All categories

3 years ago

HELP!!!!! WILL MARK BRAINLIEST!!!!

Mathematics

1 answer:

julia-pushkina [17]3 years ago

6 0

Answer:

D: Use the zero function

You might be interested in

I need this question sovled asap y=200x+35,000

Svetradugi [14.3K]

Answer: look at the picture

Step-by-step explanation: Hope this help :D

6 0

3 years ago

stion 1 OT 5A researcher recorded the number of swans and the number of ducks in a lake every month. Function s represents the n

Anon25 [30]

Given functions are

$s(n)=2(1.1)^n+5_{}$ $d(n)=4(1.08)^n+3$

The total number of ducks and swans in the lake after n months can be determined by adding the functions s(n) and d(n).

$t(n)=s(n)+d(n)$

$t(n)=(2(1.1)^n^{}+5)+(4(1.08)^n+3)$

$t(n)=2(1.1)^n+5+4(1.08)^n+3$

$t(n)=2(1.1)^n+4(1.08)^n+5+3$

$t(n)=2(1.1)^n+4(1.08)^n+8$

Taking 2 as common, we get

$t(n)=2\lbrack(1.1)^n+2(1.08)^n+4\rbrack$

Hence The total number of ducks and swans in the lake after n months is

$t(n)=2\lbrack(1.1)^n+2(1.08)^n+4\rbrack$

8 0

2 years ago

What is the image point of (1, -2) after a translation right 4 units and down 4 units?

zaharov [31]

It would be 4 over -4

3 0

3 years ago

Help<br> I neeed to get this question right

Irina18 [472]

Answer:

6/10 is greater than 6/100

Step-by-step explanation:

6/10 = 3/5

6/100 = 3/50

3 fifties is lot more than 3 with 5 parts

hope it makes sense

5 0

3 years ago

Kamal wrote the augmented matrix below to represent a system of equations.

Sergio039 [100]

Answer:

$\left[\begin{array}{cccc}1&0&1&|-1\\-3&-9&3&|27\\3&2&0&|-2\end{array}\right]$

Step-by-step explanation:

Given [Missing from the question]

$\left[\begin{array}{cccc}1&0&1&|-1\\1&3&-1&|-9\\3&2&0&|-2\end{array}\right]$

Required

$R_2 \to -3R_2$

This implies that, we form a new matrix where the second row of the new matrix is a product of -3 and the second row of the previous matrix.

So, we have:

$Initial =\left[\begin{array}{cccc}1&0&1&|-1\\1&3&-1&|-9\\3&2&0&|-2\end{array}\right]$

$New =\left[\begin{array}{cccc}1&0&1&|-1\\-3*1&-3*3&-3*-1&|-3*-9\\3&2&0&|-2\end{array}\right]$

$New =\left[\begin{array}{cccc}1&0&1&|-1\\-3&-9&3&|27\\3&2&0&|-2\end{array}\right]$

7 0

3 years ago

Read 2 more answers