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3 years ago

I need help with my math problems

Mathematics

2 answers:

Oxana [17]3 years ago

7 0

13 16 15 19 20 15 19 20 14 15 17 18 and that's all i see hopes that helped :)

makvit [3.9K]3 years ago

6 0

13, 16, 15, 19,20,15,19,20,14,15,17,19,

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☺️☺️does anyone know☺️

Artemon [7]

The standard form of a circle:
$(x-h)^2+(y-k)^2=r^2$
Where:
(h; k) - the coordinates of a centerr - the radius
We have: $(x+5)^2+(y-10)^2=121$
therefore
$(-5;\ 10)$ - the center
$r^2=121\to r=\sqrt{121}\to r=11$ - the radius
Answer: 11

4 0

4 years ago

Read 2 more answers

Which of these is not a method to solve systems of

Allushta [10]

Answer:

Fourth option

Step-by-step explanation:

The answer is the fourth option "difference of squares method." It's not the first option because the substitution method is when you pick one of the equation in a system of equations, solve it for a variable and then plug the variable into the other equation to help solve. It's not the second option because the graphing method is used to solve a system of equations by graphing the equations on the same coordinate plane, it's not the third option because the addition method is also known as the elimination method which is the third method to solving a system of equations. Therefore your answer is the fourth option.

Hope this helps.

5 0

3 years ago

Can y’all help me on question 23?!

Wittaler [7]

Answer:

Option D. 5x + 12y + 4z is the corrct one.

5 0

3 years ago

Read 2 more answers

Answer this correctly for brainliest pls!!

Molodets [167]

A function that depicts depreciation can be represented as equations, graphs and tables.

<h3>The yearly value of each car for the first five years</h3>

The functions are given as:

$V(t) =C(1 - r^t)$

$V(t) =C(1 - r)^t$

Where:

C = 5895 --- the initial value of the car

r = 22% --- the rate of depreciation

For the first 5 years, the values of t = 1, 2, 3, 4 and 5

For the first function, we have:

$V(1) =5895(1 - 22\%^1) = 4598.1$

$V(2) =5895(1 - 22\%^2) = 5609.7$

$V(3) =5895(1 - 22\%^3) = 5832.2$

$V(4) =5895(1 - 22\%^4) = 5881.2$

$V(5) =5895(1 - 22\%^5) = 5892.0$

For the second function, we have

$V(1) =5895(1 - 22\%)^1 = 4598.1$

$V(2) =5895(1 - 22\%)^2 = 3586.5$

$V(3) =5895(1 - 22\%)^3 = 2797.5$

$V(4) =5895(1 - 22\%)^4 =2182.0$

$V(5) =5895(1 - 22\%)^5 = 1702.0$

So, the table of values is:

$\left[\begin{array}{ccc}Years&Car\ 1&Car\ 2\\1&4598.1&4598.1\\2&5609.7&3586.5\\3&5832.2&2797.5\\4&5881.2&2182.0\\5&5892.0&1702.0\end{array}\right]$

See attachment for the graphs of the two functions

<h3>Interpreting the graphs</h3>

The graph of car 1 increases up to C(t) = 5895, and then remains constant as the value of t increases while the graph of car 2 follows an exponential pattern.

This means that, the graph 2 that follows an exponential pattern is more realistic

<h3>The worth of the cars after 5 years</h3>

From the table, the worth of car 1 is $5892.0, while the worth of car 1 is $1702.0

Read more about graphs and functions at:

brainly.com/question/13473114

5 0

2 years ago

When the function f(x) = 5•2x is evaluated for x = 3, the output is:

In-s [12.5K]

The answer is 30

2 x 3 = 6
5 x 6 = 30

3 0

3 years ago