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scoray [572]
3 years ago
13

FIND THE LENGTH OF THE HYPOTENUS OF THE TRAINGLE ?HURRY¿ DO QUESTIONS 1 AND 2

Mathematics
2 answers:
kakasveta [241]3 years ago
7 0
I agree with screamingzebra420
beks73 [17]3 years ago
3 0

Answer:

1.) C = 17 ft.

2.) C = 0.5 in. or 1/2 in.

Step-by-step explanation:

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Sveta_85 [38]

Answer:6/4

Step-by-step explanation:

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3 years ago
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Find the perimeter. Simplify your answer.<br> 2-5<br> 2-5<br> Z-8
lukranit [14]

2(z - 5) + (z - 8) =

= 2z - 10 + z - 8 =

= 2z + z - 10 - 8 = <u>3</u><u>z</u><u> </u><u>-</u><u> </u><u>1</u><u>8</u> ← the end

4 0
3 years ago
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Consider a series circuit consisting of a resistor of R ohms, an inductor of L henries, and variable voltage source of V(t) volt
ser-zykov [4K]

Answer:

I(t)=\frac{1}{3}(1-e^{-30t})

Step-by-step explanation:

We are given that

\frac{dI}{dt}+\frac{R}{L}I=\frac{V(t)}{L}

R=150 ohm

L=5 H

V(t)=10 V

P=\frac{R}{L}=\frac{150}{5}=30

I.F=e^{\int Pdt}=e^{\int 30 dt}=e^{30 t}

I(t)\times I.F=\int e^{30 t}\times 10 dt+C

I(t)\times e^{30 t}=\frac{10}{30}e^{30 t}+C

I(t)=\frac{1}{3}+Ce^{-30 t}

I(0)=0

Substitute t=0

0=\frac{1}{3}+C

C=-\frac{1}{3}

Substitute the values

I(t)=\frac{1}{3}-\frac{1}{3}e^{-30 t}

I(t)=\frac{1}{3}(1-e^{-30t})

5 0
3 years ago
You have a wire that is 20 cm long. You wish to cut it into two pieces. One piece will be bent into the shape of a square. The o
Aleksandr [31]

Answer:

Therefore the circumference of the circle is =\frac{20\pi}{4+\pi}

Step-by-step explanation:

Let the side of the square be s

and the radius of the circle be r

The perimeter of the square is = 4s

The circumference of the circle is =2πr

Given that the length of the wire is 20 cm.

According to the problem,

4s + 2πr =20

⇒2s+πr =10

\Rightarrow s=\frac{10-\pi r}{2}

The area of the circle is = πr²

The area of the square is = s²

A represent the total area of the square and circle.

A=πr²+s²

Putting the value of s

A=\pi r^2+ (\frac{10-\pi r}{2})^2

\Rightarrow A= \pi r^2+(\frac{10}{2})^2-2.\frac{10}{2}.\frac{\pi r}{2}+ (\frac{\pi r}{2})^2

\Rightarrow A=\pi r^2 +25-5 \pi r +\frac{\pi^2r^2}{4}

\Rightarrow A=\pi r^2\frac{4+\pi}{4} -5\pi r +25

For maximum or minimum \frac{dA}{dr}=0

Differentiating with respect to r

\frac{dA}{dr}= \frac{2\pi r(4+\pi)}{4} -5\pi

Again differentiating with respect to r

\frac{d^2A}{dr^2}=\frac{2\pi (4+\pi)}{4}    > 0

For maximum or minimum

\frac{dA}{dr}=0

\Rightarrow \frac{2\pi r(4+\pi)}{4} -5\pi=0

\Rightarrow r = \frac{10\pi }{\pi(4+\pi)}

\Rightarrow r=\frac{10}{4+\pi}

\frac{d^2A}{dr^2}|_{ r=\frac{10}{4+\pi}}=\frac{2\pi (4+\pi)}{4}>0

Therefore at r=\frac{10}{4+\pi}  , A is minimum.

Therefore the circumference of the circle is

=2 \pi \frac{10}{4+\pi}

=\frac{20\pi}{4+\pi}

4 0
3 years ago
Do jakiej szkoły po podstawówce można iść bez matematyki (mniej ważna​
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Answer:

Gimnazjum?

Również twoje pytanie jest trudne dla ludzi, którzy mówią po angielsku ...

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