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SIZIF [17.4K]
3 years ago
11

Which expression is equivalent to the expression below? m+3 m2-16 m2-9 m+4

Mathematics
1 answer:
Dvinal [7]3 years ago
6 0
To answer this problem, we group the first the like terms, in this case, the terms 3/m2 and 16/m2. The difference between these two terms -13/m2. Hence, the overall expression is m - 13/m2 - 9/m + 4.
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Help wanted pls math
Serjik [45]

Answer:

39 ft²

Step-by-step explanation:

Area of a triangle = 1/2bh

Let:

base = b = 4 ft

height = h = 3 ft

Solve for the area of the triangle:

A = 1/2(4)(3)

A = 2 * 3

A = 6 ft²

Area of a rectangle = bh

Let:

base = b = 9 ft

height = h = 5 ft

Solve for the area of the rectangle:

A = (9)(5)

A = 9 * 5

A = 45 ft²

Subtract the area of the triangle from the area of the rectangle:

45 - 6 = 39 ft²

39 ft² is your answer.

~

3 0
3 years ago
Read 2 more answers
Awnser in the picture <br><br> btw this question is hard so work carefully
Triss [41]

Answer:

5       10

2         4

Step-by-step explanation:

8 0
3 years ago
WILL GIVE BRAINLIEST!!<br><br> Which equations are correct?<br><br> Select each correct answer.
solong [7]

To multiply these expression, multiply the numeric parts, and sum the exponents of the literal parts (because the variable is the same: we're using the power property a^b\cdot a^c = a^{b+c}).

So, you have:

  • 3z^7 \cdot (4z^2) = 3\cdot 4 \cdot z^{7+2} = 12z^9
  • 6a^5 \cdot (6a^5) = 6\cdot 6 \cdot z^{5+5} = 36z^{10}
  • 6b^4 \cdot (-3b^4) = 6\cdot (-3) \cdot b^{4+4} = -18b^8
  • 5x^4 \cdot (4x^2) = 5\cdot 4 \cdot x^{4+2} = 20x^6
3 0
3 years ago
What is the area of the square adjacent to the third side of the triangle?
lana66690 [7]

Answer:

The area of the square adjacent to the third side of the triangle is 11 units²

Step-by-step explanation:

We are given the area of two squares, one being 33 units² the other 44 units². A square is present with all sides being equal, and hence the length of the square present with an area of 33 units² say, should be x² = 33 - if x = the length of one side. Let's make it so that this side belongs to the side of the triangle, to our convenience,

x² = 33,

x = \sqrt{33} .... this is the length of the square, but also a leg of the triangle. Let's calculate the length of the square present with an area of 44 units². This would also be the hypotenuse of the triangle.

x² = 44,

x = \sqrt{44} .... applying pythagorean theorem we should receive the length of a side of the unknown square area. By taking this length to the power of two, we can calculate the square's area, and hence get our solution.

Let x = the length of the side of the unknown square's area -

\sqrt{(44)}^2 = x^2 + \sqrt{33}^2,

x = \sqrt{11} ... And \sqrt{11} squared is 11, making the area of this square 11 units².

3 0
4 years ago
Read 2 more answers
Fit a quadratic function to these three points: (-2,8)(0,-4),and (4,68)
aev [14]
Given three points
P1(-2,8)
P2(0,-4)
P3(4,68)
We need the quadratic equation that passes through all three points.

Solution:
We first assume the final equation to be
f(x)=ax^2+bx+c .............................(0)

Observations:
1. Points are not symmetric, so cannot find vertex visually.
2. Using the point (0,-4) we substitute x=0 into f(x) to get
f(0)=0+0+c=-4, hence c=-4.
3. We will use the two other points (P1 & P3) to set up a system of two equations to find a and b.
f(-2)=a(-2)^2+b(-2)-4=8 => 4a-2b-4=8.................(1)
f(4)=a(4^2)+b(4)-4=68 =>  16a+4b-4=68.............(2)
4. Solve system
2(1)+(2) => 24a+0b-12=84 => 24a=96 => a=96/24 => a=4 ......(3)
substitute (3) in (2) => 16(4)+4b-4=68 => b=8/4 => b=2  ..........(4)
5. Put values c=-4, a=4, b=2 into equation (0) to get

f(x)=4x^2+2x-4

Check:
f(-2)=4((-2)^2)+2(-2)-4=16-4-4=8
f(0)=0+0-4 = -4
f(4)=4(4^2)+2(4)-4=64+8-4=68
So all consistent, => solution ok.

6 0
4 years ago
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