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1 year ago

Given the triangle below, what is m ZA, rounded to the nearest tenth?

Mathematics

1 answer:

quester [9]1 year ago

5 0

Answer:

Step-by-step explanation:

By the Law of Sines,

(sin A)/(11)=(sin 30°)/18
sin A = (11 sin 30°)/18
A = arcsin((11 sin 30°)/18), which is about 17.8 degrees.

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Divide 180 in a ratio of 4:5:9

Amiraneli [1.4K]

Ratio in which 180 should be divided is 4:5:9
firstly we need to find the total number of portions
this we can find by adding all the numbers making up the ratio
4 + 5 + 9 = 18
180 is made of 18 portions
size of 1 portion = 180 / 18 = 10
when dividing 180 into the respective ratio 4:5:9 the portions are as follows
4 x 10 = 40
5 x 10 = 50
9 x 10 = 90
therefore when we divide 180 in the ratio 4:5:9
the answer we get is
40 :50: 90

7 0

3 years ago

Read 2 more answers

if the farm needs to pack 12,70,224 litchis in boxes that hold 144 litchis ,how many boxes would be required .

uysha [10]

For this, we will divide 1,270,224 by 144.

1,270,224 ÷ 144 = 8,821.

For this result, we can conclude that the number of boxes required to pack 1,270,224 litchis is 8,821 boxes.

7 0

3 years ago

Read 2 more answers

Use the image from problem 11 to answer this problem.

guajiro [1.7K]

Answer:

34.

Step-by-step explanation:

All angles of a triangle combined equal 180, so if 1 angle measures 146, subtract 146 from 180. 180-146=34

4 0

2 years ago

PLEASE PLEASE ANSWER!

muminat

The area of the shaded region is $$8(\pi \ -\sqrt{3})\ \text{cm}^2$ .

Solution:

Given radius = 4 cm

Diameter = 2 × 4 = 8 cm

Let us first find the area of the semi-circle.

Area of the semi-circle = $\frac{1}{2}\times \pi r^2$

$$=\frac{1}{2}\times \pi\times 4^2$

$$=\frac{1}{2}\times \pi\times 16$

Area of the semi-circle = $$8\pi$ cm²

Angle in a semi-circle is always 90º.

∠C = 90°

So, ABC is a right angled triangle.

Using Pythagoras theorem, we can find base of the triangle.

$AC^2+BC^2=AB^2$

$AC^2+4^2=8^2$

$AC^2=64-16$

$AC^2=48$

$AC=4\sqrt{3}$ cm

Base of the triangle ABC = $4\sqrt{3}$ cm

Height of the triangle = 4 cm

Area of the triangle ABC = $\frac{1}{2}\times b \times h$

$$=\frac{1}{2}\times 4\sqrt{3} \times 4$

Area of the triangle ABC = $8\sqrt{3}$ cm²

Area of the shaded region

= Area of the semi-circle – Area of the triangle ABC

= $$8\pi \ \text{cm}^2-8\sqrt{3}\ \text{cm}^2$

= $$8(\pi \ -\sqrt{3})\ \text{cm}^2$

Hence the area of the shaded region is $$8(\pi \ -\sqrt{3})\ \text{cm}^2$ .

3 0

2 years ago

Claire is looking at the equation of a conic section where A=1, B=0, C=0. Which type of conic section is represented by the equa

Kitty [74]

The general equation of second degree is
Ax²+By²+Gx+Fy+C=0 ------1

Now the given conditions are
A=1, B=0, C=0
Now,
putting in eq. 1
x²+Gx+Fy=0
x²+Gx= -Fy
For completing square, add both side (G/2)²
x²+Gx+(G/2)²=-Fy+(G/2)²
(x+G/2)²=-F(y-G²/4F)
Or
X²=-FY where X=x+G/2, Y=y-G²/4F
which is the standard eq. of parabola in XY-coordinate.

4 0

2 years ago