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3 years ago

Help me with the triangle area please

Mathematics

2 answers:

dexar [7]3 years ago

7 0

Answer:

15 m ^2

Step-by-step explanation:

3 x 10 is 30

divide that by 2 since the triangle can fit in half of the rectangle shape

you will get 15

Sindrei [870]3 years ago

5 0

Answer:

Step-by-step explanation:

<h2>Area of a triangle</h2><h3>formula of area: </h3>

1/2 * b*h

b is the base

h is the height

= 1/2 * 10 * 3

= 15

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Triangle ABC is a right triangle and cos(22.6o)=StartFraction b Over 13 EndFraction. Solve for b and round to the nearest whole

Art [367]

Answer:

a = 5 and b = 12

Step-by-step explanation:

Step 1: Find angle B

Angle C = 90°

Angle A = 22.6°

Angle B = B

All angles in a triangle are equal to 180°.

Angle A + Angle B + Angle C = 180°

22.6 + 90 + B = 180°

B = 180 - 112.6

B = 67.4°

Step 2: Find the value of side AC 'b'

Hypotenuse = 13

Adjacent = b

Angle A = 22.6°

Cos (Angle) = Adjacent/Hypotenuse

Cos (22.6) = b/13

b = 12

Step 3: Find the value of side CB 'a'

Hypotenuse = 13

Opposite = a

Angle A = 22.6°

Sin (angle) = Opposite/Hypotenuse

Sin (22.6°) = a/13

a = 4.99 rounded off to 5

Therefore, the value of a=5 and b=12.

8 0

4 years ago

Read 2 more answers

Please answer for 15 points Thank you

DedPeter [7]

Answer:3

Step-by-step explanation:

The answer is 3 because we know were adding

5 0

3 years ago

Read 2 more answers

A child wanders slowly down a circular staircase from the top of a tower. With x,y,zx,y,z in feet and the origin at the base of

babymother [125]

Answer:

a) The tower is 90 feet tall

b) She reaches the bottom at t = 18 minutes.

c) Her speed at time t is 5 \sqrt[]{5} ft/minute

d) Her acceleration at time t is 10 ft/minute^2

Step-by-step explanation:

Consider the path described by the child as going down the tower to have the following parametrization $\gamma(t) = (10\cos t, 10 \sin t, 90-5t)$

a) Assuming that the child is at the top of the tower when she starts going down, we have that at the initial time (t=0) we will have the value of the height of the tower. That is z = 90-5*0 = 90 ft.

b) The child reaches the bottom as soon as z =0. We want to find the value of t that does that. Then we have 0 = 90-5t, which gives us t = 18 minutes.

c) Given the parametrization we are given, the velocity of the child at time t is given by $\frac{d\gamma}{dt}= (\frac{d}{dt}(10\cos t), \frac{d}{dt} (10 \sin t ), \frac{d}{dt}(90-5t)) = (-10 \sin t, 10 \cos t, -5)$ . The speed is defined as the norm of the velocity vector,

so, the speed at time t is given by $v = \sqrt[]{(-10 \sin t)^2+(10 \cos t)^2+(-5)^2} = \sqrt[]{100(\sin^2 t + \cos^2 t)+25} = \sqrt[]{125}= 5 \sqrt[]{5}$