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3 years ago

3) A box of tile contains 12 tiles. If

Mathematics

1 answer:

vichka [17]3 years ago

3 0

Answer:

3 tiles left

Step-by-step explanation:

You can only use a maximum of 9 tiles to make a perfect square. You would need 7 more tiles to make a larger square. Having used 9 tiles, you will have 3 leftover. (12 - 9 = 3)

Hope I helped!

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SOLVE THE QUESTION BELOW ASAP

qwelly [4]

Answer:

Part A) The graph in the attached figure (see the explanation)

Part B) 16 feet

Part C) see the explanation

Step-by-step explanation:

Part A) Graph the function

Let

h(t) ----> the height in feet of the ball above the ground

t -----> the time in seconds

we have

$h(t)=-16t^{2}+98$

This is a vertical parabola open downward (the leading coefficient is negative)

The vertex is a maximum

To graph the parabola, find the vertex, the intercepts, and the axis of symmetry

<em>Find the vertex</em>

The function is written in vertex form

The vertex is the point (0,98)

Find the y-intercept

The y-intercept is the value of the function when the value of t is equal to zero

For t=0

$h(t)=-16(0)^{2}+98$

$h(0)=98$

The y-intercept is the point (0,98)

Find the t-intercepts

The t-intercepts are the values of t when the value of the function is equal to zero

For h(t)=0

$-16t^{2}+98=0$

$t^{2}=\frac{98}{16}$

square root both sides

$t=\pm\frac{\sqrt{98}}{4}$

$t=\pm7\frac{\sqrt{2}}{4}$

therefore

The t-intercepts are

$(-7\frac{\sqrt{2}}{4},0), (7\frac{\sqrt{2}}{4},0)$

$(-2.475,0), (2.475,0)$

Find the axis of symmetry

The equation of the axis of symmetry in a vertical parabola is equal to the x-coordinate of the vertex

$x=0$ ----> the y-axis

To graph the parabola, plot the given points and connect them

we have

The vertex is the point $(0,98)$

The y-intercept is the point $(0,98)$

The t-intercepts are $(-2.475,0), (2.475,0)$

The axis of symmetry is the y-axis

The graph in the attached figure

Part B) How far is the artifact fallen from the time t=0 to time t=1

we know that

For t=0

$h(t)=-16(0)^{2}+98$

$h(0)=98\ ft$

For t=1

$h(t)=-16(1)^{2}+98$

$h(1)=82\ ft$

Find the difference

$98\ ft-82\ ft=16\ ft$

Part C) Does the artifact fall the same distance from time t=1 to time t=2 as it does from the time t=0 to time t=1?

we know that

For t=1

$h(t)=-16(1)^{2}+98$

$h(1)=82\ ft$

For t=2

$h(t)=-16(2)^{2}+98$

$h(2)=34\ ft$

Find the difference

$82\ ft-34\ ft=48\ ft$

The artifact fall 48 feet from time t=1 to time t=2 and fall 16 feet from time t=0 to time t=1

therefore

The distance traveled from t=1 to t=2 is greater than the distance traveled from t=0 to t=1

8 0

3 years ago

5n + 3n<br> simplify each expression

Setler79 [48]

Answer: 8n

Step-by-step explanation: 5n + 3n

3 0

3 years ago

Which algebraic expression is equivalent to the expression below?

Anarel [89]

Answer:

A. 4x²+20x+25

Step-by-step explanation:

Because of PEMDAS, you calculate the value inside of the parenthesis first, so you expand [2x+1-(4x+6)]:

2x+1-4x-6 = -2x-5.

Then apply the exponent:

(-2x-5)² = (-2x-5)(-2x-5) = 4x²+10x+10x+25 = 4x²+20x+25

4 0

3 years ago

Read 2 more answers

100 POINTSbrainlest to the person who get it right

Zina [86]

Answer:

[1] x² - 4/2x - 4

3² - 4/2 × 3 - 4 = 9 - 4/2
5/2

[2] 4 - 3x/2x + 4

4 - 3 × 2/2 × 2 + 4 = 2/8
1/4

[3] x²/2x² - 2

0²/2 × 0² - 2
0

[4] x² + 5/5 - x

- 2² + 5/5 + 2 = 4 + 5/7
9/7

[5] - x³/x² - 4

- 1³/-1² - 4 = - 1/1 - 4
1/3

[6] x + 3/x³ - 1

- 3 + 3/- 3³ - 1 = 0/- 27 - 1
0

[7] 5 - x²/3x²

5 - (- 4)²/3 × - 4² = 5 - 16/3 × 16
- 11/48

[8] - 2x + 6/2x³ - 3

- 2 × - 2 + 6/2 × - 2³ - 3
10/- 19

[9] 4x²/- 3x² + 1

4 × 1²/- 3 × 1² + 1 = 4/- 2
- 2

[10] 4x - 1/6x + 1

4 × 3 - 1/6 × 3 + 1
11/19

[11] x² - 1/x² + 1

1² - 1/1² + 1 = 0/2
0

8 0

3 years ago

Read 2 more answers

Please write the solution

GrogVix [38]

Answer:

Step-by-step explanation:

that is the answer to the question above

3 0

2 years ago