All categories

1 year ago

Given: △AFD, m∠F=90° DF=7√3 , m∠D=30° Find: Area of △AFD

Mathematics

1 answer:

Aliun [14]1 year ago

4 0

Using the definition of the tangent function, you can find ...

... FA = DF·tan(30°) = (7√3)·(1/√3) = 7

Then the usual formula for triangle area can be used:

... Area = (1/2)bh = (1/2)·(7√3)(7)

... Area = (49/2)√3 square units

... ≈ 42.44 square units

You might be interested in

Solve for m:<br> -3(1 – 5m) = — 38 + 8m

Solnce55 [7]

Answer:

m = - 5

Step-by-step Explanation:

$-3(1-5m)=-38+8m \\ \\ - 3 + 15m = - 38 + 8m \\ \\ 15m - 8m = 3 - 38 \\ \\ 7m = - 35 \\ \\ m = \frac{ - 35}{7} \\ \\ \huge \purple{ \boxed{m = - 5}}$

3 0

1 year ago

Read 2 more answers

Megan consumed 3.5 gallons of water in one day. How many milliliters are equal to 3.5 gallons, if 1 liter = 1,000 milliliters an

baherus [9]

1 gallon = 3.785 liters

3.5 * 3.785 = 13.2475 total liters

1 liter = 1000 milliliters

13.2475 * 1000 = 13,247.5 milliliters

Answer is B

3 0

1 year ago

Read 2 more answers

Solve the following quadratic equations by extracting square roots.Answer the questions that follow.

Vikentia [17]

Answer:

1. x=±4

2. t=±9

3. r=±10

4. x=±12

5. s=±5

Step-by-step explanation:

1. x^2 = 16

Taking square root on both sides

$\sqrt{x^2}=\sqrt{16}\\\sqrt{x^2}=\sqrt{(4)^2}\\$

x=±4

2. t^2=81

Taking square root on both sides

$\sqrt{t^2}=\sqrt{81}\\\sqrt{t^2}=\sqrt{(9)^2}$

t=±9

3. r^2-100=0

$r^{2}-100=0\\r^2 =100\\Taking\ Square\ root\ on\ both\ sides\\\sqrt{r^2}=\sqrt{100}\\\sqrt{r^2}=\sqrt{(10)^2}$

r=±10

4. x²-144=0

x²=144

Taking square root on both sides

$\sqrt{x^2}=\sqrt{144}\\\sqrt{x^2}=\sqrt{(12)^2}$

x=±12

5. 2s²=50

$\frac{2s^2}{2} =\frac{50}{2}\\s^2=25\\Taking\ Square\ root\ on\ both\ sides\\\sqrt{s^2}=\sqrt{25}\\\sqrt{s^2}=\sqrt{(5)^2}$

s=±5 ..

4 0

1 year ago

Read 2 more answers

For a normally distributed random variable x with m = 75 and s = 4, find the probability that 69 < x < 79 Use the table to

nalin [4]

10-14-19-12-14-18-10-15-15

6 0

1 year ago

An upscale resort has built its circular swimming pool around a central area that contains a restaurant. The central area is a r

Elena L [17]

The problem is modelled in the diagram below

Question 1:
The largest area of the pool is half of the area of the circle

Area of circle = πr², where r is the radius given by 1/2 of the diameter
Area of circle = π(60)² = 11309.73355 feet

Area of half of the circle is 11309.73355/2 = 5654.866... ≈ 5654.87 feet (2dp)

Question b:

The area of the pool is the area of the circle subtracts the area of the triangle.

The area of the circle is 11309.73²

The area of the triangle is 1/2 (60×103.92) = 3117.6 feet²

The area of the pool is 11309.73 - 3117.6 = 7922.13 feet²

The volume of the pool = 7922.13 × 4 = 31688.52 feet³

Note: there isn't any information on the fish tank part so the answer above is assumed for the whole pool.

6 0

1 year ago