1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Pepsi [2]
3 years ago
9

Plzzzz help!!!! I don't understand

Mathematics
2 answers:
Kazeer [188]3 years ago
6 0

Answer:

x = 1

Step-by-step explanation:

nirvana33 [79]3 years ago
6 0

Answer:

x=1

Step-by-step explanation:

You might be interested in
Cody bought two burgers and two
8_murik_8 [283]

Answer:

$5

Step-by-step explanation:

Let the cost of one burger be $b while the cost of one small fries be s

2 burgers and 2 small fries cost $14

This means that;

2b + 2s = 14

divide both sides by 2

b + s = 7 •••••••• (i)

Three burgers and four small fries cost $23

Mathematically;

3b + 4s = 23 •••••••• (ii)

From i , we can see that s = 7- b

we can now substitute this into equation ii

3b + 4(7-b) = 23

3b + 28 -4b = 23

4b -3b = 28 -23

b = $5

one burger costs $5

7 0
3 years ago
Simply the attached question​
Finger [1]

Answer:

=x⁴−x³−14x²

Step-by-step explanation:

<h3>Let's simplify step-by-step.</h3>

x²(3x²+5x−4)−2x²(x²+3x+5)

<h3>Distribute:</h3>

=(x²)(3x²)+(x²)(5x)+(x²)(−4)+−2x⁴+−6x³+−10x²

=3x⁴+5x³+−4x²+−2x⁴+−6x³+−10x²

<h3>Combine Like Terms:</h3>

=3x⁴+5x³+−4x²+−2x⁴+−6x³+−10x²

=(3x⁴+−2x⁴)+(5x³+−6x³)+(−4x²+−10x²)

=x⁴+−x³+−14x²

<h3>ANS-</h3>

=x⁴−x³−14x²

6 0
2 years ago
Somebody help me please i will mark you the brainliest
Reptile [31]
I’m 100% sure it’s the 3rd one :)
5 0
2 years ago
Read 2 more answers
Prove that
Pani-rosa [81]
Let's start from what we know.

(1)\qquad\sum\limits_{k=1}^n1=\underbrace{1+1+\ldots+1}_{n}=n\cdot 1=n\\\\\\&#10;(2)\qquad\sum\limits_{k=1}^nk=1+2+3+\ldots+n=\dfrac{n(n+1)}{2}\quad\text{(arithmetic  series)}\\\\\\&#10;(3)\qquad\sum\limits_{k=1}^nk\ \textgreater \ 0\quad\implies\quad\left|\sum\limits_{k=1}^nk\right|=\sum\limits_{k=1}^nk

Note that:

\sum\limits_{k=1}^n(-1)^k\cdot k^2=(-1)^1\cdot1^2+(-1)^2\cdot2^2+(-1)^3\cdot3^2+\dots+(-1)^n\cdot n^2=\\\\\\=-1^2+2^2-3^2+4^2-5^2+\dots\pm n^2

(sign of last term will be + when n is even and - when n is odd).
Sum is finite so we can split it into two sums, first S_n^+ with only positive trems (squares of even numbers) and second S_n^- with negative (squares of odd numbers). So:

\sum\limits_{k=1}^n(-1)^k\cdot k^2=S_n^+-S_n^-

And now the proof.

1) n is even.

In this case, both S_n^+ and S_n^- have \dfrac{n}{2} terms. For example if n=8 then:

S_8^+=\underbrace{2^2+4^2+6^2+8^2}_{\frac{8}{2}=4}\qquad\text{(even numbers)}\\\\\\&#10;S_8^-=\underbrace{1^2+3^2+5^2+7^2}_{\frac{8}{2}=4}\qquad\text{(odd numbers)}\\\\\\

Generally, there will be:

S_n^+=\sum\limits_{k=1}^\frac{n}{2}(2k)^2\\\\\\S_n^-=\sum\limits_{k=1}^\frac{n}{2}(2k-1)^2\\\\\\

Now, calculate our sum:

\left|\sum\limits_{k=1}^n(-1)^k\cdot k^2\right|=\left|S_n^+-S_n^-\right|=&#10;\left|\sum\limits_{k=1}^\frac{n}{2}(2k)^2-\sum\limits_{k=1}^\frac{n}{2}(2k-1)^2\right|=\\\\\\=&#10;\left|\sum\limits_{k=1}^\frac{n}{2}4k^2-\sum\limits_{k=1}^\frac{n}{2}\left(4k^2-4k+1\right)\right|=\\\\\\

=\left|4\sum\limits_{k=1}^\frac{n}{2}k^2-4\sum\limits_{k=1}^\frac{n}{2}k^2+4\sum\limits_{k=1}^\frac{n}{2}k-\sum\limits_{k=1}^\frac{n}{2}1\right|=\left|4\sum\limits_{k=1}^\frac{n}{2}k-\sum\limits_{k=1}^\frac{n}{2}1\right|\stackrel{(1),(2)}{=}\\\\\\=&#10;\left|4\dfrac{\frac{n}{2}(\frac{n}{2}+1)}{2}-\dfrac{n}{2}\right|=\left|2\cdot\dfrac{n}{2}\left(\dfrac{n}{2}+1\right)-\dfrac{n}{2}\right|=\left|n\left(\dfrac{n}{2}+1\right)-\dfrac{n}{2}\right|=\\\\\\&#10;

=\left|\dfrac{n^2}{2}+n-\dfrac{n}{2}\right|=\left|\dfrac{n^2}{2}+\dfrac{n}{2}\right|=\left|\dfrac{n^2+n}{2}\right|=\left|\dfrac{n(n+1)}{2}\right|\stackrel{(2)}{=}\\\\\\\stackrel{(2)}{=}&#10;\left|\sum\limits_{k=1}^nk\right|\stackrel{(3)}{=}\sum\limits_{k=1}^nk

So in this case we prove, that:

 \left|\sum\limits_{k=1}^n(-1)^k\cdot k^2\right|=\sum\limits_{k=1}^nk

2) n is odd.

Here, S_n^- has more terms than S_n^+. For example if n=7 then:

S_7^-=\underbrace{1^2+3^2+5^2+7^2}_{\frac{n+1}{2}=\frac{7+1}{2}=4}\\\\\\&#10;S_7^+=\underbrace{2^2+4^4+6^2}_{\frac{n+1}{2}-1=\frac{7+1}{2}-1=3}\\\\\\

So there is \dfrac{n+1}{2} terms in S_n^-, \dfrac{n+1}{2}-1 terms in S_n^+ and:

S_n^+=\sum\limits_{k=1}^{\frac{n+1}{2}-1}(2k)^2\\\\\\&#10;S_n^-=\sum\limits_{k=1}^{\frac{n+1}{2}}(2k-1)^2

Now, we can calculate our sum:

\left|\sum\limits_{k=1}^n(-1)^k\cdot k^2\right|=\left|S_n^+-S_n^-\right|=&#10;\left|\sum\limits_{k=1}^{\frac{n+1}{2}-1}(2k)^2-\sum\limits_{k=1}^{\frac{n+1}{2}}(2k-1)^2\right|=\\\\\\=&#10;\left|\sum\limits_{k=1}^{\frac{n+1}{2}-1}4k^2-\sum\limits_{k=1}^{\frac{n+1}{2}}\left(4k^2-4k+1\right)\right|=\\\\\\=&#10;\left|\sum\limits_{k=1}^{\frac{n-1}{2}-1}4k^2-\sum\limits_{k=1}^{\frac{n+1}{2}}4k^2+\sum\limits_{k=1}^{\frac{n+1}{2}}4k-\sum\limits_{k=1}^{\frac{n+1}{2}}1\right|=\\\\\\

=\left|\sum\limits_{k=1}^{\frac{n-1}{2}-1}4k^2-\sum\limits_{k=1}^{\frac{n+1}{2}-1}4k^2-4\left(\dfrac{n+1}{2}\right)^2+\sum\limits_{k=1}^{\frac{n+1}{2}}4k-\sum\limits_{k=1}^{\frac{n+1}{2}}1\right|=\\\\\\=&#10;\left|-4\left(\dfrac{n+1}{2}\right)^2+4\sum\limits_{k=1}^{\frac{n+1}{2}}k-\sum\limits_{k=1}^{\frac{n+1}{2}}1\right|\stackrel{(1),(2)}{=}\\\\\\&#10;\stackrel{(1),(2)}{=}\left|-4\dfrac{n^2+2n+1}{4}+4\dfrac{\frac{n+1}{2}\left(\frac{n+1}{2}+1\right)}{2}-\dfrac{n+1}{2}\right|=\\\\\\

=\left|-n^2-2n-1+2\cdot\dfrac{n+1}{2}\left(\dfrac{n+1}{2}+1\right)-\dfrac{n+1}{2}\right|=\\\\\\=&#10;\left|-n^2-2n-1+(n+1)\left(\dfrac{n+1}{2}+1\right)-\dfrac{n+1}{2}\right|=\\\\\\=&#10;\left|-n^2-2n-1+\dfrac{(n+1)^2}{2}+n+1-\dfrac{n+1}{2}\right|=\\\\\\=&#10;\left|-n^2-n+\dfrac{n^2+2n+1}{2}-\dfrac{n+1}{2}\right|=\\\\\\=&#10;\left|-n^2-n+\dfrac{n^2}{2}+n+\dfrac{1}{2}-\dfrac{n}{2}-\dfrac{1}{2}\right|=\left|-\dfrac{n^2}{2}-\dfrac{n}{2}\right|=\left|-\dfrac{n^2+n}{2}\right|=\\\\\\

=\left|-\dfrac{n(n+1)}{2}\right|=|-1|\cdot\left|\dfrac{n(n+1)}{2}\right|=\left|\dfrac{n(n+1)}{2}\right|\stackrel{(2)}{=}\left|\sum\limits_{k=1}^nk\right|\stackrel{(3)}{=}\sum\limits_{k=1}^nk

We consider all possible n so we prove that:

\forall_{n\in\mathbb{N}}\quad\left|\sum\limits_{k=1}^n(-1)^k\cdot k^2\right|=\sum\limits_{k=1}^nk
7 0
3 years ago
Which of the following functions is not a linear
kicyunya [14]

By looking at the degree of the polynomials, we conclude that the option that is not linear is option C.

<h3>Which of the following functions is not a linear function?</h3>

A polynomial is an expression of the form:

p(x) = a_n*x^n + a_{n-1}*x^{n-1} + ... + a_0

Where all the exponents are natural numbers.

We define the degree of a polynomial as the largest exponent, and we define a linear function as a polynomial of degree 1.

So a linear function is of the form:

f(x) = a_1*x + a_0

Now, if you look at option C, you can see that the degree of that polynomial is 2. So that is not a linear function.

Then the correct option is C.

If you want to learn more about linear equations:

brainly.com/question/1884491

#SPJ1

7 0
1 year ago
Other questions:
  • Draw 8 lines that are between 1 inch and three inches long measure each line to the nearest fourth inch and make a line plot
    13·2 answers
  • Your great aunt sally loans you $5000 for three years and asks that you repay it with annually compounding interest at the rate
    11·1 answer
  • If you save $4,500 at an interest rate of 13 percent per year how much will you have at the end of seven years?
    5·1 answer
  • Helppppppppppppppppppppppppp
    8·1 answer
  • Being timmed please help me
    15·1 answer
  • Oliver needs to save at least $1500 to buy a computer. he already has saved $650. how solve the inequality
    13·2 answers
  • Given: △ABC, D∈ AC m∠ABC=m∠BDA AB=2, AC=4 Find: AD and DC
    10·2 answers
  • 1.
    15·1 answer
  • Help me please!!!!!!!!!!!!!!!!!!!
    9·2 answers
  • Evaluate 8.3p+7.9r when p =3 and r=4. 8.3( ) +7.9( ) = +31.6 help!!!!! ​
    12·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!