All categories

2 years ago

D) What is the smallest odd number that he can make? Overview Minute Version

Mathematics

2 answers:

Luda [366]2 years ago

8 0

Answer:

1 is the smallest odd number that he can make

never [62]2 years ago

6 0

Its 0, smallest number

You might be interested in

Find angle <QPS in the diagram

marin [14]

Answer:

40°

Step-by-step explanation:

Because triangle QSR is isosceles ∠SQR=∠SRQ=35°. The sum of the angles in a triangle is 180°, so ∠QSR=180°-35°-35°=110°. The measure of a straight line is 180°, so ∠PSQ=180°-110°=70°. Because triangle PSQ is also isosceles ∠PSQ=∠PQS=70°. Then, ∠QPS=180°-70°-70°=40°.

4 0

2 years ago

Look at the street map. What street intersects with Polk Street?

Alenkinab [10]

Answer:

Geary street :D

they intersect here

click on the photo. it's not a link it's a screenshot of where they intersect

4 0

2 years ago

Brainlist<br> Please help me

creativ13 [48]

Answer:

Yes she made an error. Explanation is below.

Step-by-step explanation:

Let the sample space for a six sided number cube be S

$n(S) = \{1,2,3,4,5,6\}\\n(S) = 6$

Let the event of a number getting greater than two be B

$n(B) = \{3,4,5,6\}\\n(B) = 4$

To find:

the probability of complement of B that is

p(~B) = ?

Solution"

First we need to solve probability for number getting greater than two

$p(B) = \frac{n(B)}{n(S)}$

Now substituting the values we get

$p(B) = \frac{4}{6}$

$p(B) = \frac{2}{3}$

but we need complement of B so

$p(\textrm{complement of B)} = 1 - p(B)\\p(\~B) = 1 - p(B)\\p(\~B) = 1 - \frac{2}{3}\\ =\frac{1}{3} \\$

P(complement of rolling a number greater than two) = $\frac{1}{3}$

3 0

3 years ago

A tire on a toy car has a diameter of 8 cm if the toy car travels 12,560 cm approximately how many times does the tire rotate? U

Marina86 [1]

Answer:

The number of rotations completed by tire = 500.

Step-by-step explanation:

The diameter of the tire of toy car = 8 cm

Now, DIAMETER = 2 x RADIUS

So, radius of the tire = D/2 = 8/2 = 4 cm

Now, Circumference of the Circle = 2πr

= 2 x 3.14 x 4 = 25.12 cm

Now, $\textrm{Total number of rotations} = \frac{\textrm{Total Area covered}}{\textrm{Area covered in 1 rotation}}$

= $\frac{12,560}{25.12} = 500$

The number of rotations completed by circle = 500.

6 0

3 years ago

Prove that the sum of the squares of the lengths of the medians of a triangle is three fourths the sum of the squares of the len

aleksandr82 [10.1K]

Answer:

3[|AB|² + |BC|² + |AC|²] = 4[|AD|² + |BE|² + |CF|²]

Step-by-step explanation:

I've attached an image showing a triangle divided into it's medians.

Now, from the attached image, we see that AB, BC & CD are the lengths of sides of triangle while AD, BE & CF are lengths of the 3 medians of the triangle.

Now, to prove our question, we will use appolonius theorem which states that "the sum of the squares of any of the two sides of a triangle is equal to twice the square on half the third side including twice the square on the median that bisects the third side.

Applying this theorem to the image attached, we have the following;

|AB|² + |AC|² = 2[|AD|² + |BC/2|²]

|AB|² + |BC|² = 2[|BE|² + |AC/2|²]

|AC|² + |BC|² = 2[|CF|² + |AB/2|²]

Adding the 3 equations above gives us;

2|AB|² + 2|BC|² + 2|AC|² = 2|AD|² + |BC|²/2 + 2|BE|² + |AC|²/2 + 2|CF|² + |AB|²/2

Collecting like terms;

(2|AB|² - |AB|²/2) + (2|BC|² - |BC|²/2) + (2|AC|² - |AC|²/2) = 2|AD|² + 2|BE|² + 2|CF|²

Thus gives;

(3/2)[|AB|² + |BC|² + |AC|²] = 2[|AD|² + |BE|² + |CF|²]

Multiply both sides by 2 to give;

3[|AB|² + |BC|² + |AC|²] = 4[|AD|² + |BE|² + |CF|²]

6 0

3 years ago