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KonstantinChe [14]
3 years ago
7

If f(x) = 6x - 4, evaluate – -forx=8. A.-2 B.-11 C.44 D.11

Mathematics
1 answer:
nignag [31]3 years ago
8 0
C.44 8x6 is 48-4 is 44
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Find four consecutive even integers whose sum is 84. Find the integers.
mina [271]
Suppose the integers are n , n+2 , n+4 and n+6.
84=n+(n+2)+(n+4)+(n+6)=4n+12.
Subtract 12 from both ends to get.
72=4n.
Divide both ends by 4 to get.
n=18.
So the integers are: 18 , 20 , 22 , 24.
4 0
3 years ago
Please can anyone help what is AC
Nesterboy [21]

The answer is C.

You need you use Pythagoras theorem 2 times to have 2 equations with the same sides i.e. for example a2 + b2 = 81 and a2 - b2 = 9.

You can do it as you have different triangles but with the same sides.

7 0
4 years ago
Copy and complete the table on the previous page into a Word document. Title your journal "Pythagorean Theorem". Discuss any rel
scoundrel [369]
In trigonometry, the right triangle is considered a special triangle because there are derived equations solely for this type. It is really convenient when dealing right triangle problems because it is more simplified courtesy of the Pythagorean theorems. It is derived that the square of the hypotenuse (longest side of the triangle) is equal to the sum of the squares of the other two legs. In equation, that would be c² = a² + b². For this activity, all you have to do is find the sum of the squares in columns a and b. Then, see if this is equal to the square of the values in column c. Let's calculate each row:

Row 1:
3² + 4² ? 5²
25 ? 25
25 = 25

Row 2:
5² + 12² ? 13²
169 ? 169
169 = 169

Row 3:
9² + 12² ? 15²
225 ? 225
225 = 225

Therefore, all of the given values conform to a² + b² = c².
7 0
3 years ago
PWEASE HELP, AREA OF A TRAPEZOID
sdas [7]

Answer:

34 m²

Step-by-step explanation:

4 0
3 years ago
Please help me<br> Show your work
GrogVix [38]

<u>ANSWER TO PART A</u>

The given triangle has vertices J(-4,1), K(-4,-2),L(-3,-1)


The mapping for rotation through 90\degree counterclockwise has the mapping


(x,y)\rightarrow (-y,x)


Therefore

J(-4,1)\rightarrow J'(-1,-4)


K(-4,-2)\rightarrow K'(2,-4)


L(-3,-1)\rightarrow L'(1,-3)


We plot all this point and connect them with straight lines.


ANSWER TO PART B


For a reflection across the y-axis we negate the x coordinates.


The mapping is



(x,y)\rightarrow (-x,y)


Therefore

J(-4,1)\rightarrow J''(4,1)


K(-4,-2)\rightarrow K''(4,-2)


L(-3,-1)\rightarrow L''(3,-1)


We plot all this point and connect them with straight lines.


See graph in attachment







7 0
3 years ago
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