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4 years ago

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(50 points) Could someone help me with this? Part A: Solve –mk – 90 > 85 for m. Part B: Solve 3c – 5f = 55 for f. Show your w

ork for both but please don't make it too complicated.

2 answers:

PilotLPTM [1.2K]4 years ago

6 0

A:

-mk - 90 > 85

-mk - 90 + 90 > 85 + 90

-mk > 175

-mk / k > 175 / k

-m > 175/k

-m * -1 < 175/k * -1

m < -175/k

Part B:

3c - 5f = 55

3c - 5f - 3c = 55 - 3c

-5f = 55 - 3c

-5f/5 = (55 - 3c) / 5

-f = 55/5 - 3c/5

labwork [276]4 years ago

3 0

Answer:

90>85

Step-by-step explanation:

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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

4 years ago

Natasha_Volkova [10]

ITS NUMBER c!!!hope it helps I did the test

7 0

2 years ago

Mrs. Duggin walked every day for seven days. If she walked 9/10 of a mile each day for seven days, how many total miles did she

Molodets [167]

Answer:

Mrs. Duggin walked a total of $\frac{63}{10} \ miles.$ .

Step-by-step explanation:

Given:

Number of miles walked on each day = $\frac{9}{10} \ mile$

Number of days she walked = 7

We need to find the Total miles she walked using repeated addition method.

Solution:

Now we know that Repeated Addition means adding each group together and it is also known as Multiplication.

So we can Find Total miles she walked by adding Number of miles walked on each day with number of day times.

OR

we can Find Total miles she walked by Multiplying Number of miles walked on each day with number of days.

framing in equation form we get;

Total miles she walked = $\frac{9}{10}+\frac{9}{10}+\frac{9}{10}+\frac{9}{10}+\frac{9}{10}+\frac{9}{10}+\frac{9}{10}= \frac{9+9+9+9+9+9+9}{10} = \frac{63}{10}\ miles$

OR

Total miles she walked = $\frac{9}{10}\times7= \frac{63}{10} \ miles.$

Hence Mrs. Duggin walked a total of $\frac{63}{10} \ miles.$ .

4 0

4 years ago

Is 35% more than 2/5

eduard

If you convert 2/5 into a decimal which is 0.4 and multiply it by 100
you'll get 40 which means 2/5 is 40%
so 35% is less than 2/5

6 0

3 years ago

Read 2 more answers

Can someone check my answer?

ivanzaharov [21]

The measure of EC is 1 ft.

Solution:

Given data:

AD = 8 ft, DB = 2 ft, AE = 4 ft

To find the measure of EC:

Using triangle proportionality theorem,

<em>If a line is parallel to one side of a triangle intersects the other two sides, then it divides the two side proportionally.</em>

$$\frac{AD}{DB} =\frac{AE}{EC}$

$$\frac{8}{2} =\frac{4}{EC}$

Do cross multiplication.

$8EC=2\times 4$

8 EC = 8

Divide by 8 on both side of the equation, we get

EC = 1

EC = 1 ft

Hence the measure of EC is 1 ft.

7 0

3 years ago