All categories

3 years ago

First down = 25 yards.

Mathematics

1 answer:

valentinak56 [21]3 years ago

3 0

This is the question.

At the start of the first down, the football was 25 yards from the Captains' end zone. During three downs, the ball moved 6 yards closer to their end zone, then 10 yards farther from it, and then 8 yards closer to it. How many yards from their end. zone was the ball at the end of the third down?

You might be interested in

Help anyone can help me do 16 and 17 question,I will mark brainlest.The no 16 question is find the area of the shaded region

yanalaym [24]

Answer:

Question 16 = 22

Question 17 = 20 cm²

Step-by-step explanation:

Concepts:

Area of Square = s²

s = side

Area of Triangle = bh/2

b = base
h = height

Diagonals of the square are congruent and bisect each other, which forms a right angle with 90°

Segment addition postulate states that given 2 points A and C, a third point B lies on the line segment AC if and only if the distances between the points satisfy the equation AB + BC = AC.

Solve:

Question # 16

Step One: Find the total area of two squares

Large square: 5 × 5 = 25

Small square: 2 × 2 = 4

25 + 4 = 29

Step Two: Find the area of the blank triangle

b = 5 + 2 = 7

h = 2

A = bh / 2

A = (7) (2) / 2

A = 14 / 2

A = 7

Step Three: Subtract the area of the blank triangle from the total area

Total area = 29

Area of Square = 7

29 - 7 = 22

-----------------------------------------------------------

Question # 17

Step One: Find the length of PT

Given:

PR = 4 cm
RT = 6 cm

PT = PR + RT [Segment addition postulate]

PT = (4) + (6)

PT = 10 cm

Step Two: Find the length of S to PT perpendicularly

According to the diagonal are perpendicular to each other and congruent. Therefore, the length of S to PT perpendicularly is half of the diagonal

Length of Diagonal = 4 cm

4 ÷ 2 = 2 cm

Step Three: Find the area of ΔPST

b = PT = 10 cm

h = S to PT = 2 cm

A = bh / 2

A = (10)(2) / 2

A = 20 / 2

A = 10 cm²

Step Four: Find the length of Q to PT perpendicularly

Similar to step two, Q is the endpoint of one diagonal, and by definition, diagonals are perpendicular and congruent with each other. Therefore, the length of Q to PT perpendicularly is half of the diagonal.

Length of Diagonal = 4 cm

4 ÷ 2 = 2 cm

Step Five: Find the area of ΔPQT

b = PT = 10 cm

h = Q to PT = 2 cm

A = bh / 2

A = (10)(2) / 2

A = 20 / 2

A = 10 cm²

Step Six: Combine area of two triangles to find the total area

Area of ΔPST = 10 cm²

Area of ΔPQT = 10 cm²

10 + 10 = 20 cm²

Hope this helps!! :)

Please let me know if you have any questions

6 0

3 years ago

Which equation has the same solution as this equation?

lesya [120]

Answer:

3rd Option is correct.

Step-by-step explanation:

Given Equation:

x² - 16x + 12 = 0

First We need to find solution of the given equation.

x² - 16x + 12 = 0

here, a = 1 , b = -16 & c = 12

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

$x=\frac{-(-16)\pm\sqrt{(-16)^2-4(12)}}{2}$

$x=\frac{16\pm\sqrt{256-48}}{2}$

$x=\frac{16\pm\sqrt{208}}{2}$

$x=\frac{16\pm4\sqrt{13}}{2}$

$x=8+2\sqrt{13}\:\:andx=8-2\sqrt{13}$

Now,

Option 1).

( x - 8 )² = 144

x - 8 = ±√144

x - 8 = ±12

x = 8 + 12 = 20 and x = 8 - 12 = -4

Thus, This is not correct Option.

Option 2).

( x - 4 )² = 4

x - 4 = ±√4

x - 4 = ±2

x = 4 + 2 = 6 and x = 4 - 2 = 2

Thus, This is not correct Option.

Option 3).

( x - 8 )² = 52

x - 8 = ±√52

x - 8 = ±2√13

x = 8 + 2√13 and x = 8 - 2√13

Thus, This is correct Option.

Option 4).

( x - 4 )² = 16

x - 4 = ±√116

x - 4 = ±4

x = 4 + 4 = 8 and x = 4 - 4 = 0

Thus, This is not correct Option.

Therefore, 3rd Option is correct.

8 0

3 years ago

What is the length of the diagonal (d) is the solid below

elena-14-01-66 [18.8K]

420!!!!!!!!!!!!!!!!!!

8 0

3 years ago

2.375 in expanded form

kari74 [83]

Answer:

2375^-3

Step-by-step explanation:

Move the decimal place from 2.375 to the back of the number so it should look like this, 2375.

Than move the decimal place back to its original spot and count the amount of spaces you move. Remember, if you move forward, your exponent should be negative. For example, if the problem is 6.321, you would move the decimal place to the back like this, 6321. And then move it back to where it originally was, 6.321, the amount of spaces I moved was 3 spaces and I moved forward so It would be 6.321^-3

6 0

3 years ago

7 ∙ 15 can be written as _____. 5(7 + 10) 10(7 + 5) 7(10 + 5)

Bogdan [553]

Both side multiply by one number.

8 0

3 years ago

Read 2 more answers