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3 years ago

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Javier teaches yoga classes at his community center. In a 60-minute class last night, he taught 30 poses. Today, he has a 30-min

ute class.
If he teaches at the same rate, how many poses will Javier teach in today's class?

1 answer:

marshall27 [118]3 years ago

6 0

Answer:

15 Poses

Step-by-step explanation:

SInce he can teach 30 poses in 60 minutes and if the class time is cut by half then the poses are cut by half. Look below.

Minutes : 60/2 = 30 minutes

Poses : 30/2 = 15 poses

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Help me please. <br>Help me please. <br>Help me please.

Gnesinka [82]

A)
To be similar triangles have to have equal angles
triangle ZDB'
1)angle Z=90 degrees
triangle B'CQ
1) angle C 90 degrees

angle A'B'Q=90
DB'Z+A'B'Q+CB'Q=180, straight angle
DB'Z+90+CB'Q=180
DB'Z+CB'Q=90

triangle ZDB'
DZB'+DB'Z=180-90=90

DB'Z+CB'Q=90
DZB'+DB'Z=90
DB'Z+CB'Q=DZB'+DB'Z
2)CB'Q=DZB' (these angles from two triangles ZDB' and B'CQ )
3)so,angles DB'Z and B'QC are going to be equal because of sum of three angles in triangles =180 degrees and 2 angles already equal.
so this triangles are similar by tree angles
b)
B'C:B'D=3:4
B'D:DZ=3:2
CQ-?
DC=AB=21
DC=B'C+B'D (3+4= 7 parts)
21/7=3
B'C=3*3=9
B'D=3*4=12
B'D:DZ=3:2
12:DZ=3:2
DZ=12*2/3=8
B'D:DZ=CQ:B'C
3:2=CQ:9
CQ=3*9/2=27/2

c)
BC=BQ+QC=B'Q+QC
BQ' can be found by pythagorean theorem

3 0

3 years ago

52 cards contains 4 aces, you randomly pick a card. X= your net of winnings

nirvana33 [79]

Chance of topdecking an ace is 1/13

8 0

3 years ago

Determine whether LINE AB and LINE CD are parallel, perpendicular, or neither. A(−1, −4) , B(2, 11) , C(1, 1) , D(4, 10) ( write

Luba_88 [7]

Step-by-step explanation:

Hey there!

The points of line AB are; (-1,-4) and (2,11).

Note:

Use double point formula and simplify it to get two eqaution.
Use condition of parallel lines, perpendicular lines to know whether the lines are parallel or perpendicular or nothing.

~ Use double point formula.

$(y - y1) = \frac{y2 - y1}{x2 - x1} (x - x1)$

~ Keep all values.

$(y + 4) = \frac{11 + 4}{2 + 1} (x + 1)$

~ Simplify it.

$y + 4 = \frac{15}{3} (x + 1)$

$y + 4 = 5x + 5$

$5x - y + 1 = 0$

Therefore this is the equation of line AB.

Now, Finding the equation of line CD.

Given;

The points of line CD are; (1,1) and (4,10).

~ Using formula.

$(y - y1) = \frac{y2 - y1}{x2 - x1}(x - x1)$

~ Keep all values.

$(y - 1) = \frac{10 - 1}{4 - 1} (x - 1)$

~ Simplify it.

$y - 1 = 3 x - 3$

$3x - y - 2 = 0$

Therefore, 3x - y- 2 = 0 is the eqaution of line CD.

Use condition of parallel lines.

m1= m2

Slope of equation (i)

$m1 = \frac{ - coeff. \: of \: x}{coeff \: of \: y}$

$m1 = \frac{ - 5}{ - 1}$

Therefore, m1 = 5

Slope of second equation.

$m2 = \frac{ - coeff \: .of \: x}{coeff \: .of \: y}$

$m2 = \frac{ - 3}{ - 1}$

Therefore, m2 = 3.

Now, m1≠m2.

So, the lies are not parallel.

Check for perpendicular.

m1*m2= -1

3*5≠-1.

Therefore, they aren't perpendicular too.

So, they are neither.

<em><u>Hope </u></em><em><u>it</u></em><em><u> helps</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em>

6 0

3 years ago

The outside temperature can be estimated based on how fast crickets chirp. At 104 chirps per minute, the temperature is 63 degre

Rasek [7]

Answer:

For 124 chirps per minute the temperature is 68 ºF.

For 68 chirps per minute the temperature is 54 ºF.

Step-by-step explanation:

Linear functions are those whose graph is a straight line. A linear function has the following form

$f(x)=b+mx$

b is the constant term or the y intercept. It is the value of the dependent variable when x = 0.

m is the coefficient of the independent variable. It is also known as the slope and gives the rate of change of the dependent variable.

We know that

At 104 chirps per minute, the temperature is 63 ºF.

At 176 chirps per minute, the temperature is 81 ºF.

This information can be converted to Cartesian coordinates (x, y). Where x = the number of chirps per minute and y = the temperature in ºF.

To find a linear function that let us find the outside temperature from how fast crickets chirp we must:

Find the slope:

$m=\frac{y_2-y_1}{x_2-x_1}=\frac{81-63}{176-104}=\frac{1}{4}$

Find the equation:

$81=\frac{1}{4}\cdot 104+b$

Solving for b

$b=81-\frac{1}{4} (176)=37$

Therefore, the linear function is

$y=\frac{1}{4} \cdot x+37$

Now, using this linear function we can know the temperature when we know the chirps per minute:

For 124 chirps per minute the temperature is:

$y=\frac{1}{4} \cdot (124)+37=68$

For 68 chirps per minute the temperature is:

$y=\frac{1}{4} \cdot (68)+37=54$

6 0

2 years ago

Find the surface area of each figure to the nearest tenth.

Stells [14]

For the cylinder on the left-hand-side,

$\bf \textit{surface area of a cylinder}\\\\
S=2\pi r(h+r)\quad 
\begin{cases}
r=radius\\
h=height\\
------\\
h=10\\
r=5
\end{cases}\implies S=2\pi (5)(10+5)$

now, for the triangular prism on the right-hand-side,

notice is really just 2 triangles and 3 rectangles, stacked up to each other at the edges.

the triangles have a base of 4, and a height of 1.5.

the rectangle on the left and the one on the right is a 6x2.5 rectangle.

the rectangle at the bottom, is a 4x6 rectangle.

adding their areas, is the area of the prism,

$\bf \stackrel{two~triangles}{2\left[\cfrac{1}{2}(4)(1.5) \right]}+\stackrel{left-right~rectangles}{2(6\cdot 2.5)}+\stackrel{bottom~rectangle}{4\cdot 6}
\\\\\\
6~~+~~30~~+~~24$

7 0

3 years ago