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harkovskaia [24]

3 years ago

12

Plz Help!!

1 answer:

DIA [1.3K]3 years ago

3 0

I think the answer is 3/8 pounds. You multiply the 3/4 pounds by 1/2 because the recipie has been cut in half. Hope this helps!

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Simplify 5y^2 + 5y + 5y + 5x^2.

Ierofanga [76]

Answer:

5y^2 + 10y + 5x^2

Step-by-step explanation:

there's only two like terms that can be combined.

4 0

3 years ago

Hiii please help i’ll give brainliest if you give a correct answer tysm!

ANTONII [103]

Answer:

7

Step-by-step explanation:

because 105/15=7

84/12=7

28/4=7

so 1x7=7

8 0

2 years ago

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Find the value of k so that (x + 3) a factor of H(x) = 3x ^ 3 - 2x ^ 2 + kx - 3

gavmur [86]

Step-by-step explanation:

By Factor Theorem, (x + 3) is a factor of H(x) if and only if H(-3) = 0.

3(-3)³ - 2(-3)² + (-3)k - 3 = 0

-81 - 18 - 3k - 3 = 0

3k = -102

k = -34.

4 0

2 years ago

Mathematical logic for this game to make the best

slega [8]

Answer:

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6 0

2 years ago

A large storage tank, open to the atmosphere at the top and filled with water, develops a small hole in its side at a point 10.4

vlabodo [156]

Answer:

Therefore the diameter of the hole is $1.94 \times 10^{-3}$ m.

Step-by-step explanation:

Bernoulli's equation,

$P_1+\frac12 \rho v^2_1+\rho g h_1= P_2+\frac12 \rho v^2_2+\rho g h_2$

P₁ = P₂= atmospheric presser

$\rho$ = density

$\frac12 \rho v^2_1+\rho g h_1= \frac12 \rho v^2_2+\rho g h_2$ [since P₁ = P₂]

$\Rightarrow\rho (\frac12 v^2_1+ g h_1)= \rho(\frac12 v^2_2+ g h_2)$

$\Rightarrow\frac12 v^2_1+ g h_1= \frac12 v^2_2+ g h_2$

$\Rightarrow\frac12 v^2_2-\frac12 v^2_1=g h_1- g h_2$

$\Rightarrow v^2_2- v^2_1=2g h$ [ $h_1-h_2=h$ ]

Here $v_1\approx 0$

$\Rightarrow v^2_2=2g h$

$\therefore v_2=\sqrt {2gh$

Here g= 9.8 m/s² , h = 10.4 m

The velocity of water that leaves from the hole $v_2$ = $\sqrt {2\times 9.8\times 10.4}$ m/s

=14.28 m/s.

Given, the rate of flow from the leak is $2.53\times 10^{-3}$ $m^3/min$

$=\frac{2.53\times 10^{-3}}{60}$ $m^3/s$

Let the diameter of the hole be d.

Then the cross section area of the hole is $=\pi (\frac d2)^2$

We know that,

The rate of flow = Cross section area × speed

$\Rightarrow \frac{2.53\times 10^{-3}}{60} =\pi (\frac d2)^2\times 14.28$

$\Rightarrow (\frac d2)^2=\frac{2.53\times 10^{-3}}{60\times 14.28\times \pi}$

$\Rightarrow d= 1.94 \times 10^{-3}$

Therefore the diameter of the hole is $1.94 \times 10^{-3}$ m.

4 0

3 years ago