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3 years ago

∠A=6x−2

Mathematics

2 answers:

Alex3 years ago

7 0

Like my comment like my comment

Lemur [1.5K]3 years ago

4 0

Xx el sns ska so 55 . 66

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Determine the quotient: (22.8 × 10–3) ÷ (5.7 × 10–6)

lana66690 [7]

Answer:

0.399

Step-by-step explanation:

The key to doing this problem properly lies in knowing and following order of operations rules.

Here we must perform mult. and div. before addition and subtr., but even before that we must do all work enclosed in parentheses first.

(22.8 × 10–3) is evaluated by doing the mult. first, and then subtracting 3:

(228-3) = 225

and

(5.7 × 10–6) is evaluated by doing the mult. first, then subtracting 6:

(570-6) = 564.

Finally, we divide 225 by 564, obtaining 0.399 (after rounding off to three decimal places).

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3 years ago

HELP PLSS WILL GIVE BRAINLIEST<br> Solve N = 3rt^4 - 5rz <br> (solve for z)

Whitepunk [10]

Answer:

Step-by-step explanation:

N = 3rt^4 - 5rz

now factor r out

N = r(3t^4-5z)

divide by r on both sides

N/r = (3t^4 - 5z)

subtract (3t^4)

N/r - 3t^4 = -5z

now divide by -5 to isolate the variable (z)

$\frac{(n/r)-3t^4}{-5}$ = z

7 0

3 years ago

Read 2 more answers

What are the solution and the answer?

dedylja [7]

Answer:

I think its 40 if i did it right Ill show you how i did it

Step-by-step explanation:

First I added 12+12 and the is 24.

The I added 8+8 and that was 16.

Lastly I added 16+24= 40

7 0

3 years ago

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Si sumamos el cuadrado de un número más el cuadruple del siguiente resulta 225 ¿De que número se trata?

Talja [164]

Answer:

What does it say in english

Step-by-step explanation:

ill answer it in comments if you tell me what it says in english

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3 years ago

The circular stream of water from a faucet is observed to taper from a diameter of 20 mm to 10 mm in a distance of 50 cm. Determ

steposvetlana [31]

Solution:

From Bernoulli equation

$\frac{1}{2}$ ρ $V_{1}^{2}$ + ρgh = $\frac{1}{2}$ ρ $V_{2}^{2}$ , where ρ is density of water, h – height difference and $V_{1}$ and $V_{2}$ are the velocities in upper and lower cross sections correspondingly. We take into account that the pressure in both cross-sections is the same and equal to the atmospheric one.

From the continuity and assuming water incompressible:

$\pi r_{1}^{2}$ $V_{1}$ = $\pi r_{2}^{2}$ $V_{2}$ , where $A_{1}$ and $A_{2}$ are the corresponding cross-sections. As the lower diameter is twice as low as the upper one, we can conclude that $V_{1} = \frac{V_{2}}{4}$

Inserting it back into Bernoulli equations produces:

$V_{1} = \sqrt{\frac{2}{15}gh} = 0.26m/s$ and the flow rate is

$\pi r_{1}^{2}V_{1} = \pi r_{2}^{2}V_{2} = 8.10^{-5} \frac{m^{3} }{s}$

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3 years ago