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3 years ago

What is 98.86-85.15 show me answers

Mathematics

1 answer:

tiny-mole [99]3 years ago

7 0

98.86-85.15= 13.71 This is the answer to the problem. It's really easy. U just subtract and that's it.

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The sum of the ages of Nicole and Rachel is 30. Currently Nicole is four years older than Rachel. How old is Rachel?

Lunna [17]

Rachel is 13. since Nicole is 17.
13+17=30

3 0

2 years ago

Read 2 more answers

Find the length of line segment UV.

pshichka [43]

Answer:

The length of the line segment UV is 76 units

Step-by-step explanation:

In a triangle, the line segment joining the mid-points of two sides is parallel to the third side and equal to half its length

In Δ ONT

∵ U is the mid-point of ON

∵ V is the mid-point of TN

→ That means UV is joining the mid-points of two sides

∴ UV // OT

∴ UV = $\frac{1}{2}$ OT

∵ UV = 7x - 8

∵ OT = 12x + 8

∴ 7x - 8 = $\frac{1}{2}$ (12x + 8)

→ Multiply the bracket by $\frac{1}{2}$

∵ $\frac{1}{2}$ (12x + 8) = $\frac{1}{2}$ (12x) + $\frac{1}{2}$ (8) = 6x + 4

∴ 7x - 8 = 6x + 4

→ Add 8 to both sides

∴ 7x - 8 + 8 = 6x + 4 + 8

∴ 7x = 6x + 12

→ Subtract 6x from both sides

∴ 7x - 6x = 6x - 6x + 12

∴ x = 12

→ Substitute the value of x in the expression of UV to find it

∵ UV = 7(12) - 8 = 84 - 8

∴ UV = 76

∴ The length of the line segment UV is 76 units

8 0

2 years ago

Answer to this question please

fredd [130]

Answer:

Volume is 21

Explanation

Base length is 4.5

Widgth is 3.5

and height is 4 (I can't tell if its 2 because picture is blurry)

V=lwh/3=4.5·3.5·4= 21

8 0

2 years ago

I need help solving this equation<br> 4n-9=2(5+2n)

Nastasia [14]

First, use distributive property on the right half.

2 * 5 = 10

2 * 2n = 4n

4n - 9 = 10 + 4n

Add 9 to both sides

4n = 19 + 4n

Subtract 4n from both sides

0 = 19

But thats not true. Therefore, there is no solution.

4 0

3 years ago

A ship leaves port at noon and has a bearing of S29oW. The ship sails at 20 knots. How many nautical miles south and how many na

ira [324]

Answer:

Approximately $58.2\; \text{nautical miles}$ (assuming that the bearing is ${\rm S$29^{\circ}$W}$ .)

Step-by-step explanation:

Let $v$ denote the speed of the ship, and let $t$ denote the duration of the trip. The magnitude of the displacement of this ship would be $v\, t$ .

Refer to the diagram attached. The direction ${\rm S$29^{\circ}$W}$ means $29^{\circ}$ west of south. Thus, start with the south direction and turn towards west (clockwise) by $29^{\circ}$ to find the direction of the displacement of the ship.

The hypothenuse of the right triangle in this diagram represents the displacement of the ship, with a length of $v\, t$ . The dashed horizontal line segment represents the distance that the ship has travelled to the west (which this question is asking for.) The angle opposite to that line segment is exactly $29^{\circ}$ .

Since the hypotenuse is of length $v\, t$ , the dashed line segment opposite to the $\theta = 29^{\circ}$ vertex would have a length of:

$\begin{aligned}& \text{opposite (to $\theta$)} \\ =\; & \text{hypotenuse} \times \frac{\text{opposite (to $\theta$)}}{\text{hypotenuse}} \\ =\; & \text{hypotenuse} \times \sin (\theta) \\ =\; & v\, t \, \sin(\theta) \\ =\; & v\, t\, \sin(29^{\circ})\end{aligned}$ .

Substitute in $\begin{aligned} v &= 20\; \frac{\text{nautical mile}}{\text{hour}}\end{aligned}$ and $t = 6\; \text{hour}$ :

$\begin{aligned} & v\, t\, \sin(29^{\circ}) \\ =\; & 20\; \frac{\text{nautical mile}}{\text{hour}} \times 6\; \text{hour} \times \sin(29^{\circ}) \\ \approx\; & 58.2\; \text{nautical mile}\end{aligned}$ .

7 0

2 years ago