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DIA [1.3K]
2 years ago
9

(y+3):(y+5)=k:1 show that y=(5k-3)/(1-k) hence y=(5k-2)/(1-k)

Mathematics
1 answer:
luda_lava [24]2 years ago
4 0

hi,

Express the ratios in fractional form, that is

= ( cross- multiply )

y + 3 = k(y + 5) ← distribute

y + 3 = ky + 5k ( subtract ky from both sides )

y - ky + 3 = 5k ( subtract 3 from both sides )

y - ky = 5k - 3 ← factor out y from each term on the left side

y(1 - k) = 5k - 3 ← divide both sides by (1 - k)

y = 5k- 3

____

1 - k

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-18 - 3 /4 v = 3<br> I need help with this who is smart enough to solve it
Blababa [14]

Answer:

v =-28

Step-by-step explanation:

-18 - 3 /4 v = 3

Add 18 to each side

-18+18 - 3 /4 v = 3+18

-3/4 v = 21

Multiply each side by -4/3 to isolate v

-4/3 *-3/4v = 21*-4/3

v =-28

8 0
3 years ago
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Find the dimensions of the open rectangular box of maximum volume that can be made from a sheet of cardboard 11 in. by 7 in. by
11111nata11111 [884]

Answer:

The dimension of the open rectangular box is 8.216\times 4.216\times 1.392.

The volume of the box is 8.217 cubic inches.

Step-by-step explanation:

Given : The open rectangular box of maximum volume that can be made from a sheet of cardboard 11 in. by 7 in. by cutting congruent squares from the corners and folding up the sides.

To find : The dimensions and the volume of the open rectangular box ?

Solution :

Let the height be 'x'.

The length of the box is '11-2x'.

The breadth of the box is '7-2x'.

The volume of the box is V=l\times b\times h

V=(11-2x)\times (7-2x)\times x

V=4x^3-36x^2+77x

Derivate w.r.t x,

V'(x)=4(3x^2)-2(36x)+77

V'(x)=12x^2-72x+77

The critical point when V'(x)=0

12x^2-72x+77=0

Solve by quadratic formula,

x=\frac{18+\sqrt{93}}{6},\frac{18-\sqrt{93}}{6}

x=4.607,1.392

Derivate again w.r.t x,

V''(x)=24x-72

Now, V''(4.607)=24(4.607)-72=38.568>0 (+ve)

V''(1.392)=24(1.392)-72=-38.592 (-ve)

So, there is maximum at x=1.392.

The length of the box is l=11-2x

l=11-2(1.392)=8.216

The breadth of the box is b=7-2x

b=7-2(1.392)=4.216

The height of the box is h=1.392.

The dimension of the open rectangular box is 8.216\times 4.216\times 1.392.

The volume of the box is V=l\times b\times h

V=8.216\times 4.216\times 1.392

V=48.217\ in.^3

The volume of the box is 8.217 cubic inches.

5 0
3 years ago
8x+8(c+8)-6 show work its for my homework i get it but im just making sure i did it right:)
ioda
8x+8c+164-6
=8x+8c+158
7 0
3 years ago
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From experience, it is known that on average 10% of welds performed by a particular welder are defective. if this welder is requ
bulgar [2K]
Binomial distribution can be used because the situation satisfies all the following conditions:1. Number of trials is known and remains constant (n)2. Each trial is Bernoulli (i.e. exactly two possible outcomes) (success/failure)3. Probability is known and remains constant throughout the trials (p)4. All trials are random and independent of the othersThe number of successes, x, is then given byP(x)=C(n,x)p^x(1-p)^{n-x}whereC(n,x)=\frac{n!}{x!(n-x)!}
Here we're given
p=0.10  [ success = defective ]
n=3

(a) x=0
P(x)=C(n,x)p^x(1-p)^{n-x}
=C(3,0)0.1^0(1-0.1)^{3-0}
=1(1)(0.729)
=0.729

(b) x=2
P(x)=C(n,x)p^x(1-p)^{n-x}
=C(3,2)0.1^2(1-0.1)^{3-2}
=3(0.01)(0.9)
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(c) x &ge; 2
P(x)=\sum_{x=2}^3C(n,x)p^x(1-p)^{n-x}
=P(2)+P(3)
=C(n,2)p^2(1-p)^{n-2}+C(n,3)p^3(1-p)^{n-3}
=C(3,2)0.1^2(1-0.1)^{3-2}+C(3,3)0.1^3(1-0.1)^{3-3}
=3(0.01)(0.9)+1(0.001)1
=0.027+0.001
=0.028


8 0
3 years ago
What is the value of x?<br><br><br> x+15=8<br><br> x=
melisa1 [442]
8-15= -7

x= -7

I think this is correct..
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3 years ago
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