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stich3 [128]
3 years ago
13

PLEASE DONT SKIP THIS I NEED HELP Can someone please explain how to solve this problem?

Mathematics
1 answer:
Setler [38]3 years ago
7 0
I would say 168 because if you throw a dice with 6 sides a thousand 1000 you will most likely get one certain side 168 times.
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What is 5 divided by 3/4
3241004551 [841]

Answer: 20/3 or 6.67

Step-by-step explanation:

5/1÷3/4

While dividing fractions, flip to the other side to get 5/1×4/3

=20/3

5 0
4 years ago
At the school carnival, the sixth graders are making directional arrows, as shown below.
victus00 [196]
They would need A. 26 square inches of red paint. I hope this helped ^^
8 0
3 years ago
Read 2 more answers
The question is in the picture please help
ahrayia [7]

Answer:

Claim 2

Step-by-step explanation:

The Inscribed Angle Theorem* tells you ...

... ∠RPQ = 1/2·∠ROQ

The multiplication property of equality tells you that multiplying both sides of this equation by 2 does not change the equality relationship.

... 2·∠RPQ = ∠ROQ

The symmetric property of equality says you can rearrange this to ...

... ∠ROQ = 2·∠RPQ . . . . the measure of ∠ROQ is twice the measure of ∠RPQ

_____

* You can prove the Inscribed Angle Theorem by drawing diameter POX and considering the relationship of angles XOQ and OPQ. The same consideration should be applied to angles XOR and OPR. In each case, you find the former is twice the latter, so the sum of angles XOR and XOQ will be twice the sum of angles OPR and OPQ. That is, angle ROQ is twice angle RPQ.

You can get to the required relationship by considering the sum of angles in a triangle and the sum of linear angles. As a shortcut, you can use the fact that an external angle is the sum of opposite internal angles of a triangle. Of course, triangles OPQ and OPR are both isosceles.

4 0
3 years ago
Simplify the expression shown below: (a^3b^12c^2)^2(a^5c^2)^3(b^5c^4)^0
NemiM [27]
= a^21b^24c^10 
use the product rules for each set of () then multiply them together. when multiplying variables with exponents you just add them together like for a^3 * a^5 = a^8.     (a^3 b^12 c^2)^2 = (a^6 b^24 c^4)
4 0
3 years ago
Find the area of quadrilateral ABCD whose vertices are A(1,1) B(7,-3) C(12,2) D(7,21)
sasho [114]

Answer:

The area of quadrilateral ABCD is 139 unit^2.

Step-by-step explanation:

Given:

Quadrilateral ABCD whose vertices are A(1,1) B(7,-3) C(12,2) D(7,21).

Now, to find the area.

The coordinates of the quadrilateral are A(1,1), B(7,-3), C(12,2), D(7,21).

So, to find the area we need to bisect the quadrilateral ABCD and get the triangles ABC and ADC and then calculate their areas:

In Δ ABC:

A(x_1,y_1)=(1,1)\:,\:B(x_2,y_2)=(7,-3)\:and\:C(x_3,y_3)=(12,2)

Now, to get the area of triangle ABC:

Area\,of\,triangle\,=\,\frac{1}{2}\left|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right|

Area\,of\,triangle\,=\,\frac{1}{2}\left|1(-3-2)+(7)(2-1)+12(1--3)\right|

Area\,of\,triangle\,=\,\frac{1}{2}\left|1(-5)+(7)(1)+12(4)\right|

On solving we get:

Area\,of\,triangle\,=25.

In Δ ADC:

A(x_1,y_1)=(1,1)\:,\:D(x_2,y_2)=(7,21)\:and\:C(x_3,y_3)=(12,2)

Now, to get the area of triangle ADC:

Area\,of\,triangle\,=\,\frac{1}{2}\left|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right|

Area\,of\,triangle\,=\,\frac{1}{2}\left|1(21-2)+(7)(2-1)+12(1-21)\right|

On solving it by same process as above we get:

Area\,of\,triangle\,=114

Now, to get the area of the quadrilateral we add the areas of the triangles ABC and ADC:

25+114\\=25+114\\=139\ unit^2

Therefore, area of quadrilateral ABCD is 139 unit^2.

4 0
3 years ago
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