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anastassius [24]
3 years ago
10

How do I make 7/15 into a decimal.

Mathematics
1 answer:
nordsb [41]3 years ago
5 0

Well, this is really simple math.

Divide 7 by 15.

7/15 = 0.46666666666

To make it simple, round to the nearest hundredths.

7/15 = 0.47

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What is the equation of this line in slope-intercept form?<br> (-1,5) (1,-1)
enot [183]

Answer:

y=-3x+2

Step-by-step explanation:

m=(y2-y1)/(x2-x1)

m=(-1-5)/(1-(-1))

m=-6/(1+1)

m=-6/2

m=-3

y-y1=m(x-x1)

y-5=-3(x-(-1))

y-5=-3(x+1)

y=-3(x+1)+5

y=-3x-3+5

y=-3x+2

4 0
2 years ago
There are 5.32 of face paint in each paint pot she used 13 of them while painting forces how many milliliter of face paint
Goshia [24]

Answer:

69.16

Step-by-step explanation:

5 0
2 years ago
<img src="https://tex.z-dn.net/?f=5%20-%203x%20%3D%2011" id="TexFormula1" title="5 - 3x = 11" alt="5 - 3x = 11" align="absmiddle
const2013 [10]

Hi there! :)

<u>Answer:</u>

x = -2


Step-by-step explanation:

<u>5</u> - 3x = 11

Subtract "5" from each side of the equation → 11 - 5 = 6

<u>-3</u>x = 6

Divide each side of the equation by "-3" → 6 ÷ -3 = -2

<u>x = -2</u>


There you go! I really hope this helped, if there's anything just let me know! :)

5 0
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}ρV_{1}^{2} + ρgh = \frac{1}{2}ρV_{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}

6 0
3 years ago
Find the distance between the points (4,3) and (2,-1). Express your answer in simplest radical form.
Kamila [148]

Answer : B

the distance between the points (4,3) and (2,-1)

We apply distance formula

d^2= (x_2-x1)^2 + (y_2-y_1)^2

(x1,y1) is (4,3)

(x2,y2) is (2,1)

d^2= (2-4)^2 + (-1-3)^2

d^2= (-2)^2 + (-4)^2

d^2 = 4+16 = 20

Take square root on both sides

So  d=2\sqrt{5}

the distance between the points (4,3) and (2,-1) is 2sqrt(5)

3 0
3 years ago
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