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3 years ago

Pls help me its due sooon

Mathematics

2 answers:

stira [4]3 years ago

3 0

Answer:

nobody

Step-by-step explanation:

fomenos3 years ago

3 0

Somebody ㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤ

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PLZ HELP ASAP!!!!!!!!

andreyandreev [35.5K]

Answer:

Step-by-step explanation:

formula for cylinder:V=πr2h

shape of the base: circle

formula for circle: A=πr2

radius: 6.5cm

height: 16cm

V=3.14x6.5^2x16

2122.64cm^3

4 0

3 years ago

Read 2 more answers

What is the lenght of the hypothesis in the triangle?

seraphim [82]

Answer:

ok in this case will represent the height with a the bells with b and the hypotenuse with c so it will be c squared is equals to a squared plus b squared 12 squared + 5 squared is equals to c squared 12 square is 144 while 5 squared is 25 when added it is 169 so csquared is equals to 169 to get c we find √ 169 source is equals to 13

6 0

3 years ago

The area of a triangular flag is 132 square centimetres. Its altitude is 2 centimetres longer than twice its base. Find the leng

Anika [276]

Answer:

24 cm and 11 cm

Step-by-step explanation:

GIVEN: The area of a triangular flag is $132 \text{ cm}^2$ . Its altitude is $2\text{ cm}$ longer than twice its base.

TO FIND: lengths of the altitude and the base.

SOLUTION:

Let the altitude and the base of flag be $a$ and $b$

According to the question

$a=2+2b$

area of triangle $=\frac{1}{2}\times base \times alltitude=\frac{1}{2}a\times b$

$\implies \frac{1}{2}(2b+2)b=132 \Rightarrow 2b^2+2b-264$

Solving equation we get

$b=11$ then $a=24$

Hence the lengths of altitude and base of triangular flag is 24 cm and 11 cm respectively.

8 0

3 years ago

(1 point) Find the length traced out along the parametric curve x=cos(cos(4t))x=cos⁡(cos⁡(4t)), y=sin(cos(4t))y=sin⁡(cos⁡(4t)) a

Mazyrski [523]

The length of a curve $C$ given parametrically by $(x(t),y(t))$ over some domain $t\in[a,b]$ is

$\displaystyle\int_C\mathrm ds=\int_a^b\sqrt{\left(\frac{\mathrm dx}{\mathrm dt}\right)^2+\left(\frac{\mathrm dy}{\mathrm dt}\right)^2}\,\mathrm dt$

In this case,

$x(t)=\cos(\cos4t)\implies\dfrac{\mathrm dx}{\mathrm dt}=-\sin(\cos4t)(-\sin4t)(4)=4\sin4t\sin(\cos4t)$

$y(t)=\sin(\cos4t)\implies\dfrac{\mathrm dy}{\mathrm dt}=\cos(\cos4t)(-\sin4t)(4)=-4\sin4t\cos(\cos4t)$

So we have

$\displaystyle\left(\frac{\mathrm dx}{\mathrm dt}\right)^2+\left(\frac{\mathrm dy}{\mathrm dt}\right)^2=16\sin^24t\sin^2(\cos4t)+16\sin^24t\cos^2(\cos4t)=16\sin^24t$

and the arc length is

$\displaystyle\int_0^1\sqrt{16\sin^24t}\,\mathrm dt=4\int_0^1|\sin4t|\,\mathrm dt$

We have

$\sin(4t)=0\implies4t=n\pi\implies t=\dfrac{n\pi}4$

where $n$ is any integer; this tells us $\sin(4t)\ge0$ on the interval $\left[0,\frac\pi4\right]$ and $\sin(4t)$ on $\left[\frac\pi4,1\right]$ . So the arc length is