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vekshin1
3 years ago
9

Ava has 80 pictures to put in her photo album . She can fit 4 photos on each page of the album . If the album has 15 pages , doe

s Ava have room for all her photos? yes no Why or why not?
Mathematics
1 answer:
lorasvet [3.4K]3 years ago
8 0

Answer:

No Ava can only put 60 pictures in the photo album.

Step-by-step explanation:

4 x 15 = 60

80 - 60 = 20

Ava needs 5 more pages.

You might be interested in
Drag the tiles to the correct boxes to complete the pairs. Not all tiles will be used. Match each verbal description of a sequen
galben [10]

Answer:

I think the question is wrong so, I will try and explain with some right questions

Step-by-step explanation:

We are give 6 sequences to analyse

1. an = 3 · (4)n - 1

2. an = 4 · (2)n - 1

3. an = 2 · (3)n - 1

4. an = 4 + 2(n - 1)

5. an = 2 + 3(n - 1)

6. an = 3 + 4(n - 1)

1. This is the correct sequence

an=3•(4)^(n-1)

If this is an

Let know an+1, the next term

an+1=3•(4)^(n+1-1)

an+1=3•(4)^n

There fore

Common ratio an+1/an

r= 3•(4)^n/3•(4)^n-1

r= (4)^(n-n+1)

r=4^1

r= 4, then the common ratio is 4

Then

First term is when n=1

an=3•(4)^(n-1)

a1=3•(4)^(1-1)

a1=3•(4)^0=3.4^0

a1=3

The first term is 3 and the common ratio is 4, it is a G.P

2. This is the correct sequence

an=4•(2)^(n-1)

Therefore, let find an+1

an+1=4•(2)^(n+1-1)

an+1= 4•2ⁿ

Common ratio=an+1/an

r=4•2ⁿ/4•(2)^(n-1)

r=2^(n-n+1)

r=2¹=2

Then the common ratio is 2,

The first term is when n =1

an=4•(2)^(n-1)

a1=4•(2)^(1-1)

a1=4•(2)^0

a1=4

It is geometric progression with first term 4 and common ratio 2.

3. This is the correct sequence

an=2•(3)^(n-1)

Therefore, let find an+1

an+1=2•(3)^(n+1-1)

an+1= 2•3ⁿ

Common ratio=an+1/an

r=2•3ⁿ/2•(3)^(n-1)

r=3^(n-n+1)

r=3¹=3

Then the common ratio is 3,

The first term is when n =1

an=2•(3)^(n-1)

a1=2•(3)^(1-1)

a1=2•(3)^0

a1=2

It is geometric progression with first term 2 and common ratio 3.

4. I think this correct sequence so we will use it.

an = 4 + 2(n - 1)

Let find an+1

an+1= 4+2(n+1-1)

an+1= 4+2n

This is not GP

Let find common difference(d) which is an+1 - an

d=an+1-an

d=4+2n-(4+2(n-1))

d=4+2n-4-2(n-1)

d=4+2n-4-2n+2

d=2.

The common difference is 2

Now, the first term is when n=1

an=4+2(n-1)

a1=4+2(1-1)

a1=4+2(0)

a1=4

This is an arithmetic progression of common difference 2 and first term 4.

5. I think this correct sequence so we will use it.

an = 2 + 3(n - 1)

Let find an+1

an+1= 2+3(n+1-1)

an+1= 2+3n

This is not GP

Let find common difference(d) which is an+1 - an

d=an+1-an

d=2+3n-(2+3(n-1))

d=2+3n-2-3(n-1)

d=2+3n-2-3n+3

d=3.

The common difference is 3

Now, the first term is when n=1

an=2+3(n-1)

a1=2+3(1-1)

a1=2+3(0)

a1=2

This is an arithmetic progression of common difference 3 and first term 2.

6. I think this correct sequence so we will use it.

an = 3 + 4(n - 1)

Let find an+1

an+1= 3+4(n+1-1)

an+1= 3+4n

This is not GP

Let find common difference(d) which is an+1 - an

d=an+1-an

d=3+4n-(3+4(n-1))

d=3+4n-3-4(n-1)

d=3+4n-3-4n+4

d=4.

The common difference is 4

Now, the first term is when n=1

an=3+4(n-1)

a1=3+4(1-1)

a1=3+4(0)

a1=3

This is an arithmetic progression of common difference 4 and first term 3.

5 0
3 years ago
Please help!! Question is down below and I’m kinda confused
Arada [10]
It would be the last one
4 0
2 years ago
Which number is irrational?<br> 16.5%<br> 0.0675<br> 8 3/4<br> 3.14
eimsori [14]

Answer:

3.14

Step-by-step explanation:

it will go on foreever

6 0
3 years ago
Solve the formula for C thx if u answer!!!
kati45 [8]

Answer: The answer is C = -P + R

3 0
1 year ago
Read 2 more answers
2y+3x² +5+y+2x+x²+2 what are the coefficients?
saw5 [17]

Answer: 200

The quadratic function f(x) = a(x - h)2 + k, a not equal to zero, is said to be in standard form. If a is positive, the graph opens upward, and if a is negative, then it opens downward. The line of symmetry is the vertical line x = h, and the vertex is the point (h,k).

Any quadratic function can be rewritten in standard form by completing the square. (See the section on solving equations algebraically to review completing the square.) The steps that we use in this section for completing the square will look a little different, because our chief goal here is not solving an equation.

Note that when a quadratic function is in standard form it is also easy to find its zeros by the square root principle.

Example 3.

Write the function f(x) = x2 - 6x + 7 in standard form. Sketch the graph of f and find its zeros and vertex.

f(x) = x2 - 6x + 7.

= (x2 - 6x )+ 7.        Group the x2 and x terms and then complete the square on these terms.

= (x2 - 6x + 9 - 9) + 7.

We need to add 9 because it is the square of one half the coefficient of x, (-6/2)2 = 9. When we were solving an equation we simply added 9 to both sides of the equation. In this setting we add and subtract 9 so that we do not change the function.

= (x2 - 6x + 9) - 9 + 7. We see that x2 - 6x + 9 is a perfect square, namely (x - 3)2.

f(x) = (x - 3)2 - 2. This is standard form.

From this result, one easily finds the vertex of the graph of f is (3, -2).

To find the zeros of f, we set f equal to 0 and solve for x.

(x - 3)2 - 2 = 0.

(x - 3)2 = 2.

(x - 3) = ± sqrt(2).

x = 3 ± sqrt(2).

To sketch the graph of f we shift the graph of y = x2 three units to the right and two units down.

If the coefficient of x2 is not 1, then we must factor this coefficient from the x2 and x terms before proceeding.

Example 4.

Write f(x) = -2x2 + 2x + 3 in standard form and find the vertex of the graph of f.

f(x) = -2x2 + 2x + 3.

= (-2x2 + 2x) + 3.

= -2(x2 - x) + 3.

= -2(x2 - x + 1/4 - 1/4) + 3.

We add and subtract 1/4, because (-1/2)2 = 1/4, and -1 is the coefficient of x.

= -2(x2 - x + 1/4) -2(-1/4) + 3.

Note that everything in the parentheses is multiplied by -2, so when we remove -1/4 from the parentheses, we must multiply it by -2.

= -2(x - 1/2)2 + 1/2 + 3.

= -2(x - 1/2)2 + 7/2.

The vertex is the point (1/2, 7/2). Since the graph opens downward (-2 < 0), the vertex is the highest point on the graph.

Exercise 2:

Write f(x) = 3x2 + 12x + 8 in standard form. Sketch the graph of f ,find its vertex, and find the zeros of f. Answer

Alternate method of finding the vertex

In some cases completing the square is not the easiest way to find the vertex of a parabola. If the graph of a quadratic function has two x-intercepts, then the line of symmetry is the vertical line through the midpoint of the x-intercepts.

The x-intercepts of the graph above are at -5 and 3. The line of symmetry goes through -1, which is the average of -5 and 3. (-5 + 3)/2 = -2/2 = -1. Once we know that the line of symmetry is x = -1, then we know the first coordinate of the vertex is -1. The second coordinate of the vertex can be found by evaluating the function at x = -1.

Example 5.

Find the vertex of the graph of f(x) = (x + 9)(x - 5).

Since the formula for f is factored, it is easy to find the zeros: -9 and 5.

The average of the zeros is (-9 + 5)/2 = -4/2 = -2. So, the line of symmetry is x = -2 and the first coordinate of the vertex is -2.

The second coordinate of the vertex is f(-2) = (-2 + 9)(-2 - 5) = 7*(-7) = -49.

Therefore, the vertex of the graph of f is (-2, -49).

8 0
2 years ago
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