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1 year ago

Which construction is illustrated above?

Mathematics

1 answer:

ioda1 year ago

3 0

Answer:

a perpendicular to a given line from a point on the line

Step-by-step explanation:

The line which is being constructed in the image is a line parallel to a given line through a point not on the line. It is clear that the line being constructed is neither perpendicular nor congruent but parallel to the given line because the two lines never intersect and are always the same distance apart.

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Which inequality is equivalent to |x-4|<9

Lorico [155]

Answer with explanation:

The Given Inequality is

|x-4|<9

Modulus of a number yields always a positive value.

→|x-4|= x-4, when, x-4≥0

That is , x≥4.

= -(x-4), when, x-4<0

means, x<4.

it can be Written as

→ -9<x-4<9

→ -9+4 < x -4+4 < 9+4

→ -5<x<13

→x∈(-5,13)

8 0

3 years ago

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I need help with this.

Dimas [21]

0.5-2(0)^2
0.5-0
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7 0

3 years ago

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Given the function f(x)=-5x^2+2x+9,find f(1) and f(2) Choose the statement that is true concerning these two values

SIZIF [17.4K]

F(1) = -5(1)^2+2(1)+9 = 6
f(2) = -5(2)^2+2(2)+9 = -9

So, because f(1) = 6 is positive and f(2) = -9 is negative, f(1) > f(2).

f(1) is greater than f(2)!

6 0

2 years ago

Read 2 more answers

Can anyone solve this x + 2 ( x + 2 + ( 2 x − 8 ) ) = x + 1 + ( 2 x − 5 ) x + 2 ( x + 2 + ( 2 x - 8 ) ) = x + 1 + ( 2 x - 5 )

r-ruslan [8.4K]

Answer:

simplify the equation to solve it, remove unnessecary parentheses and spaces.

5 0

2 years ago

Find the error in the following sample of work, if one exists.

ch4aika [34]

Answer:

D. There is no mistake.

Step-by-step explanation:

The following lines show the process of factorization by using common factor.

Line 1:

In line 1, the equation is given and is completely fine.

$y^{2}+8xy+2y+16x=0$

The only thing missing was equate to zero, but the options below talk about correct factors only, therefore this can't be considered as a mistake and can be ignored completely.

Line 2:

In line 2, the terms are grouped, from which we can factor out common terms.

$(y^{2}+8xy)+(2y+16x)=0$

This is also fine.

Line 3:

In line 3, the common term y is taken out from group 1 and 2 from other group.

$y(y+8x)+2(y+8x)=0$

which is exactly what is given in line 3.

Line 4:

In line 4 the common factors can be seen and easily split into 2 factors.

$(y+2)(y+8x)=0$

which is exactly what is given in line 4.

Options:

A. The grouping is correct in line 2. So this option is does not hold.

B. Common factor was factored correctly from group 1. So this option does not hold.

C. Common factor was factored correctly from group 2. So this option does not hold.

D. There is no mistake. This is correct. Thus we choose this option as correct answer.

3 0

2 years ago