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zalisa [80]
3 years ago
9

Pls help me with these 2 questions

Mathematics
1 answer:
Zigmanuir [339]3 years ago
7 0
  • xy-4y^2-(x-4y^2)^2
  • y(x-4y){y-(x-4y)}
  • (x-4y)(y-x+4y)
  • (x-4y)(5y-x)
You might be interested in
What of the angle is the vertex?
Gala2k [10]

Answer:

Vertex: The common end point at which the two rays meet to form an angle is called the vertex. Here, the point O is the vertex of ∠AOB. We can find angles in various things around us, such as in a pair of scissors, a hockey stick, a chair.

8 0
3 years ago
BRAINLIEST!!! PLEASE HELP..!!
amm1812

Answer:

2nd choice:

x int: 11/6

y int: 11/2

Step-by-step explanation:

put 6x + 2y = 11 into y = mx+b form to find y intercept. (b stands for the y int)

you get y = -3x + 11/2.

now you can graph with this.

plug 0 in for y and solve for x to find x int

0 = -3x + 11/2 then solve

7 0
3 years ago
A bridge 48m tall.a scale model of the bridge 37cm long and 12cm tall .how long is the actual bridge .
Roman55 [17]
\frac{48m}{x}=\frac{37cm}{12cm}

37x=48*12

37x=576

x=<span>15.57m long.

The bridge is about 16 meters long.
</span>
6 0
4 years ago
Math (will give Brainliest) 1
Zielflug [23.3K]

Answer: Okay! Here is your answer!

Exact form : 41/6

Decimal form : 6.83 (line over the 3)

Mixed number form : 6 5/6

Step-by-step explanation: Substitute the value of the variable into the equation and simplify.

Hope this is what you were looking for and I hope it helps you out! ☺

7 0
3 years ago
A rhombus ABCD has AB = 10 and m∠A = 60°. Find the lengths of the diagonals of ABCD.
melisa1 [442]
Three important properties of the diagonals of a rhombus that we need for this problem are:
1. the diagonals of a rhombus bisect each other
2. the diagonals form two perpendicular lines
3. the diagonals bisect the angles of the rhombus

First, we can let O be the point where the two diagonals intersect (as shown in the attached image). Using the properties listed above, we can conclude that ∠AOB is equal to 90° and ∠BAO = 60/2 = 30°. 

Since a triangle's interior angles have a sum of 180°, then we have ∠ABO = 180 - 90 - 30 = 60°. This shows that the ΔAOB is a 30-60-90 triangle.

For a 30-60-90 triangle, the ratio of the sides facing the corresponding anges is 1:√3:2. So, since we know that AB = 10, we can compute for the rest of the sides.

\overline{OB}:\overline{AB} = 1:2
\overline {OB}:10 = 1:2
\overline{OB} = \frac{1}{2}(10) = 5

Similarly, we have

\overline{AO}:\overline{AB} = \sqrt{3}:2
\overline {AO}:10 = \sqrt{3}:2
\overline{AO} = \frac{\sqrt{3}}{2}(10) = 5\sqrt{3}

Now, to find the lengths of the diagonals, 

\overline{AD} = 2(\overline{AO}) = 10\sqrt{3}
\overline{BC} = 2(\overline{OB}) = 10

So, the lengths of the diagonals are 10 and 10√3.

Answer: 10 and 10√3 units

8 0
3 years ago
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