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Alenkasestr [34]
3 years ago
8

I need help on this question plz

Mathematics
1 answer:
soldi70 [24.7K]3 years ago
8 0

Cylinders are 3D figures, as they are solids with volume and take up space.

2D figures are flat figures that you can draw, like a square or circle.

1D figures are actually just a line, a segment, or a point, etc.

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Dean's Diner offers its clients a choice of regular and diet soda. Last night, the diner served 50 sodas in all, 90% of which we
Montano1993 [528]

Answer:

The diner served 45 regular sodas

Step-by-step explanation:

90% of 50 is 45 so the answer is 45

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2 years ago
Suzanne wants to put a fence around her square garden. If the garden covers an area of 64 ft^2, how many feet of fencing does sh
Art [367]

Answer:

32 feet of fence

Step-by-step explanation:

The garden is a square, so the length and width are equal.

A = L x W

W = L

A = W²

A = 64

W² = 64

W = √64 = ±8 -- (-8) would be extraneous

W = 8 ft

L = 8ft

Since she is putting the fence around the garden, she needs to find the perimeter.

P = 2L + 2W

P = 2(8) + 2(8)

P = 16 + 16

P = 32 ft

32 feet of fence

8 0
3 years ago
Which table represents a nonlinear function? x y 2 -9 4 1 6 11 x y 2 -14 4 -16 6 -18 x y 2 0 4 6 6 16 x y 2 -9 4 -6 6 -3
OverLord2011 [107]
The Third one, Y jumps from 0 to 6 then 16
8 0
3 years ago
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True or False: 3D figures have volume.
Ksivusya [100]

Answer:

True

Step-by-step explanation:

7 0
3 years ago
Prove that the roots of x2+(1-k)x+k-3=0 are real for all real values of k​
masha68 [24]

Answer:

Roots are not real

Step-by-step explanation:

To prove : The roots of x^2 +(1-k)x+k-3=0x

2

+(1−k)x+k−3=0 are real for all real values of k ?

Solution :

The roots are real when discriminant is greater than equal to zero.

i.e. b^2-4ac\geq 0b

2

−4ac≥0

The quadratic equation x^2 +(1-k)x+k-3=0x

2

+(1−k)x+k−3=0

Here, a=1, b=1-k and c=k-3

Substitute the values,

We find the discriminant,

D=(1-k)^2-4(1)(k-3)D=(1−k)

2

−4(1)(k−3)

D=1+k^2-2k-4k+12D=1+k

2

−2k−4k+12

D=k^2-6k+13D=k

2

−6k+13

D=(k-(3+2i))(k+(3+2i))D=(k−(3+2i))(k+(3+2i))

For roots to be real, D ≥ 0

But the roots are imaginary therefore the roots of the given equation are not real for any value of k.

6 0
3 years ago
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