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sineoko [7]
3 years ago
9

The right triangle above has height h centimeters and hypotenuse t centimeters. What is the area, in square centimeters, of the

triangle, in terms of h and t ?
Mathematics
1 answer:
White raven [17]3 years ago
8 0

Answer:

A = \frac{h\sqrt{t^{2} - h^{2}}}{2}

Step-by-step explanation:

Right triangle:

height h, width w, hypotenuse h

Area is: A = \frac{h*w}{2}

Pytagoras theorem: h^{2} + w^{2} = t^{2}

In this question:

To find the area, in terms of h and t, we have to write w as a function of h and t. So

h^{2} + w^{2} = t^{2}

w^{2} = t^{2} - h^{2}

w = \sqrt{t^{2} - h^{2}}

So the area is:

A = \frac{h*w}{2}

A = \frac{h\sqrt{t^{2} - h^{2}}}{2}

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ludmilkaskok [199]

Answer:

953.78 in^3 of grain.

Step-by-step explanation:

The volume of a cylinder = π r^2 h.

For this cylinder r =  radius of the base = 9/2

= 4.5 in  and the height h = 15 in

So the volume is 3.14 * 4.5^2 * 15

= 953.78 in^3.

8 0
3 years ago
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I NEED HELP ASP 20 points
leva [86]
Circumference
C = pi d
C = 3.14 x 10
C = 31.4 cm

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A = pi r^2
A = 3.14 x 5^2
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Y=2x+3 in ordered pairs
Iteru [2.4K]

Answer:

(0,3) , (1,5) , (2,7)

Step-by-step explanation:

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3 years ago
A bag contains 4 white,5 blacks and 2 blue balls. 3 balls are drawn one after the other without replacement from the bag .what i
Doss [256]

Answer:

24/99

Step-by-step explanation:

From the question given above, the following data were obtained:

White (W) balls = 4

Black (B) balls = 5

Blue (Bl) balls = 2

Probability that they are of different color =?

Next, we shall determine the total number of balls in the bag. This can be obtained as follow:

White (W) balls = 4

Black (B) balls = 5

Blue (Bl) balls = 2

TOTAL = 4 + 5 + 2 = 11 balls

Next, we shall determine the possible outcome of draw. This can be obtained as follow.

The possible outcome could be:

WBLB or WBBL or BBLW or BWBL or BLBW or BLWB

Next we shall determine the probability of each of outcomes.

Since the ball is drawn without replacement, it means the total number of ball will reduce after each draw.

White (W) balls = 4

Black (B) balls = 5

Blue (Bl) balls = 2

TOTAL = 11

P(WBLB) = 4/11 × 2/10 × 5/9 = 40/990

P(WBLB) = 4/99

P(WBBL) = 4/11 × 5/10 × 2/9 = 40/990

P(WBBL) = 4/99

P(BBLW) = 5/11 × 2/10 × 4/9 = 40/990

P(BBLW) = 4/99

P(BWBL) = 5/11 × 4/10 × 2/9 = 40/990

P(BWBL) = 4/99

P(BLBW) = 2/11 × 5/10 × 4/9 = 40/990

P(BLBW) = 4/99

P(BLWB) = 2/11 × 4/10 × 5/9 = 40/990

P(BLWB) = 4/99

Finally, we shall determine the probability that they are of different color. This can be obtained as follow:

P(WBLB) = 4/99

P(WBBL) = 4/99

P(BBLW) = 4/99

P(BWBL) = 4/99

P(BLBW) = 4/99

P(BLWB) = 4/99

Probability that they are of different color =?

Probability that they are of different color = P(WBLB) + P(WBBL) + P(BBLW) + P(BWBL) + P(BLBW) + P(BLWB)

= 4/99 + 4/99 + 4/99 + 4/99 + 4/99 + 4/99

= (4 + 4 + 4 + 4 + 4 + 4)/99

= 24/99

Thus, the probability that they are of different color is 24/99

4 0
3 years ago
In rhombus ABCD, diagonals AC and BD meet at point E. If the measure of angle DAB is 46 degrees, find the length of EB.
slamgirl [31]

Answer:

  EB ≈ 1.563 in

Step-by-step explanation:

The diagonals of a rhombus divide the figure into four congruent right triangles. Angle DAB is bisected by EA, so angle EAB is 46°/2 = 23°. EB is the side opposite, so the relevant trig relation is ...

  Sin = Opposite/Hypotenuse

  sin(EAB) = EB/AB

  EB = (4 in)sin(23°) . . . . . . multiply by the hypotenuse

  EB ≈ 1.563 in

3 0
2 years ago
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