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Nonamiya [84]
3 years ago
9

What is another way to group the factors (3×2)×5

Mathematics
2 answers:
SVEN [57.7K]3 years ago
6 0

I believe the answer is  30

olasank [31]3 years ago
3 0
<h2>3*(2*5)</h2><h2>2*(3*5) is the correct groupings </h2><h2 />
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Solve for x. <br> 25x=65<br> x = 1225 <br> x = 125<br> x = 3 <br> x = 6
masya89 [10]

25x=65

We divide 25 to cancel it out.

65/25=2.6

I would assume 3, because 2.6 is 3 when rounded up.

---

hope it helps

3 0
3 years ago
Helpp!!!!! i will mark as brainliest !!<br><br>solve question i and ii ​
melomori [17]

Answer:

(i) Lowest Temperature: Warsaw

Highest Temperature: Cairo

(ii) 27.1 + 35.2 = 62.3°C

4 0
3 years ago
Read 2 more answers
In the diagram below AB is parallel to CD what is the value of y?
Ilya [14]
Y° = y₁° (opposite or Vertical angles), but y₁ = 45° corresponding angles,


then y = 45°
5 0
3 years ago
Read 2 more answers
Please, can someone help?
LekaFEV [45]
The area of that figure is 2453.25 square feet

First, you would need to find the area of the rectangle

(L)(W)
(70)(30)
2100

Then since it is only a semi circle, the formula is pi*r^2/2

You find the area of a full circle first

(3.14)(15^2)
(3.14)(225)
706.5

Then you divide that by 2 since it’s only half a circle

706.5/2 = 353.25

Finally you add that by the area of the rectangle

2100 + 353.25 = 2453.25 square feet
3 0
3 years ago
Find all the zeros for each function P(x)=x^4-4x^3-x^2+20x-20
ivanzaharov [21]

Answer:

The zeros of the given polynomial function are

2,2,\pm\sqrt{5}

Step-by-step explanation:

Given polynomial is P(x)=x^4-4x^3-x^2+20x-20

To find the zeros equate the given polynomial to zero

ie., P(x)=0

P(x)=x^4-4x^3-x^2+20x-20=0

By using synthetic division we can solve the polynomial:

2_|   1     -4     -1      20      -20

       0      2     -4     -10       20

   _____________________

       1     -2      -5      10      |_0

Therefore x-2=0

x=2 is a zero of P(x)

Now we can write the cubic equation as below:

x^3-2x^2-5x+10=0

Again using synthetic division

2_|   1     -2     -5     10      

       0      2      0    -10    

    ______________

       1      0      -5     |_0

Therefore x-2=0

x=2 is also a zero of P(x).

Now we have x^2+0x-5=0

x^2-5=0

x^2=5

x^=\pm\sqrt{5} is a zero of P(x)

Therefore the zeros are 2,2,\pm\sqrt{5}

8 0
3 years ago
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