I have 3 hundreds,17 tens,13 ones,1 tenths and 2 hundreths what number am I?

it's a week before the maths festival a maths exhibition costa £75.40 and so far 12 have been bought. what's the total income so

motikmotik

£75.40*12= £904.80

A quick multiplication

6 0

3 years ago

Part A: Find the LCM of 5 and 12. Show your work. (3 points)

diamong [38]

Answer:

Part A: 60

5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60.

12, 24, 36, 48, 60.

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Part B: 9

81 / 72 = 9

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Part C: 9 (8 + 9)

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GCF of 72 and 81 is 9...

Divide 72 by 9 and get 8.

Divide 81 by 9 and get 9.

To check multiply 9 with 8, and multiply 9 by 9.

5 0

2 years ago

If the diagonals of a parallelogram are congruent and are perpendicular bisectors of each other than what is it

alexira [117]

The parallelogram is a square

5 0

2 years ago

A ramp to a building has a height of 4 feet and the angle of elevation is 33 defrees. How long is the ramp?

sladkih [1.3K]

Height=4 ft
angle=33 degrees
My tip first is to draw it out. Draw an acute angle opening up to a building forming a triangle. Inside of the vertex of the acute angle is the degree 33 and the height of the building is 4 ft. So, use trig! Do you have a calculator? Opposite over hypotenuse is Sine. (SOH CAH TOA). (4)/(sin(33)). which equals 7.34 ft as your height! Hope this helped!

5 0

2 years ago

If <img src="https://tex.z-dn.net/?f=%20300cm%5E%7B2%7D%20" id="TexFormula1" title=" 300cm^{2} " alt=" 300cm^{2} " align="absmid

Artist 52 [7]

Check the picture below. Recall, is an open-top box, so, the top is not part of the surface area, of the 300 cm². Also, recall, the base is a square, thus, length = width = x.

$\bf \textit{volume of a rectangular prism}\\\\
V=lwh\quad 
\begin{cases}
l = length\\
w=width\\
h=height\\
-----\\
w=l=x
\end{cases}\implies V=xxh\implies \boxed{V=x^2h}\\\\
-------------------------------\\\\
\textit{surface area}\\\\
S=4xh+x^2\implies 300=4xh+x^2\implies \cfrac{300-x^2}{4x}=h
\\\\\\
\boxed{\cfrac{75}{x}-\cfrac{x}{4}=h}\\\\
-------------------------------\\\\
V=x^2\left( \cfrac{75}{x}-\cfrac{x}{4} \right)\implies V(x)=75x-\cfrac{1}{4}x^3$

so.. that'd be the V(x) for such box, now, where is the maximum point at?

$\bf V(x)=75x-\cfrac{1}{4}x^3\implies \cfrac{dV}{dx}=75-\cfrac{3}{4}x^2\implies 0=75-\cfrac{3}{4}x^2
\\\\\\
\cfrac{3}{4}x^2=75\implies 3x^2=300\implies x^2=\cfrac{300}{3}\implies x^2=100
\\\\\\
x=\pm10\impliedby \textit{is a length unit, so we can dismiss -10}\qquad \boxed{x=10}$

now, let's check if it's a maximum point at 10, by doing a first-derivative test on it. Check the second picture below.

so, the volume will then be at $\bf V(10)=75(10)-\cfrac{1}{4}(10)^3\implies V(10)=500 \ cm^3$

6 0

2 years ago