Please help me please i really need help please

Mathematics

2 answers:

Artyom0805 [142]3 years ago

6 0

Answer:

Step-by-step explanation:

We are adding together, so b and d can't be the answer. Fran has 2 fewer hats than elena , so Fran cannot be h or else she would have more hats.

So, a would have to be the answer.

DanielleElmas [232]3 years ago

3 0

Answer:

this is very hard, I do not understand such math, I don't know

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Roberto wants to reduce a drawing that is 12 inches long by 9 wide. If his new drawing is 8 inches long, how wide will it be?

Novosadov [1.4K]

1.) ¹²/₉ = ⁴/₃
Simplify the fraction so it’s easier to work with.
2.) ⁴/₃ = ⁸/x
Make ⁴/₃ equal to ⁸/x because since you want to find the new width, you need to make it equal to a new fraction.
3.) 4x = 24
Cross multiply.
4.) x = 6
Divide each side by 4.

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3 years ago

What is 3n-2/5=7/10 solve please

Tems11 [23]

Answer: $\bold{n=\dfrac{11}{6}}$

<u>Step-by-step explanation:</u>

$\dfrac{3n-2}{5}=\dfrac{7}{10}\\\\\\10(3n-2)=5(7)\qquad \longrightarrow \quad \text{(cross multiplied)}\\\\30n-20=35\qquad \longrightarrow \quad \text{(distributed)}\\\\30n=55\qquad \longrightarrow \quad \text{(added 20 to both sides)}\\\\n=\dfrac{55}{30}\qquad \longrightarrow \quad \text{(divided 30 from both sides)}\\\\n=\dfrac{11}{6}\quad \longrightarrow \quad \text{(simplified fraction)}\\\\\\\large\boxed{n=\dfrac{11}{6}}\\$

8 0

3 years ago

How to prove this???

swat32

$\cos^3 2A + 3 \cos 2A \\
\Rightarrow \cos 2A (\cos^2 2A + 3) \\
\Rightarrow (\cos^2 A - \sin^2 A) (\cos^2 2A + 3) \\
\Rightarrow (\cos^2 A - \sin^2 A) (1 - \sin^2 2A + 3) \\
\Rightarrow (\cos^2 A - \sin^2 A) (4 - \sin^2 2A) \\
\Rightarrow (\cos^2 A - \sin^2 A) (4 - (2\sin A \cos A)(2\sin A \cos A)) \\
\Rightarrow (\cos^2 A - \sin^2 A) (4 - 4\sin^2 A \cos^2 A) \\ 
\Rightarrow 4(\cos^2 A - \sin^2 A) (1 - \sin^2 A \cos^2 A) 
$

go to right side now

$4( \cos^6 A - \sin^6 A)\\
\Rightarrow 4( \cos^3 A - \sin^3 A)(\cos^3 A + \sin^3 A)$

use $x^3 - y^3 = (x-y)(x^2 + xy + y^2)$ and $x^3 + y^3 = x^2 - xy + y^2$

$4( \cos^6 A - \sin^6 A)\\ \Rightarrow 4( \cos^3 A - \sin^3 A)(\cos^3 A + \sin^3 A) \\
\Rightarrow 4(\cos A - \sin A)(\cos^2 A + \cos A \sin A + \sin^2 A) \\
~\quad \quad\cdot ( \cos A + \sin A)(\cos^2 A - \cos A \sin A + \cos^2 A)$

so $\sin^2 A + \cos^2 A = 1$

$4( \cos^6 A - \sin^6 A)\\ \Rightarrow 4(\cos A - \sin A)(\cos^2 A + \cos A \sin A + \sin^2 A) \\ ~\quad \quad\cdot ( \cos A + \sin A)(\cos^2 A - \cos A \sin A + \cos^2 A) \\ \Rightarrow 4(\cos^2 A - \sin^2 A)(1 + \cos A \sin A )(1- \cos A \sin A ) \\ \Rightarrow 4(\cos^2 A - \sin^2 A)(1 - \cos^2 A \sin^2 A )\\ \Rightarrow 4(\cos^2 A - \sin^2 A)(1 - \sin^2 A \cos^2 A ) \\
 \Rightarrow Left hand side$