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Anna71 [15]
3 years ago
5

carter lives on a street where all the house numbers are multiples of 6. Name two possible house numbers between 601 and 650. ex

plain
Mathematics
1 answer:
Mamont248 [21]3 years ago
4 0

Answer: The answer is 606 and 612.


Step-by-step explanation:  Given that Carter lives on a street where all the house numbers are multiples of 6. We are given to name any two possible house numbers between 601 and 650.

The numbers lying between 601 and 650 are

602, 603, 604, 605, . . . ,648, 649.

And the numbers among these which are multiples of 6 are

606, 612, 618, 624, 630, 636, 642, 648.

So, any two of these eight numbers can be possible.

Thus, the numbers are 606 and 650.

You might be interested in
In ΔPQR, the measure of ∠R=90°, the measure of ∠P=26°, and PQ = 8.5 feet. Find the length of QR to the nearest tenth of a foot.
Vitek1552 [10]

Answer:

3.7feet

Step-by-step explanations

Using the sin rule

A/sin a = B/sin b

Let A = PQ = 8.5feet

B = QR = x feet

a = R = 90°

b = P = 26°

Substitute the values into the Sin rule

8.5/sin90 = x/sin26

8.5×sin 26 = x × sin 90

8.5×0.4383 = x× 1

3.7255 = x

Hence the length of QR to the nearest tenth 3.7feet

5 0
2 years ago
Read 2 more answers
7) If angle K is defined by the expression "8x+25" and angle M is defined by the expression "6x+15" find the measure of angle K.
castortr0y [4]

Answer:

i looked it up in math ay and i got nothing i put 6x + 15 in to it and it said ***Factor 3 out of 6x + 15**   **3 2x + 15**

Step-by-step explanation:

hope this helps

6 0
3 years ago
Does anybody know the answer to this?
faust18 [17]

Answer:

C (136)

Step-by-step explanation:

The triangle is only rotated, the measure of <D doesn't change

4 0
3 years ago
A ship is stationary at sea. A tugboat is 28 km away at a bearing of 210°, and a yacht is 24 km from the tugboat at a bearing of
3241004551 [841]

Answer:

The distance between the yacht from the ship is about 33.6 km.

Step-by-step explanation:

The diagram is attached below.

Point A is the shipe, Point B is the tugboat, and Point C is the yacht.

Since the tugboat is 28 km away, and the scale being used is 1 cm : 4 km, the tugboat is 7 cm away from the ship at a 210° bearing on the diagram.

Likewise, the yacht is 6 cm away from the tugboat at a 210° bearing.

To find the distance between the yacht to the ship, we need to

yacht to the ship, we need to determine the value of x.

To do so, we can use the Law of Cosines:

b²a² + c²2ac cos(B) =

First, we need to determine ZB.

A has a bearing of 210°. 180° is covered by QI and IV. So, the small angle in QIII is 30°.

By Alternate Interior Angles, the angle in QI (inside the triangle) of Point B must also be 30°.

And since Point B has a bearing of 310°, 270° is covered in total by QI, QIV, and QIII. So, 40° is left in QII.

Therefore, 50° is the measure of the angle in QIII within the triangle of Point B.

Thus, ZB measures 50° + 30° = 80°.

So, ZB measures 80°. We also know that a = 6 and c= 7. Substitute:

b² = (6)² + (7)² - 2(6)(7) cos(80)

Simplify and take the square root of both sides. So:

b = √85-84 cos(80)

Approximate using a calculator:

b = 8.3912... ≈ 8.39 cm

Since 1 cm: 4km, to find the distance in km, we simply need to multiply by four. So:

⇒b= 4(8.3912...) = 33.5651 ≈

33.6 km

The distance between the yacht from the ship is about 33.6 km.

Step-by-step explanation:

hope it helps you itz admirer

4 0
2 years ago
CAN SOMEONE PLEASE HELP ME ITS KHAN ACADEMY
Vlad [161]
It’s definitely B I believe
8 0
2 years ago
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