All categories

3 years ago

Help im stupid plzzzzz

Mathematics

1 answer:

Marianna [84]3 years ago

4 0

Answer:

the third option: x + 2(2x - 4) = 10

Step-by-step explanation:

the question tells you that y = 2x - 4.

so, instead of writing y in the equation, just replace it with 2x - 4.

x - 2y = 10

becomes

x - 2(2x-4) = 10

You might be interested in

Will give u brainlist

dusya [7]

The value of x equals 30.

Step by Step explanation:

All sides are equal so you know the angle lengths are equal. When you multiple 2x by 3 you get 6x. All angles of a triangle add up to 180 so you divide 180 by 6 and you get 30. Hope This Helps!

5 0

3 years ago

Determine if the ordered pair (6, 4) is a solution to the inequality

kvasek [131]

Answer:

Step-by-step explanation:

all we need to do is to plug the point (6,4) in the inequality and see if it satisfies it :

pay attention that here we have x=6 y=4

4< $\frac{3}{4} (6) - 3$

we simplify we get :

4<4.5-3

4<1.5 which is incorrect so (6,4) is not a solution. moreover

notice that 4 is > than 1.5 so the point lies above the line

thus the answer is : A

you can also solve this problem by graphing the line $y=$ $\frac{3}{4} x-3$ and plotting the point (6,4) and hence you will notice that the point is above the line

5 0

3 years ago

Read 2 more answers

What is <img src="https://tex.z-dn.net/?f=%20%5Cfrac%7B8m%7D%7Bn%7D%20%5E7" id="TexFormula1" title=" \frac{8m}{n} ^7" alt=" \fra

kifflom [539]

$\bf \cfrac{8m^7}{n}\implies 8m^7n^{-1}\implies 2^3m^{3+3+1}n^{-1}\implies 2^3m^3m^3m^1n^{-1}
\\\\\\
2^3(m^2)^3mn^{-1}\implies (2m^2)^3mn^{-1}$

7 0

3 years ago

Which pair of complex factors results in a real-number product?

Tcecarenko [31]

Answer:

Step-by-step explanation:

6 0

3 years ago

For each of 2 ≤ n ≤ 5, compute c(n, 0) + c(n, 2) + ⋅⋅⋅ + c(n, 2k), where 2k is largest even integer ≤ n, and compute c(n, 1) + c

Ksivusya [100]

Denote $C(n,m)=\dbinom nm$ . Then

$n=2\implies\displaystyle\binom20+\binom22=1+1=2$
$n=3\implies\displaystyle\binom30+\binom32=1+3=4$
$n=4\implies\displaystyle\binom40+\binom42+\binom44=1+6+1=8$
$n=5\implies\displaystyle\binom50+\binom52+\binom54=1+10+5=16$

In general, it would appear that

$\displaystyle\sum_{k=0}^{2\lfloor\frac n2\rfloor}\binom n{2k}=2^{n-1}$

On the other hand,

$n=2\implies\displaystyle\binom21=2$
$n=3\implies\displaystyle\binom31+\binom33=3+1=4$
$n=4\implies\displaystyle\binom41+\binom43=4+4=8$
$n=5\implies\displaystyle\binom51+\binom53+\binom55=5+10+1=16$

so that in general, we also get

$\displaystyle\sum_{k=0}^{2\lfloor\frac{n-1}2\rfloor+1}\binom n{2k+1}=2^{n-1}$

7 0

3 years ago