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2 years ago

Which solid does this net form?

Mathematics

1 answer:

polet [3.4K]2 years ago

7 0

Answer:

hexagonal pyramid

Step-by-step explanation:

i don't know exactly

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Masja [62]

The perimeter of RSTU will be C.22m if you look at both of them u will see they are both the same and tell you how long the sides are,........hope i was right:))

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3 years ago

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What is another way to name ∠MNO ?

Serhud [2]

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3 years ago

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Consider the initial value problem:

Inessa [10]

Answer:

$\left \{ {{u'=v} \atop {v'=5v-6u-5\sin(2t)}} \right. \\u(0)=-5;u'(0)=2$

Step-by-step explanation:

The given initial value problem is;

$y''-5y'+6y=-5\sin(2t)---(1)\\y(0)=-5,y'(0)=2$

Let

$u(t)=y(t)----(2)\\v(t)=y'(t)----(3)$

Differentiating both sides of equation (1) with respect to $t$ , we obtain:

$u'(t)=y'(t)---(4)\\ \\\implies u'(t)=v(t)---(5)$

Differentiating both sides of equation (2) with respect to $t$ gives:

$v'(t)=y''(t)----(6)$

From equation (1),

$y''=5y'-6y-5\sin(2t)\\y''=5v(t)-6u(t)-5\sin(2t)\\\implies v'(t)=5v(t)-6u(t)-5\sin(2t)----(7)$

Putting t=0 into equation (2) yields

$u(0)=y(0)\\\implies u(0)=-5$

Also putting t=0 into equation (3)

$u'(0)=y'(0)\\u'(0)=2$

The system of first order equations is:

$\left \{ {{u'=v} \atop {v'=5v-6u-5\sin(2t)}} \right. \\u(0)=-5;u'(0)=2$

3 0

3 years ago

Danny charges $30 for 3 hours of swimming lessons, Martin charges $44 for 4 hours of swimming lessons. Who offers a better deal?

emmasim [6.3K]

Answer:

Danny

Step-by-step explanation:

danny has a better offer

8 0

3 years ago

An inverted pyramid is being filled with water at a constant rate of 25 cubic centimeters per second. The pyramid, at the top, h

UNO [17]

Answer:

$\frac{225}{16} cm/s$

Step-by-step explanation:

We are given that

$\frac{dV}{dt}=25cm^3/s$

Side of base=4 cm

l=w=4 cm

Height,h=12 cm

We have to find the rate at which the water level rising when the water level is 4 cm.

Volume of pyramid= $\frac{1}{3}lwh=\frac{1}{3}l^2h$

$\frac{l}{h}=\frac{4}{12}=\frac{1}{3}$

$l=\frac{1}{3}h$

Substitute the value

$V=\frac{1}{27}h^3$

Differentiate w.r.t t

$\frac{dV}{dt}=\frac{3}{27}h^2\frac{dh}{dt}$

Substitute the values

$25=\frac{1}{9}(4^2)\frac{dh}{dt}$

$\frac{dh}{dt}=\frac{25\times 9}{16}=\frac{225}{16} cm/s$

5 0

3 years ago