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3 years ago

13

what are the new coordinates for (1,4) when rotated 180 degrees and 90 degrees counterclockwise, and 90 degrees clockwise?

1 answer:

Lady bird [3.3K]3 years ago

7 0

90 degrees counterclockwise and 90 degrees cancel each other out. The coordinates of the point are only reflected over the x-axis. The point becomes (1, -4).

Hope I helped :)

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Thee questionn is beloww

Phantasy [73]

Given:

$m\angle ABC=60$

To find:

The $m\angle ADC$ .

Solution:

In circle B, $\angle ABC$ is central angle and $\angle ADC$ is inscribed angle from two points A and C.

According to central angle theorem, central angle is always twice of inscribed angle.

$m\angle ABC=2(m\angle ADC)$ [Central angle theorem]

$60=2(m\angle ADC)$

Divide both sides by 2.

$\dfrac{60}{2}=m\angle ADC$

$30=m\angle ADC$

Therefore, $m\angle ADC=30$ .

6 0

3 years ago

What is the value of b?

labwork [276]

The value of b is 61°.

you solve by doing this:
add up the two given angles, 58° and 61° to get 119°. then you subtract from 180° to get 61.

4 0

3 years ago

Read 2 more answers

Write the equation of the circle centered at (3, 5) and tangent to x = -1. I need help pleasse

Annette [7]

Answer:

(x - 3)² + (y - 5)² = 16

Step-by-step explanation:

the equation of a circle in standard form is

(x - h)² + (y - k)² = r²

where (h, k ) are the coordinates of the centre and r the radius

given (h, k ) = (3, 5 ) we require to find r

r is the distance from the centre to a point on the circle

given x = - 1 is a tangent the r is the distance between the x- coordinates

r = | 3 - (- 1) | = | 3 + 1 | = | 4 | = 4

then equation of circle is

(x - 3)² + (y - 5)² = 4² , that is

(x - 3)² + (y - 5)² = 16

7 0

2 years ago

Initially a tank contains 10 liters of pure water. Brine of unknown (but constant) concentration of salt is flowing in at 1 lite

zhenek [66]

Answer:

Therefore the concentration of salt in the incoming brine is 1.73 g/L.

Step-by-step explanation:

Here the amount of incoming and outgoing of water are equal. Then the amount of water in the tank remain same = 10 liters.

Let the concentration of salt be a gram/L

Let the amount salt in the tank at any time t be Q(t).

$\frac{dQ}{dt} =\textrm {incoming rate - outgoing rate}$

Incoming rate = (a g/L)×(1 L/min)

=a g/min

The concentration of salt in the tank at any time t is = $\frac{Q(t)}{10}$ g/L

Outgoing rate = $(\frac{Q(t)}{10} g/L)(1 L/ min)$ $\frac{Q(t)}{10} g/min$

$\frac{dQ}{dt} = a- \frac{Q(t)}{10}$

$\Rightarrow \frac{dQ}{10a-Q(t)} =\frac{1}{10} dt$

Integrating both sides

$\Rightarrow \int \frac{dQ}{10a-Q(t)} =\int\frac{1}{10} dt$

$\Rightarrow -log|10a-Q(t)|=\frac{1}{10} t +c$ [ where c arbitrary constant]

Initial condition when t= 20 , Q(t)= 15 gram

$\Rightarrow -log|10a-15|=\frac{1}{10}\times 20 +c$

$\Rightarrow -log|10a-15|-2=c$

Therefore ,

$-log|10a-Q(t)|=\frac{1}{10} t -log|10a-15|-2$ .......(1)

In the starting time t=0 and Q(t)=0

Putting t=0 and Q(t)=0 in equation (1) we get

$- log|10a|= -log|10a-15| -2$

$\Rightarrow- log|10a|+log|10a-15|= -2$

$\Rightarrow log|\frac{10a-15}{10a}|= -2$

$\Rightarrow |\frac{10a-15}{10a}|=e ^{-2}$

$\Rightarrow 1-\frac{15}{10a} =e^{-2}$

$\Rightarrow \frac{15}{10a} =1-e^{-2}$

$\Rightarrow \frac{3}{2a} =1-e^{-2}$

$\Rightarrow2a= \frac{3}{1-e^{-2}}$

$\Rightarrow a = 1.73$

Therefore the concentration of salt in the incoming brine is 1.73 g/L

8 0

3 years ago

Place the grouping symbols to make this equation true. 84-48/8-4=72

vlabodo [156]

84-6-4=72
48/4=12. 84-12= 72
84 - (48/(8-4)) = 84 - (48/4) -4 = 72
84 - 12 = 72

4 0

3 years ago