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1 year ago

Alice is making an octahedron out of origami. The last step requires her to blow into the octahedron to fill it up.

Mathematics

1 answer:

sergejj [24]1 year ago

5 0

C because a negitive would equal positive

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An hourglass composed of two identical cones is 18 cm tall.the radius of each cone is 4cm if you want to fill the bottom half of

Kitty [74]

Check the picture below.

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1 year ago

Match the following items.

Margarita [4]

Answer:

$1. m\angle ECB=50\\2. m\textrm{ arc BC}=100\\3. m\textrm{ arc CD}=80\\4. m\angle DCF=40$

Step-by-step explanation:

Given:

$\angle DBC=40$ °

From the triangle, using the theorem that center angle by an arc is twice the angle it subtend at the circumference.

$m\textrm{ arc CD}=2\times \angle DBC\\m\textrm{ arc CD}=2\times 40=80$

Also, the diameter of the circle is BD. As per the theorem that says that angle subtended by the diameter at the circumference is always 90°,

$m\angle BCD=90$

From the Δ BCD, which is a right angled triangle,

$m\angle DBC+m\angle BDC=90\textrm{ (right angled triangle)}\\40+m\angle BDC=90\\m\angle BDC=90-40=50$

Now, using the theorem that angle between the tangent and a chord is equal to the angle subtended by the same chord at the circumference.

Here, chords CD and BC subtend angles 40 and 50 at the circumference as shown in the diagram by angles $m\angle DBC\textrm{ and }m\angle BDC$ and EF is a tangent to the circle at point C.

Therefore, $m\angle DCF=m\angle DBC=40\\m\angle ECB=m\angle BDC=50$

Again, using the same theorem as above,

$m\angle DCF=50\\\therefore m\textrm{ arc BC}=2\times m\angle DCF=2\times 50=100$

Hence, all the angles are as follows:

$1. m\angle ECB=50\\2. m\textrm{ arc BC}=100\\3. m\textrm{ arc CD}=80\\4. m\angle DCF=40$

6 0

1 year ago

Read 2 more answers

If a set xy-axes has 20 grid units in each direction and the greatest value of the data to be placed on the horizontal axis is 1

Alborosie

Use 7 scale point.
132/20 = 6.6
to its better to use more than 6.6 scale. to make it easier use 7, so to avoid decimals

4 0

1 year ago

∆ABC and ∆PQR are similar. ∆ABC is dilated by a scale factor of 1.25 and rotated 45° counterclockwise about point B to form ∆PQR

Andrej [43]

This is the concept of transformation of figures, given that ΔABC is similar to ΔPQR, the sides of ABC are 5 units, 4.2 units and 4 units. Since the two triangles are similar and PQR is the image of ABC under the dilation 1.25, the sides of PQR will be:
(5*1.25),(4.2*1.25),(4*1.25)
this will give us:
6.25, 5.25, 5
The length of the sides of PQR are 6.25 units, 5.25 units and 5 units

7 0

1 year ago

A consulting firm has received 2 Super Bowl playoff tickets from one of its clients. To be fair, the firm is randomly selecting

Soloha48 [4]

Answer:

Therefore the only statement that is not true is b.)

Step-by-step explanation:

There employees are 6 secretaries, 5 consultants and 4 partners in the firm.

a.) The probability that a secretary wins in the first draw

= $\frac{number \hspace{0.1cm} of \hspace{0.1cm} secreataries}{total \hspace{0.1cm} number \hspace{0.1cm} of \hspace{0.1cm} employees} = \frac{6}{15}$

b.) The probability that a secretary wins a ticket on second draw. It has been given that a ticket was won on the first draw by a consultant.

p(secretary wins on second draw | consultant wins on first draw)

= $\frac{p((consultant \hspace{0.1cm} wins \hspace{0.1cm} on \hspace{0.1cm} first \hspace{0.1cm}draw)\cap( secretary\hspace{0.1cm} wins\hspace{0.1cm} on second \hspace{0.1cm}draw))}{p(consultant \hspace{0.1cm} wins \hspace{0.1cm} on \hspace{0.1cm} first \hspace{0.1cm}draw)}$

= $\frac{\frac{5}{15} \times \frac{6}{14}}{\frac{5}{15} } = \frac{6}{14}$ .

The probability that a ticket was won on the first draw by a consultant a secretary wins a ticket on second draw = $\frac{6}{15}$ is not true.

The probability that a secretary wins on the second draw = $\frac{number \hspace{0.1cm} of \hspace{0.1cm} secretaries \hspace{0.1cm} remaining } { number \hspace{0.1cm} of \hspace{0.1cm} employees \hspace{0.1cm} remaining} = \frac{6 - 1}{15 - 1} = \frac{5}{14}$

c.) The probability that a consultant wins on the first draw =

$\frac{number \hspace{0.1cm} of \hspace{0.1cm} consultants \hspace{0.1cm} } { number \hspace{0.1cm} of \hspace{0.1cm} employees \hspace{0.1cm} } = \frac{5 }{15} = \frac{1}{3}$

d.) The probability of two secretaries winning both tickets

= (probability of a secretary winning in the first draw) × (The probability that a secretary wins on the second draw)

= $\frac{6}{15} \times \frac{5}{14} = \frac{1}{7}$

Therefore the only statement that is not true is b.)

5 0

1 year ago