Find the missing values in these equivalent fractions 1/2 2/? 3/? 4/?

Greg is designing a clock face. Using the center of the clock face as the origin, he keeps its diameter at 10 units. Match the p

Nat2105 [25]

The center of the clock is taken as the origin.The clock is a circle with a diameter 10 units.Radius is half the diameter .Radius = 10 ÷2= 5 units.

The clock is divided in four quadrants .On x axis y=0 and on y axis x=0.

When it is 12 o'clock the hour hand is on positive of y axis.Coordinates of the point at 12 o'clock=(0,5)

When it is 3 o 'clock the hour hand is on positive of x axis .Coordinate of the point at 3o'clock is (5,0)

When it is 6 o'clock the hour hand is on negative of y axis .The coordinates of the point at 6o'clock is (0,-5)

At 9o'clock the hour hand is on negative of x axis .The coordinate of the point at 6o'clock is(-5,0)

3 0

3 years ago

Read 2 more answers

Please help and thank you

nalin [4]

Answer:

Your answer is D. Hoped i helped

Step-by-step explanation:

plot each coordinate, Then see which has the most/all in the shaded area.

4 0

3 years ago

HELP ASAP//BRAINLEST+THANKS: Subj:gravity formula stuff

Y_Kistochka [10]

Answer:

6.46

Step-by-step explanation:

h(t) = (1/2) $att^{2}$ + v0t + h0

Origin is at surface of water (h = 0), up is positive.

a = -g = -9.8 m/ $s^{2}$

v0 = 25 m/s

h0 = 50 m

h(t) = -4.9t2 + 25t + 50

Find t when

7 = $-4.9t^{2}$ + 25t + 50

$4.9t^{2}$ - 25t - 43 = 0

t = [1/(2*4.9)] [25 ±√( $25^{2}$ + 4*4.9*43)]

t = (0.102)(25 ± 38.312)

t > 0 ⇒ use only the positive square root

t = (0.102)(25 + 38.312) ≅ 6.46 s

It is 7m above the water at about 6.46 s

7 0

3 years ago

Find the gradient of the line segment between the points (2,2) and (0,6).

Alexeev081 [22]

Answer:

slope-intercept formula -----> Y2-Y1/X2-X1

Step-by-step explanation:

X1 = 2 X2= 0

Y1 = 2 Y2=6

6-2/0-2= -2

G= -2

5 0

3 years ago

The points ????(0,0), ????(−4,1), ????(−3,5), and ????(1,4) are the vertices of parallelogram OABC Is this parallelogram a

gavmur [86]

Answer:

Yes, parallelogram OABC is a rectangle.

Step-by-step explanation:

The vertices of a parallelogram are O(0,0), A(-4,1), B(-3,5) and C(1,4).

A parallelogram is a rectangle if its any two adjacent sides are perpendicular to each other.

product of slopes of two perpendicular lines is -1.

Formula for slope is

$Slope=\dfrac{y_2-y_1}{x_2-x_1}$

Using the formula we get

$m_{OA}=\dfrac{1-0}{-4-0}=-\dfrac{1}{4}$

$m_{AB}=\dfrac{5-1}{-3-(-4)}=-\dfrac{4}{1}=4$

$m_{OA}\times m_{AB}=-\dfrac{1}{4}\times 4=-1$

It means OA is perpendicular to AB.

Since two adjacent sides are perpendicular, therefore the given parallelogram OABC is a rectangle.

6 0

3 years ago