How do you evaluate the expression to this problem 24 divided by (-4)+(-2) times (-5)

Mathematics

2 answers:

Tju [1.3M]3 years ago

8 0

24/-30 would be the answer

balu736 [363]3 years ago

7 0

I think it's 24/- 36

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Can somebody help please!????

stiv31 [10]

A and b proves that they are parallel

4 0

3 years ago

Do any of you guys have an idea how to do this?

fiasKO [112]

It would be the last one

Explanation

A vowel out of apple so they are two vowels in apple out of 5 letters so that’s 2/5

And there are two consonants in the word book all that’s 2/4

2/5 + 2/4 = 4/9

6 0

2 years ago

Help me please I don’t understand what to do

Norma-Jean [14]

Answer:

i dont know please dont hate me

Step-by-step explanation:

<h2>i dont know please dont hate me</h2>

6 0

3 years ago

Solve 4p^2 – p=0.<br> p=____, p=_____

Anon25 [30]

Answer:

p= 0

p= 1/4

............

8 0

2 years ago

If x^3 + 1/x^3 = 110 then find x + 1/x

Svetlanka [38]

$Use:(a+b)^3=a^3+3a^2b+3ab^2+b^3\ (*)\\\\x^3+\left(\dfrac{1}{x}\right)^3\\=\underbrace{x^3+3\cdot x^2\cdot\dfrac{1}{x}+3\cdot x\cdot\left(\dfrac{1}{x}\right)^2+\left(\dfrac{1}{x}\right)^3}_{(*)}-3\cdot x^2\cdot\dfrac{1}{x}-3\cdot x\cdot\left(\dfrac{1}{x}\right)^2\\\\=\left(x+\dfrac{1}{x}\right)^3-3x-3\cdot\dfrac{1}{x}=\left(x+\dfrac{1}{x}\right)^3-3\left(x+\dfrac{1}{x}\right)$

therefore

$x^3+\left(\dfrac{1}{x}\right)^3=110\iff \left(x+\dfrac{1}{x}\right)^3-3\left(x+\dfrac{1}{x}\right)=110\\\\subtitute\ t=x+\dfrac{1}{x}\\\\t^3-3t=110\ \ \ \ |subtract\ 110\ from\ both\ sides\\\\t^3-3t-110=0\\\\t^3-25t+22t-110=0\\\\t(t^2-25)+22(t-5)=0\ \ \ |use\ a^2-b^2=(a-b)(a+b)\\\\t(t-5)(t+5)+22(t-5)=0\\\\(t-5)[t(t+5)+22]=0\iff t-5=0\ or\ \underbrace{t^2+5t+22=0}_{no\ solution}\\\\t=5\iff x+\dfrac{1}{x}=5\\\\Answer:\boxed{x+\dfrac{1}{x}=5}$

3 0

3 years ago