Answer:
0
Step-by-step explanation:
3 <em><u>-</u></em> (<u><em>-</em></u>1) + (-1) -3 =
Minus and a negative makes it positive. In other words it would be written like this: 3 + 1 <u><em>+</em></u> (<u><em>-</em></u>1) - 3 =
Next step would be making it minus 1, because plus sign and negative makes the number to be subtracted. It would be rewritten like this: <em><u>3 + 1 </u></em>- 1 - 3 =
Then you will simply add 3 and 1, which makes it 4. So it would be rewritten likes this: <em><u>4 - 1</u></em> - 3 =
Then you will simply subtract 1 from 4, which would equal to 3. So it would be rewritten like this: <em><u>3 - 3</u></em> =
Last step, would be subtracting 3 from 3. Whatever is subtracted from itself would always be 0.
So the answer would be 0!
Answer:
40π/3 cm^2 ≈ 41.89 cm^2
Step-by-step explanation:
The centerline of the shaded region has a radius of 3cm+(4 cm)/2 = 5 cm. Its length is 1/3 that of the circumference of a circle with that radius (because 120° is 1/3 of 360°).
The area is the product of the centerline length and the width of the shaded area (4 cm).
shaded area = (1/3)(2π(5 cm))(4 cm)
shaded area = 40π/3 cm^2 ≈ 41.89 cm^2
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<em>Alternate solution</em>
The area of the donut is the difference in the areas of the outer and inner circles:
πr₂² -πr₁² = π((7 cm)^2 -(3 cm)^2) = 40π cm^2
The are of the shaded region is (120°/360°) = 1/3 of that, so is
shaded area = (40/3)π cm^2
Answer:
See proof below.
Step-by-step explanation:
True
For this case we need to use the following theorem "If 
are eigenvectors of an nxn matrix A and the associated eigenvalues 
are distinct, then 
are linearly independent". Now we can proof the statement like this:
Proof
Let A a nxn matrix and we can assume that A has n distinct real eingenvalues let's say 
From definition of eigenvector for each one 
needs to have associated an eigenvector 
for 
And using the theorem from before , the n eigenvectors 
are linearly independent since the 
are distinct so then we ensure that A is diagonalizable.
No, a triangle can't be made with sides of those lengths.
-- The 10 and the 20 must be hooked together somewhere on the triangle.
One end of the 10 is connected to one end of the 20.
-- Now imagine the distance between the OTHER ends of those sides.
If the 10 and the 20 are folded together so that they both go in the same
direction and there's no angle between them, the closest their OTHER ends
can be is 10 units apart.
-- So if you want to stick a 3rd side between them and make a triangle,
the 3rd side has to be AT LEAST 10 units long. If it's any shorter than
10 units, it can't reach the open ends of the 10 and the 20.
-- Seven is less than 10.
So it can't reach the open ends of both the 10 and the 20 at the same time.
Answer:
Here's what I get
Step-by-step explanation:
I plotted the triangles in the diagram below.
Part A
The scale factor for dilation is ½, because every coordinate has been halved.
P (8, 0) ⟶ P' (4, 0)
Q (6, 2) ⟶ Q' (3, 1)
R (-2, -4) ⟶ R' (-1, -2)
Part B
When you reflect a point (x, y) about the y-axis, the y-coordinate remains the same, but the x-coordinate gets the opposite sign. Thus,
P' (4,0) ⟶ P" (-4,0)
Q' (3,-1) ⟶ Q" (-3,-1)
R' (-1,-2) ⟶ R" (1,-2)
∆P"Q"R" has coordinates P" (-4,0), Q" (-3,-1), R"(1,-2).
Part C
∆PQR and ∆ P"Q"R" are not congruent, because corresponding sides are not equal.
