GIÚP MIK VỚI Ạ MIK CẦN GẤP !!! ÉT O ÉT
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May 1 bộ hết số mét vải là:
27 : 5 = 5,4 (m)
May 16 bộ quần áo hết số mét vải là:
5,4 x 16 = 86,4 (m)
Đáp số: 86,4 m vải
>In 100 years, there’ll be too many people on Earth. So I think we have big city space. People will live space station. Children will also go to school space station. There will be special classroom for learning more about new life in space.
>Maybe children will have robots help them go to school. And turn them get up on the bed….
>The robots will be like a best friend for the children because they will talk to them and help them with any problems.
> I think children will still go to normal school and learn from their teachers. The only different is that all children will have their own computers in the classroom. There will be a lot more machines to help us learn.
As city space becomes more squeezed, we will burrow deeper and build higher with the creation of:
- Super skyscrapers: carbon nanotubes and diamond nanothreads will help us create towering megastructures that will dwarf today’s skyscrapers
- Earth-scrapers: just as we build up, we will also dig down – huge structures will tunnel 25 storeys deep or more
- Underwater cities: are likely to become a reality – using the water itself to create breathable atmospheres and generating hydrogen fuel through the process
As technology develops, we’ll see:
- 3D printing of houses and furniture: we will be able to print exact replicas of large scale structures like houses out of local, recyclable materials
- Stepping into home medi-pods will confirm if you really are ill, providing a digital diagnosis and supplying medicine or a remote surgeon if needed, meaning ‘pulling a sickie’ could be a thing of the past
- And finally, we will:
- Colonise space: first the Moon, then Mars and then far beyond into the galaxy
4:
a: -90<a<0
=>cos a>0
cos^2a=1-(-4/5)^2=9/25
=>cosa=3/5
\(sin\left(45-a\right)=sin45\cdot cosa-cos45\cdot sina=\dfrac{\sqrt{2}}{2}\left(cosa-sina\right)\)
\(=\dfrac{\sqrt{2}}{2}\left(\dfrac{3}{5}-\dfrac{4}{5}\right)=\dfrac{-\sqrt{2}}{10}\)
b: pi/2<a<pi
=>cosa<0
cos^2a+sin^2a=0
=>cos^2a=16/25
=>cosa=-4/5
tan a=3/5:(-4/5)=-3/4
\(tan\left(a+\dfrac{pi}{3}\right)=\dfrac{tana+\dfrac{tanpi}{3}}{1-tana\cdot tan\left(\dfrac{pi}{3}\right)}\)
\(=\dfrac{-\dfrac{3}{4}+\sqrt{3}}{1-\dfrac{-3}{4}\cdot\sqrt{3}}=\dfrac{48-25\sqrt{3}}{11}\)
c: 3/2pi<a<pi
=>cosa>0
cos^2a+sin^2a=1
=>cos^2a=25/169
=>cosa=5/13
cos(pi/3-a)
\(=cos\left(\dfrac{pi}{3}\right)\cdot cosa+sin\left(\dfrac{pi}{3}\right)\cdot sina\)
\(=\dfrac{5}{13}\cdot\dfrac{1}{2}+\dfrac{-12}{13}\cdot\dfrac{\sqrt{3}}{2}=\dfrac{5-12\sqrt{3}}{26}\)
31,5 x 0,1 + x `3 = 7,65
3,15 + x 3 = 7,65
x 3 = 7,65 - 3,15
x 3 = 4,5
= 4,5 : 3
= 1,5
tick cho mik nha
`31.5xx0.1+x xx3=7.65`
`=> 3.15+x xx3=7.65`
`=> x xx3=7.65-3.15`
`=> x xx3=4.5`
`=> x=4.5:3`
`=> x=1.5`
Vậy `x=1.5`
a: A(x)=x^5+3x^4-2x^3-9x^2+11x-6
B(x)=x^5+3x^4-2x^3-10x^2+9x-8
C(x)=A(x)-B(x)
=x^5+3x^4-2x^3-9x^2+11x-6-x^5-3x^4+2x^3+10x^2-9x+8
=x^2+2x+2
b; C(x)=2x+2
=>x^2=0
=>x=0
c: C(x)=2012
=>x^2+2x-2010=0
Δ=2^2-4*1*(-2010)=8044>0
=>Phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{-2-\sqrt{8044}}{2}\simeq-45,84\left(loại\right)\\x_2=\dfrac{-2+\sqrt{8044}}{2}\simeq43,84\left(loại\right)\end{matrix}\right.\)
=>Ko có giá trị nguyên của x thỏa mãn