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1 year ago

Please Help! I will give you a lot of points if you do!

Mathematics

2 answers:

Tanzania [10]1 year ago

7 0

Answer:

Step-by-step explanation:

According to the number system 7 lines and 12 dots, a line is 5 and a dot is 1

you do 5*7+12 which is equal to 47

So the answer is 47,

Hope this helps!

Norma-Jean [14]1 year ago

7 0

First look at the chart and then convert each symbols to their respective numbers

$\\ \tt\longmapsto 6+14+13+4+0+10$

Now add

$\\ \tt\longmapsto 20+17+10$

$\\ \tt\longmapsto 37+10$

$\\ \tt\longmapsto 47$

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Yuliya22 [10]

Answer: What's up?

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2 years ago

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Please asap mathematics

Mnenie [13.5K]

<h3>Given:</h3>

w= 6 units
h= 6 units
l= 10 units

w: width
h: height
l: length

The volume of the given triangular prism.

<h3>Solution:</h3>

Triangular prism is a 3 sided prism with same area of cross section with 2 triangular bases.

$\large\boxed{Formula: V= \frac{wl}{2}×h}$

Let's solve!

First, let's multiply width and height.

$6×6$

$= 36$

Now, we'll have to divide the answer by 2.

$\frac{36}{2}$

$= 18$

Then, multiply the answer by length.

$18×10$

$\large\boxed{V= 180 \: {units}^{3}}$

Hence, the volume of the given triangular prism is 180 cubic units.

3 0

1 year ago

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Find the number of tiles required to be put in a floor with dimensions 25 m x 20 m. Each tile is a rhombus with side 25 cm altit

Romashka-Z-Leto [24]

Answer:

8000

Step-by-step explanation:

Area of floor=25×20m

=500²

Area of each tile=area if each rhombus

=25×25=625cm²

No. of tiles required= area of floor/ area of each tile

=500×104⁴/ 625

=4/5 ×10×1000

=8000

hope it helps!

6 0

2 years ago

A quadrilateral has vertices A, B, C, and D. A line of reflection is drawn so that A is 6 units away from the line, B is 4 units

Kryger [21]

Answer:

The correct option is O B'

Step-by-step explanation:

We have a quadrilateral with vertices A, B, C and D. A line of reflection is drawn so that A is 6 units away from the line, B is 4 units away from the line, C is 7 units away from the line and D is 9 units away from the line.

Now we perform the reflection and we obtain a new quadrilateral A'B'C'D'.

In order to apply the reflection to the original quadrilateral ABCD, we perform the reflection to all of its points, particularly to its vertices.

Wherever we have a point X and a line of reflection L and we perform the reflection, the new point X' will keep its original distance from the line of reflection (this is an important concept in order to understand the exercise).

I will attach a drawing with an example.

Finally, we only have to look at the vertices and its original distances to answer the question.

The vertice that is closest to the line of reflection is B (the distance is 4 units). We answer O B'