliya0001 [1]

Answer:

D

Step-by-step explanation:

x(-3x+4)=x*(-3)+x*4=-3x^2+4x

5 0

3 years ago

Read 2 more answers

Х- а<br>x-b<br>If f(x) = b.x-a÷b-a + a.x-b÷a - b<br>Prove that: f (a) + f(b) = f (a + b)

GenaCL600 [577]

Given:

Consider the given function:

$f(x)=\dfrac{b\cdot(x-a)}{b-a}+\dfrac{a\cdot(x-b)}{a-b}$

To prove:

$f(a)+f(b)=f(a+b)$

Solution:

We have,

$f(x)=\dfrac{b\cdot(x-a)}{b-a}+\dfrac{a\cdot (x-b)}{a-b}$

Substituting $x=a$ , we get

$f(a)=\dfrac{b\cdot(a-a)}{b-a}+\dfrac{a\cdot (a-b)}{a-b}$

$f(a)=\dfrac{b\cdot 0}{b-a}+\dfrac{a}{1}$

$f(a)=0+a$

$f(a)=a$

Substituting $x=b$ , we get

$f(b)=\dfrac{b\cdot(b-a)}{b-a}+\dfrac{a\cdot (b-b)}{a-b}$

$f(b)=\dfrac{b}{1}+\dfrac{a\cdot 0}{a-b}$

$f(b)=b+0$

$f(b)=b$

Substituting $x=a+b$ , we get

$f(a+b)=\dfrac{b\cdot(a+b-a)}{b-a}+\dfrac{a\cdot (a+b-b)}{a-b}$

$f(a+b)=\dfrac{b\cdot (b)}{b-a}+\dfrac{a\cdot (a)}{-(b-a)}$

$f(a+b)=\dfrac{b^2}{b-a}-\dfrac{a^2}{b-a}$

$f(a+b)=\dfrac{b^2-a^2}{b-a}$

Using the algebraic formula, we get

$f(a+b)=\dfrac{(b-a)(b+a)}{b-a}$ $[\because b^2-a^2=(b-a)(b+a)]$

$f(a+b)=b+a$

$f(a+b)=a+b$ [Commutative property of addition]

Now,

$LHS=f(a)+f(b)$

$LHS=a+b$

$LHS=f(a+b)$

$LHS=RHS$

Hence proved.

5 0

3 years ago