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MAXImum [283]
4 years ago
5

Given: AD≅AO Find: m∠OAD, m∠DBA.

Mathematics
2 answers:
disa [49]4 years ago
8 0

OA=OD=AD=R\Rightarrow \bigtriangleup OAD\: is\: an\: equilateral\: triangle  \\ \Rightarrow \widehat{AOD}=\widehat{OAD}=\widehat{ODA}=60° \\ \widehat{ABD}= \frac{\widehat{AOD}}{2}  =  \frac{60°}{2}  = 30°

MatroZZZ [7]4 years ago
5 0

Answer:

Step-by-step explanation:

Given: AD≅AO

To find: m∠OAD and m∠DBA.

Solution: It is given that AD≅AO, and also OA≅OD (Radius of circle), therefore ΔAOD is an equilateral triangle.

Hence, m∠OAD=m∠ODA=m∠AOD=60°.

Now, we know that the intercepted angle is half of the central angle, thus

m{\angle}DBA=\frac{m{\angle}AOD}{2}

⇒m{\angle}DBA=\frac{60^{\circ}}{2}

⇒m{\angle}DBA=30^{\circ}

Hence, the measures of {\angle}OAD and {\angle}DBA are 60^{\circ} and  30^{\circ} respectively.

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Julia is making pasta necklaces and wreaths. She uses n noodles in a necklace and 3/5 that number of noodles in wreaths. Which e
Fofino [41]

Answer:

the expression that shows the number of noodles Julia uses to make 9 necklaces and 20 wreaths

  = (9 \times n ) + (20 \times (3/5)n) = 9n + 12n = 21n

Step-by-step explanation:

i) there are n noodles in a necklace

ii) there are 9 necklaces

iii) therefore the number of noodles used to make the nine necklaces = 9 \times n = 9n

iv) there are 3/5 noodles in a wreath

v) there are 20 wreaths

vi) therefore the number of noodles used to make the nine necklaces

     = 20 \times (3/5)n = 12n

vii) the expression that shows the number of noodles Julia uses to make 9 necklaces and 20 wreaths

  = (9 \times n ) + (20 \times (3/5)n) = 9n + 12n = 21n

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Evaluate the expression when y = 6 and x = 9 <br> x+7y
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Step-by-step explanation:

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3 years ago
(a) The plane y + z = 13 intersects the cylinder x2 + y2 = 25 in an ellipse. Find parametric equations for the tangent line to t
klemol [59]

Answer:

Step-by-step explanation:

We have a curve (an ellipse) written as the system of equations

\begin{cases} y+z &= 13\\ x^2+y^2 &= 25\end{cases}.

And we want to calculate the tangent at the point (3,4,9).

The idea in this problem is to consider two variables as functions of the third. Usually we consider y and z as functions of x. Recall that a curve in the space can be written in parametric form in terms of only one variable. In this case we are considering the ‘‘natural’’ parametrization (x, y(x), z(x)).

Recall that the parametric equation of a line has the form

r(t)=\begin{cases} x(t) &= x_0 + v_1t \\ y(t) &= y_0 +v_2t\\ z(t) &= z_0 +v_3t \end{cases},

where (x_0,y_0,z_0) is a point on the line (in this particular case is (3,4,9)) and (v_1,v_2,v_3) is the direction vector of the line. In this case, the direction vector of the line is the tangent vector of the ellipse at the point (3,4,9).

Now, if we have the parametric equation of a curve (x, y(x), z(x)) its tangent line will have direction vector (1, y'(x), z'(x)). So, as we need to calculate the equation of the tangent line at the point (3,4,9) = (3, y(3), z(3)), we must obtain the tangent vector (1, y'(3), z'(3)). This part can be done taking implicit derivatives in the systems that defines the ellipse.

So, let us write the system as

\begin{cases} y(x)+z(x) &= 13\\ x^2+y^2(x) &= 25\end{cases}.

Then, taking implicit derivatives:

\begin{cases} y'(x)+z'(x) &= 0 \\ 2x+2y(x)y'(x) &= 0\end{cases}.

Now we substitute the values x=3 and y(3)=4, and we get the system of linear equations

\begin{cases} y'(3)+z'(3) &= 0 \\ 2\cdot 3+2\cdot 4y'(x) &= 0\end{cases},

where the unknowns are y'(3) and z'(3).

The system is

\begin{cases} y'(3)+z'(3) &= 0 \\ 6+8y'(x) &= 0\end{cases},

and its solutions are

y'(3) = -\frac{3}{4} and z'(3) = \frac{3}{4}.

Then, the direction vector of the tangent is

(1, -\frac{3}{4}, -\frac{3}{4}).

Finally, the tangent line has parametric equation

r(t)=\begin{cases} x(t) &= 3 + t \\ y(t) &= 4 -\frac{3}{4}t\\ z(t) &= 9 +\frac{3}{4}t \end{cases}

where t\in\mathbb{R}.

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4 years ago
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