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Alla [95]
4 years ago
12

Points O, A, B lie on the same line. OA = 12 cm, OB = 9 cm. Find the distance between the midpoints of segments OA and OB if: Po

int O doest lie on the segment AB .
Mathematics
1 answer:
d1i1m1o1n [39]4 years ago
6 0
Let E be the midpoint of OA  and F be the midpoint of OB. 

If point O is placed between A and B, then points E and F are placed <span>on different sides from O. The segment OE has length 12÷2=6 cm and segment OF has length 9÷2=4.5 cm. Than EF=6+4.5=10.5 cm.
</span>
If point O is placed outside the segment AB, then point B is closer to point O than point A, because OA>OB. Since OE=6 cm and OF=4.5 cm, EF=6-4.5=1.5 cm.





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yz = 7ln(x + z), (0, 0, 1) (a) the tangent plane Correct: Your answer is correct. (b) parametric equations of the normal line to
Alika [10]

Answer:

The equation of tangent plane is    7x - y + 7z - 7 = 0

Parametric equation of normal line

x = 7 t         , y=-t                 , z=1+7 t

Step-by-step explanation:

Equation of tangent

fₓ (x₀ , y₀ , z₀) (x-x₀) + fy (x₀ , y₀ , z₀) (y-y₀) +fz(x₀ , y₀ , z₀)(z-z₀)=0                (1)

From taking derivation we get

fₓ (x₀ , y₀ , z₀) = 7

fy (x₀ , y₀ , z₀)= -1

fz(x₀ , y₀ , z₀) = 7

putting these value in equation 1

 (7) (x-0) + (-1)(y-0) + 7(z-1)=0

  7x - y + 7z - 7 = 0

The equation of tangent plane is    7x - y + 7z - 7 = 0

b) Parametric equation

x=x +f ₓ (P)t , y = y₀ +f y (P) t , z=z₀ +f z (P) t

x=0 +7 t      , y =0+(-1) t        , z=1+7 t

x=7 t            ,y=-t                  , z= 1+7 t

Parametric equation of normal line

x = 7 t         , y=-t                 , z=1+7 t

7 0
4 years ago
What is the percent of increase from 10 to 127
satela [25.4K]

Answer:

27

Step-by-step explanation:

127 decrease to 10 is 27

4 0
3 years ago
What's the GCF of 20, 28, 24
zimovet [89]
The answer would be 4
8 0
3 years ago
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Rewrite each expression as a polynomial in standard form: x^2-5x+6 ÷ x-3 + x^3-1 ÷ x-1
iragen [17]

Answer:

So in standard form equation will be x^2+2x-1

Step-by-step explanation:

We have given expression \frac{x^2-5x+6}{x-3}+\frac{x^3-1}{x-1}

Let first we solve first part of the expression

So \frac{x^2-5x+6}{x-3}=\frac{x^2-3x-2x+6}{x-3}=\frac{(x-3)(x-2)}{x-3}=x-2

Now second part \frac{x^3-1}{x-1}

We know the algebraic identity a^3-b^3=(a-b)(a^2+ab+b^2)

So by using this identity

\frac{x^3-1}{x-1}=\frac{(x-1)(x^2+x+1)}{x-1}=x^2+x+1

Now adding first and second part

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4 0
3 years ago
What is the value of the expression if z = 27? StartFraction 7 over 9 EndFraction z minus 18 + one-third z
Mandarinka [93]

Answer:

The value of the given expression when z=27 is 16

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Step-by-step explanation:

Given expression is \frac{7}{9}(z-18)+\frac{1}{3}z

To find the value of the expression when z=27 :

First simplify the given expression

\frac{7}{9}(z-18)+\frac{1}{3}z

=\frac{7(z-18)}{9}+\frac{1}{3}z

=\frac{7z-126}{9}+\frac{1}{3}z

=\frac{7z-126+3z}{9}

=\frac{7z-126+3z}{9}

=\frac{10z-126}{9}

Put z=27 in the above equation we get

=\frac{10(27)-126}{9}

=\frac{270-126}{9}

=\frac{144}{9}

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=16

\frac{7}{9}(z-18)+\frac{1}{3}z=16

Therefore the value of the given expression when z=27 is 16

Therefore \frac{7}{9}(z-18)+\frac{1}{3}z=16 when z=27

4 0
4 years ago
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