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ikadub [295]
3 years ago
10

16 = d - 12 / 14 what is the answer?

Mathematics
2 answers:
Aleonysh [2.5K]3 years ago
3 0

Answer:

16=d-12/14

We simplify the equation to the form, which is simple to understand  

16=d-12/14

Simplifying:

16=d-0.857142857143

We move all terms containing d to the left and all other terms to the right.  

-1d=-0.857142857143-16

We simplify left and right side of the equation.  

-1d=-16.8571428571

We divide both sides of the equation by -01 to get d.  

d=16.8571428571

Step-by-step explanation:

PolarNik [594]3 years ago
3 0
The answer is d= 118/7.
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The area of the circle is 132.7 square centimeters. What is the diameter
VladimirAG [237]
Equating the formula  (d is the diameter)

\pi d^2/4 = 132.7
d^2 = 132.7*4/\pi
d = \sqrt{169.044586}
d = 13.0017147331
d= 13cm
7 0
4 years ago
PLEASE HELPPPPP
dsp73

Answer:

see explanation

Step-by-step explanation:

the equation of parabola in vertex form is

y = a(x - h)² + k

where (h, k ) are the coordinates of the vertex and a is a multiplier.

here (h, k ) = (3, 1 ) , then

y = a(x - 3)² + 1

to find a substitute any other point on the graph into the equation.

using (0, 7 )

7 = a(0 - 3)² + 1 ( subtract 1 from both sides )

6 = a(- 3)² = 9a ( divide both sides by 9 )

\frac{6}{9} = \frac{2}{3} = a

y = \frac{2}{3} (x - 3)² + 1 ← in vertex form

------------------------------------------------------

the equation of a parabola in factored form is

y = a(x - a)(x - b)

where a, b are the zeros and a is a multiplier

here zeros are - 1 and 3 , the factors are

(x - (- 1) ) and (x - 3), that is (x + 1) and (x - 3)

y = a(x + 1)(x - 3)

to find a substitute any other point that lies on the graph into the equation.

using (0, - 3 )

- 3 = a(0 + 1)(0 - 3) = a(1)(- 3) = - 3a ( divide both sides by - 3 )

1 = a

y = (x + 1)(x - 3) ← in factored form

3 0
2 years ago
AB, CD and EF are straight lines. AB is parallel to CD. Work out the size of angle y
mylen [45]

Answer: 28

Step-by-step explanation:

Simplifying

2x + 16 = 3x + -12

Reorder the terms:

16 + 2x = 3x + -12

Reorder the terms:

16 + 2x = -12 + 3x

Solving

16 + 2x = -12 + 3x

Solving for variable 'x'.

Move all terms containing x to the left, all other terms to the right.

Add '-3x' to each side of the equation.

16 + 2x + -3x = -12 + 3x + -3x

Combine like terms: 2x + -3x = -1x

16 + -1x = -12 + 3x + -3x

Combine like terms: 3x + -3x = 0

16 + -1x = -12 + 0

16 + -1x = -12

Add '-16' to each side of the equation.

16 + -16 + -1x = -12 + -16

Combine like terms: 16 + -16 = 0

0 + -1x = -12 + -16

-1x = -12 + -16

Combine like terms: -12 + -16 = -28

-1x = -28

Divide each side by '-1'.

x = 28

Simplifying

x = 28

5 0
4 years ago
Read 2 more answers
How the points are placed in graphing linear inequalities?
Yanka [14]

Answer:

Plot the line y = 2x - 3 and the portion above the line will satisfy the inequality.

5 0
3 years ago
Evaluate the surface integral S F · dS for the given vector field F and the oriented surface S.
sweet-ann [11.9K]

Answer:

2.794

Step-by-step explanation:

Recall that if G(x,y) is a parametrization of the surface S and F and G are smooth enough then  

\bf \displaystyle\iint_{S}FdS=\displaystyle\iint_{R}F(G(x,y))\cdot(\displaystyle\frac{\partial G}{\partial x}\times\displaystyle\frac{\partial G}{\partial y})dxdy

F can be written as

F(x,y,z) = (xy, yz, zx)

and S has a straightforward parametrization as

\bf G(x,y) = (x, y, 3-x^2-y^2)

with 0≤ x≤1 and  0≤ y≤1

So

\bf \displaystyle\frac{\partial G}{\partial x}= (1,0,-2x)\\\\\displaystyle\frac{\partial G}{\partial y}= (0,1,-2y)\\\\\displaystyle\frac{\partial G}{\partial x}\times\displaystyle\frac{\partial G}{\partial y}=(2x,2y,1)

we also have

\bf F(G(x,y))=F(x, y, 3-x^2-y^2)=(xy,y(3-x^2-y^2),x(3-x^2-y^2))=\\\\=(xy,3y-x^2y-y^3,3x-x^3-xy^2)

and so

\bf F(G(x,y))\cdot(\displaystyle\frac{\partial G}{\partial x}\times\displaystyle\frac{\partial G}{\partial y})=(xy,3y-x^2y-y^3,3x-x^3-xy^2)\cdot(2x,2y,1)=\\\\=2x^2y+6y^2-2x^2y^2-2y^4+3x-x^3-xy^2

we just have then to compute a double integral of a polynomial on the unit square 0≤ x≤1 and  0≤ y≤1

\bf \displaystyle\int_{0}^{1}\displaystyle\int_{0}^{1}(2x^2y+6y^2-2x^2y^2-2y^4+3x-x^3-xy^2)dxdy=\\\\=2\displaystyle\int_{0}^{1}x^2dx\displaystyle\int_{0}^{1}ydy+6\displaystyle\int_{0}^{1}dx\displaystyle\int_{0}^{1}y^2dy-2\displaystyle\int_{0}^{1}x^2dx\displaystyle\int_{0}^{1}y^2dy-2\displaystyle\int_{0}^{1}dx\displaystyle\int_{0}^{1}y^4dy+\\\\+3\displaystyle\int_{0}^{1}xdx\displaystyle\int_{0}^{1}dy-\displaystyle\int_{0}^{1}x^3dx\displaystyle\int_{0}^{1}dy-\displaystyle\int_{0}^{1}xdx\displaystyle\int_{0}^{1}y^2dy

=1/3+2-2/9-2/5+3/2-1/4-1/6 = 2.794

5 0
3 years ago
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