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White raven [17]

2 years ago

15

Which ordered pair is a solution to the equation?

1 answer:

Ivenika [448]2 years ago

3 0

Answer:

D

Step-by-step explanation:

none of the option satisfy the equation

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Subtract. Write your answer in simplest form.

vlabodo [156]

Answer: C

Work:

1) Change the Mixed number into an improper fraction
3 x 8 = 24
1+24 = 25/3

2) Get both fractions on equal terms. Find the Greatest Common Denominator. In this case, it’s 24. Multiply by what you need to get 24. 8x3=24 and 12x2=24

3) Multiply Accordingly
25/8 x 3 = 75/24
7/12 x 2 = 14/24

4) Subtract and change back to Mixed Number
75-14 = 61
61/24 = 2 13/24

3 0

2 years ago

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On a map ,1cm represents 30 miles. Find the actual distance between two cities if they are 5.2 cm apart on the map.

Vesna [10]

Multiply 5.2 x 30 and you get 156 miles

6 0

3 years ago

Which formula can be used to describe the sequence below?

saul85 [17]

An=A1+(n-1)d

A2=-8+(2-1)3

A2=-8+3=-5

5 0

3 years ago

Which function has a vertex on the y-axis? f(x) = (x – 2)2 f(x) = x(x + 2) f(x) = (x – 2)(x + 2) f(x) = (x + 1)(x – 2)

strojnjashka [21]

we know that

If the vertex is on the y-axis, then the x-coordinate of the vertex is equal to zero

we are going to verify the vertex of each one of the functions to determine the solution

Remember that

The equation in vertex form of a vertical parabola is equal to

$y=a(x-h)^{2} +k$

where

(h,k) is the vertex

if $a>0$ -------> the parabola open upward (vertex is a minimum)

if $a$ -------> the parabola open downward (vertex is a maximun)

<u>case A)</u> $f(x)=(x-2)^{2}$

This is a vertical parabola open upward

the vertex is the point $(2,0)$

therefore

The function $f(x)=(x-2)^{2}$ does not have a vertex on the y-axis

<u>case B)</u> $f(x)=x(x+2)$

$f(x)=x(x+2)=x^{2}+2x$

convert to vertex form

Complete the square. Remember to balance the equation by adding the same constants to each side.

$f(x)+1=x^{2}+2x+1$

Rewrite as perfect squares

$f(x)+1=(x+1)^{2}$

$f(x)=(x+1)^{2}-1$

the vertex is the point $(-1,-1)$

therefore

The function $f(x)=x(x+2)$ does not have a vertex on the y-axis

<u>case C)</u> $f(x)=(x-2)(x+2)$

$f(x)=(x-2)(x+2)=x^{2}-2^{2}$

$f(x)=x^{2}-4$

the vertex is the point $(0,-4)$

The x-coordinate of the vertex is equal to zero

therefore

The function $f(x)=(x-2)(x+2)$ has a vertex on the y-axis

<u>case D)</u> $f(x)=(x+1)(x-2)$

$f(x)=(x+1)(x-2)\\ \\f(x)= x^{2}-2x+x-2 \\ \\f(x)= x^{2} -x-2$

convert to vertex form

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$f(x)+2= x^{2} -x$

Complete the square. Remember to balance the equation by adding the same constants to each side.

$f(x)+2+0.25= x^{2} -x+0.25$

$f(x)+2.25= x^{2} -x+0.25$

Rewrite as perfect squares

$f(x)+2.25= (x-0.50)^{2}$

$f(x)=(x-0.50)^{2}-2.25$

the vertex is the point $(0.5,-2.25)$

therefore

The function $f(x)=(x+1)(x-2)$ does not have a vertex on the y-axis

<u>the answer is</u>

$f(x)=(x-2)(x+2)$

6 0

3 years ago

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Simplify the following expressions.

baherus [9]

1. 36x + 4
2. 30n + 42
3. -24p + 32
4. 24p -28
5. 15m - 30
6. 5x + 45

8 0

2 years ago