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n200080 [17]
3 years ago
12

Share £40 in the ratio 1:4 between Tim and sam

Mathematics
1 answer:
Naddika [18.5K]3 years ago
4 0
Tim : Sam : Total
1 : 4 : 1+4
1 : 4 : 5

5x = 40
x = 40 ÷ 5
x = 8

x = 8
4x = 8x4 = 32

Tim : Sam
1 : 4
£8 : £32

So Tim = £8 and Sam = £32
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A triangle has side lengths of 7 in., 9 in., and 11 in. Determine whether this is a right triangle and why.
Harlamova29_29 [7]

Answer:

Not a right triangle.  See below for my reasoning.


Step-by-step explanation:

If this is a right triangle, then the sum of the squares of the two shorter sides will equal the square of the longest side:  7^2 + 9^2 = 11^2 => 49 + 81 = 121   But this is false.

8 0
3 years ago
Graph "-3n+4<25" on a line plot
Studentka2010 [4]

Answer:The solution is in the attached file

Step-by-step explanation:

6 0
2 years ago
121 slabs . How many slabs will she need to lay in each row to make a square
netineya [11]

She needs to lay 11 slabs on each row to make a square

<h3>How to determine the number of slabs in each row?</h3>

The total number of slabs is given as:

Total = 121 slabs

Let this represent the area of the slab.

The area of a square is calculated as:

Area = Length^2

Substitute the known values in the above equation

Length^2 = 121

Take the square root of both sides

Length = 11

Hence, she needs to lay 11 slabs on each row to make a square

Read more about squares at:

brainly.com/question/24487155

#SPJ1

4 0
2 years ago
2/5 of the instruments in the marching band are brass one third or percussion and the rest are wood ones 1/8 of the brass instru
kvasek [131]
There are twelve tubas in the band. And btw you should work on your spelling. I tutor. So if you want my help just message me.
5 0
2 years ago
Find the surface area of x^2+y^2+z^2=9 that lies above the cone z= sqrt(x^@+y^2)
Mashcka [7]
The cone equation gives

z=\sqrt{x^2+y^2}\implies z^2=x^2+y^2

which means that the intersection of the cone and sphere occurs at

x^2+y^2+(x^2+y^2)=9\implies x^2+y^2=\dfrac92

i.e. along the vertical cylinder of radius \dfrac3{\sqrt2} when z=\dfrac3{\sqrt2}.

We can parameterize the spherical cap in spherical coordinates by

\mathbf r(\theta,\varphi)=\langle3\cos\theta\sin\varphi,3\sin\theta\sin\varphi,3\cos\varphi\right\rangle

where 0\le\theta\le2\pi and 0\le\varphi\le\dfrac\pi4, which follows from the fact that the radius of the sphere is 3 and the height at which the sphere and cone intersect is \dfrac3{\sqrt2}. So the angle between the vertical line through the origin and any line through the origin normal to the sphere along the cone's surface is

\varphi=\cos^{-1}\left(\dfrac{\frac3{\sqrt2}}3\right)=\cos^{-1}\left(\dfrac1{\sqrt2}\right)=\dfrac\pi4

Now the surface area of the cap is given by the surface integral,

\displaystyle\iint_{\text{cap}}\mathrm dS=\int_{\theta=0}^{\theta=2\pi}\int_{\varphi=0}^{\varphi=\pi/4}\|\mathbf r_u\times\mathbf r_v\|\,\mathrm dv\,\mathrm du
=\displaystyle\int_{u=0}^{u=2\pi}\int_{\varphi=0}^{\varphi=\pi/4}9\sin v\,\mathrm dv\,\mathrm du
=-18\pi\cos v\bigg|_{v=0}^{v=\pi/4}
=18\pi\left(1-\dfrac1{\sqrt2}\right)
=9(2-\sqrt2)\pi
3 0
2 years ago
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