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3 years ago

Which table shows a function that is decreasing over the interval (-2, 0)?

Mathematics

1 answer:

saveliy_v [14]3 years ago

4 0

Answer:

it is becoming a actual number and not a negative

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A. Solve for y.<br> B. What is the measure of the missing angles?

Alik [6]

Answer: A.) y = 15 B.) (5y + 3)° = 78° (4y + 8)° = 68° and 34°

Steps:
180° - 146° = 34°

180 = 34 + (5y + 3) + (4y + 8)
180 - 34 = (5y + 3) + (4y + 8)
146 = (5y + 3) + (4y + 8)
146 = 5y + 3 + 4y + 8
146 = 9y + 11
146 - 11 = 9y
135 = 9y
135/ 9 = y
15 = y

(5y + 3)
5(15) + 3
75 + 3
78
(5y + 3) = 78

(4y + 8)
4(15) + 8
60 + 8
68
68 = (4y + 8)

Check:
68 + 78 + 34 = 180
180 = 180 ✅

5 0

3 years ago

You have 6 green marbles, 23 blue marbles, and 18 red marbles. what is the probability of choosing a red marble. express your an

AysviL [449]

The probability of choosing a red marble is 18/47.

7 0

3 years ago

Axis of symmetry of y=x^2+6x+7

Ipatiy [6.2K]

x^2 + 6x + 7 = (x + 3)^2 - 2

So the axis of symmetry is x = -3

4 0

3 years ago

The linear combination method gives a solution of (–4, 2) for which of these systems of linear equations?

otez555 [7]

The solution (-4,2) satisfies for the system of linear equations 3x + 13y = 14; 6x + 11y = -2

<u>Step-by-step explanation:</u>

Step 1:

Given detail is the solution of the equations (-4, 2) ie, x= - 4 and y = 2

This implies that this solution should satisfy the given linear equations.

Step 2:

Substitute values of x and y in the equations and verify whether the right hand side equals the left hand side.

System 1 Eq(1) ⇒ LHS = 3(-4) + 13 (2) = -12 + 26 = 14 = RHS

System 1 Eq(2) ⇒ LHS = 6(-4) + 11(2) = -24 + 22 = -2 = RHS

Therefore, the first system of linear equations satisfy the condition.

4 0

3 years ago

Read 2 more answers

Triangle RST has vertices R(2, 0), S(4, 0), and T(1, –3). The image of triangle RST after a rotation has vertices R'(0, –2), S'(

sleet_krkn [62]

Answer with explanation:

Pre -image= Vertices of Δ R ST=R(2, 0), S(4, 0), and T(1, –3)

Image of Δ R ST after rotation= R'(0, –2), S'(0, –4), and T'(–3, –1)

Pre-Image lies in Fourth Quadrant and Image lies in Third Quadrant.

If triangle is rotated by different angles in anticlockwise direction,then

$(a,b)_{90^{\circ}}=(-b, a)\\\\(a,b)_{180^{\circ}}=(-b, -a)\\\\(a,b)_{270^{\circ}}=(b, -a)$

If triangle is rotated by different angles in Clockwise direction,then

$(a,b)_{90^{\circ}}=(b, -a)\\\\(a,b)_{180^{\circ}}=(-b, -a)\\\\(a,b)_{270^{\circ}}=(-b, a)$

⇒→So, Pre image that is ,Δ R ST having vertices ,R(2, 0), S(4, 0), and T(1, –3) when rotated by either 90° in clockwise direction or by 270°, in anticlockwise Direction to get Image Δ R' S'T' having vertices R'(0, –2), S'(0, –4), and T'(–3, –1) .

7 0

3 years ago