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3 years ago

Name the quadrilaterals

Mathematics

1 answer:

storchak [24]3 years ago

8 0

Irregular Quadrilateral? I think?

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Please help me! Need answers

RSB [31]

Ok, this definition of f is just a bunch of points (4 to be exact).

so if point (a,b) is part of f, then f(a)=b

f(1) is about point (1,0), so f(1)=0.
g(1) = 1 (due to point (1,1))
g(2/3) = 0
f(2) = 3/4
g(-2) = 3
f(π) = -2

just a lookup once you see what is happening

6 0

3 years ago

A cross between two heterozygous pea plants with yellow seeds

MAXImum [283]

Answer:

1719: 573 = 41.46: 23.93

5 0

2 years ago

[Functions] I just need yes's and no's.......a lot of them, functions just really don't click with me and I don't understand why

Semenov [28]

Answer:

uh-

Step-by-step explanation:

7 0

2 years ago

Read 2 more answers

Which of the following best describes the equation below?

ivann1987 [24]

Answer:

no real solutions

Step-by-step explanation:

2(x + 11) + 5x = 7x - 22

Distribute

2x+22 +5x = 7x -22

Combine like terms

7x +22 = 7x-22

Subtract 7x from each side

7x-7x+22 = 7x-7x-22

22 = -22

This is never true so there are no solutions

4 0

3 years ago

TRIANGLE ABC IS DILATED BY A SCALE FACTOR OF 0.5 WITH THE ORIGIN AS THE CENTER OF DILATION, RESULTING IN THE IMAGE TRIANGLE A'B'

const2013 [10]

Answer: The length of the line B'C" is 1 unit.

Step-by-step explanation:

Given: Triangle ABC is dilated by a scale factor of 0.5 with the origin as the center of dilation , resulting in the image Triangle A'B'C'.

If A (2,2), B= (4,3) and C=(6,3).

Distance between (a,b) and (c,d): $D=\sqrt{(d-b)^2+(c-b)^2}$

Then, BC $=\sqrt{(3-3)^2+(6-4)^2}$

$\\\\=\sqrt{0+2^2}\\\\=\sqrt{4}\\\\=2\text{ units}$

Length of image = scale factor x length in original figure

B'C' = 0.5 × BC

= 0.5 × 2

= 1 unit

Hence, the length of the line B'C" is 1 unit.

5 0

2 years ago