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3 years ago

15

Suppose that Cecil, the tightrope walker, can only travel lengths of 5 feet, 6 feet, and 8 feet.

1 answer:

d1i1m1o1n [39]3 years ago

7 0

Answer:

The two expressions are;

1) -8 ft. + 3 × 6 ft. = 12 feet

2) -6 ft. + 2 × 5 ft. + 8 ft. = 12 feet

Step-by-step explanation:

The given parameters are;

The lengths in feet Cecil can travel = 5 feet, 6 feet, and 8 feet

The distances Cecil can travel forward = (positive 5, 6, 8)

Backwards Cecil can travel distances of (-5, -6, -8)

Cecil can go forward past the ladder and return back

Therefore, the number expressions by which Cecil can cross a tightrope 12 feet long are;

1) -8 ft. + 3 × 6 ft. = 12 feet

Which is to go back 8 feet and then to go forward 6 feet three times

2) -6 ft. + 2 × 5 ft. + 8 ft. = 12 feet

Which is to go back 6 feet then to go forward in 5 feet two times and then in forward again in 8 feet.

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If x is a positive integer, for how many different values of x is

ValentinkaMS [17]

I think it’s 6 I’m sorry if it’s wrong

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2 years ago

Given the list of ordered pairs, what is the x intercept (8,10),(3,-4),(0,8),(4,-3),(9,0)

Naddik [55]

Answer:

The x-intercept is the ordered pair (9,0)

Step-by-step explanation:

we know that

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

so

The x-intercept is a ordered pair with a y-coordinate equal to zero

therefore

In this problem

The x-intercept is the ordered pair (9,0)

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3 years ago

P is a point (8,11). Q is a point on the y-axis so that PQ=10. Find the coordinates of Q.

Annette [7]

Answer:

The possible co-ordinates of the point Q are (0,5) and (0,17).

Step-by-step explanation:

Given:

P is a point (8,11)

Q is point on y-axis

PQ = 10 units

To find co-ordinates of point Q.

Solution:

Any point on y-axis is given as $(0,y)$ as $x=0$ at y-axis.

Let the point Q be = $(0,y)$

We use the distance formula to find length of PQ.

By distance formula:

The distance between two points $(x_1,y_1)$ and $(x_2,y_2)$ is given as:

$d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$

Thus, for the point P(8,11) and Q(0, $y$ ) the distance PQ can be given as:

$PQ=\sqrt{(8-0)^2+(11-y)^2}$

$PQ=\sqrt{(8)^2+(11-y)^2}$

$PQ=\sqrt{64+(11-y)^2}$

Substituting PQ=10 units.

$10=\sqrt{64+(11-y)^2}$

Squaring both sides.

$10^2=(\sqrt{64+(11-y)^2})^2$

$100=64+(11-y)^2$

Subtracting both sides by 64.

$100-64=64-64+(11-y)^2$

$36=(11-y)^2$

Taking square root both sides.

$\sqrt{36}=\sqrt{(11-y)^2}$

$\pm6=11-y$

So, we have two equations to solve:

$6=11-y$ and $-6=11-y$

Adding $y$ both sides.

$6+y=11-y+y$ and $-6+y=11-y+y$

$6+y=11$ and $-6+y=11$

Subtracting both sides by 6 for one equation and adding 6 both sides for the other equation.

$6-6+y=11-6$ and $-6+6+y=11+6$

∴ $y=5$ and $y=17$

Thus, the possible co-ordinates of the point Q are (0,5) and (0,17).

6 0

3 years ago

The diagonals of a rhombus are 30cm and 16cm. Find its area, the length of a side and its perimeter.

Free_Kalibri [48]

Answer:

Area = 1/2 × product of diagonalsso area = 1/2 × 30 ×16 = 240 cm²

also diagonals bisect each other at right angles so 1/2×30= 15cm,, 1/2× 16 = 8cm

therefore by pythagoras theorem,

${15 }^{2}$ + ${8}^{2}$ = ${side}^{2}$ $\sqrt{289 = side}$

side = 17

Perimeter = 4×17 =68 cm.

5 0

2 years ago

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What is the sum of all of the integers from -4 to +5?

sukhopar [10]

1 Hope this helps you

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2 years ago

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