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3 years ago

Find the missing side lengths (Use sqrt in place of a square root symbol)

Mathematics

1 answer:

Kitty [74]3 years ago

8 0

Step-by-step explanation:

Because this is a right triangle and it is given that one angle is 45 than also the ather angle is 45 so this triangle has 2 equal sides.So also the ather side is 6 .

Hypotenuse can be found by pythagorean theorem

6^2+6^2=x^2

36+36=x^2

x=√72

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In a parallelogram, consecutive angles are

goldenfox [79]

Answer:

Consecutive angles will always some to 180 degrees, that makes consecutive angles in a parallelogram supplementary

5 0

3 years ago

Solve. Round to the nearest tenth. 6 2-12 19 2x-2 HELP ASAP!!

Strike441 [17]

Answer:

$\displaystyle x = \frac{216}{7}$

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Equality Properties

Step-by-step explanation:

Step 1: Define

$\displaystyle \frac{6}{19} = \frac{x - 12}{2x - 2}$

Step 2: Solve for x

Cross-multiply: $\displaystyle 6(2x - 2) = 19(x - 12)$
Distribute: $\displaystyle 12x - 12 = 19x - 228$
Isolate x terms: $\displaystyle -12 = 7x - 228$
Isolate x term: $\displaystyle 216 = 7x$
Isolate x: $\displaystyle \frac{216}{7} = x$
Rewrite: $\displaystyle x = \frac{216}{7}$

6 0

3 years ago

Read 2 more answers

romanna [79]

Answer – Angle Measure
Generally speaking, when similarity transformations are performed on a triangle, the angle measure is preserved, whereas the length of the sides may be enlarged or reduced depending on the scale factor of the transformation, thus giving rise to similar triangles with corresponding angles that have exactly the same measure and corresponding sides that are proportional.

3 0

3 years ago

Read 2 more answers

Which statement is not true about the patter shown?

brilliants [131]

Answer:

Each fraction is greater than the previous fraction.

Step-by-step explanation:

Every statement is true except, the one that says "Each fracton is greater than the previous fraction." This statement is not true. I know it looks like it, but let's think about it. Each fraction is equivalent to 2/3, which means they are technically equal to each other. So that means that each fraction is not greater than the previous one. This means that the statement "Each fraction is equal to the previous fraction in the pattern multiplied by 2/2" is a true statement.

2/3 x 2/2 = 4/6

4/6 x 2/2 = 8/12

8/12 x 2/2 = 16/24

16/24 x 2/2 = 32/48

So that means that the next fraction in the pattern is indeed 32/48, making the last statement true.

3 0

3 years ago

A ship leaves port at 10 miles per hour, with a heading of N 35° W. There is a warning buoy located 5 miles directly north of th

Leno4ka [110]

The value of the angle subtended by the distance of the buoy from the

port is given by sine and cosine rule.

The bearing of the buoy from the is approximately 307.35°

Reasons:

Location from which the ship sails = Port

The speed of the ship = 10 mph

Direction of the ship = N35°W

Location of the warning buoy = 5 miles north of the port

Required: The bearing of the warning buoy from the ship after 7.5 hours.

Solution:

The distance travelled by the ship = 7.5 hours × 10 mph = 75 miles

By cosine rule, we have;

a² = b² + c² - 2·b·c·cos(A)

Where;

a = The distance between the ship and the buoy

b = The distance between the ship and the port = 75 miles

c = The distance between the buoy and the port = 5 miles

Angle ∠A = The angle between the ship and the buoy = The bearing of the ship = 35°

Which gives;

a = √(75² + 5² - 2 × 75 × 5 × cos(35°))

By sine rule, we have;

$\displaystyle \frac{a}{sin(A)} = \mathbf{ \frac{b}{sin(B)}}$

Therefore;

$\displaystyle sin(B)= \frac{b \cdot sin(A)}{a}$

Which gives;

$\displaystyle sin(B) = \mathbf{\frac{75 \cdot sin(35^{\circ})}{\sqrt{75^2 + 5^2 - 2 \times 75\times5\times cos(35^{\circ}) } }}$

$\displaystyle B = arcsin\left( \frac{75 \cdot sin(35^{\circ})}{\sqrt{75^2 + 5^2 - 2 \times 75\times5\times cos(35^{\circ}) } }\right) \approx 37.32^{\circ}$

Similarly, we can get;

$\displaystyle B = arcsin\left( \frac{75 \cdot sin(35^{\circ})}{\sqrt{75^2 + 5^2 - 2 \times 75\times5\times cos(35^{\circ}) } }\right) \approx \mathbf{ 142.68^{\circ}}$

The angle subtended by the distance of the buoy from the port, C is therefore;

C ≈ 180° - 142.68° - 35° ≈ 2.32°

By alternate interior angles, we have;

The bearing of the warning buoy as seen from the ship is therefore;

Bearing of buoy ≈ 270° + 35° + 2.32° ≈ 307.35°

Learn more about bearing in mathematics here:

brainly.com/question/23427938

5 0

2 years ago