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yan [13]
3 years ago
5

Are these equations equivalent?

Mathematics
2 answers:
Lostsunrise [7]3 years ago
8 0

Answer:

yes the  equations are equivalent?

Step-by-step explanation:

Tpy6a [65]3 years ago
4 0

Answer:

Yes, 2x=4 is 2 simply by dividing, and 4x+5=13 is 2 as well, because 5-13 is 8 and 8 divided by 4 is 2

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Use the FOIL method to find the product below.
Aleks04 [339]

Answer:

work is pictured and shown

8 0
3 years ago
Find the area of the figure. Round to the nearest tenth if necessary.
baherus [9]

Hello!


First of all, let's find the triangle. As we can see, it is eight units long and four units tall. We can use the triangle formula and half the product.


4(8)=32

32/2=16


Now we need to find the area of the semicircle. We can use circle formulas and half our final answer. We will find the area using A=\pir².


We know that the diameter of this circle is eight. Therefore our radius is 4.


A=4\pi²

A=16\pi


Now we multiply.


16\pi≈50.27


Now we add our triangle.


50.27+16=66.27


The nearest answer to this is B)66.2. It may have been a bit over as we used \pi, not 3.14.


I hope this helps!



3 0
3 years ago
3x+2 less or equal to 8
bearhunter [10]

Answer:

less than

Step-by-step explanation:

x = 0(number of pictures)   3 * 0 = 0

0 is less than 8 :)

6 0
3 years ago
List formulas for: Area of a triangle
Ede4ka [16]
The formula for the area of a triangle is the base times the height, divided by 2. Or in simple forms, bh/2
7 0
3 years ago
How do I solve this?
vladimir1956 [14]

Answer:

3p³ + 2p² – 3p – 11

Step-by-step explanation:

From the question given above, the following data were obtained:

Side 1 (S₁) = –1(p + 5)

Side 2 (S₂) = 2(p² – 3)

Side 3 (S₃) = 3p³ – 2p

Perimeter (P) =?

The perimeter of the triangle can be obtained as follow

P = S₁ + S₂ + S₃

P = –1(p + 5) + 2(p² – 3) + 3p³ – 2p

Clear bracket

P = –p – 5 + 2p² – 6 + 3p³ – 2p

Rearrange

P = 3p³ + 2p² – 2p – p – 6 – 5

P = 3p³ + 2p² – 3p – 11

Therefore, the perimeter of the triangle is 3p³ + 2p² – 3p – 11

4 0
3 years ago
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