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2 years ago

How far will i travel if i am driving @75kmh for 18 mins. can the distance over speed applies?

Mathematics

1 answer:

Roman55 [17]2 years ago

8 0

Answer:

22.5 km

Step-by-step explanation:

Speed = distance/time

Given speed = 75km/h

time = 18 mins

Distance = ?

Given the speed is in km/h , convert the time from minutes to hour .

18/60 = 0.3 hours

Therefore,

75 = distance/0.3

Cross multiply

75 x 0.3 = distance

22.5 = distance

Distance = 22.5 km

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The lower supports are and the area of the two supports is square meters. The upper arch can be decomposed as one semicircle wit

Galina-37 [17]

Answer:

The lower supports are Congruent Rectangles and the area of the two supports is 24 square meters.

The upper arch can be decomposed as one semicircle with radius 6 meters minus a semicircle with radius 3 meters.

The area of the archway is (13.5 π + 24) square meters.

Step-by-step explanation:

See attachment for the firgure,

In order to determine the area of archway = The area of upper support + area of lower support.

As given, the lower support are two congruent rectangles consists of dimension 3 m × 4 m

Therefore, the area of lower support can be written as,

The area of lower support = 3 × 4 + 3 × 4 = 12 + 12 = 24 square m.

Now, the upper support arch can be decomposed as the two concentric semi circles having radius 3 m and 6 m,

Hence, the area of the upper support = Area of semi circle having radius i.e 6 m - Area of semi circle having radius i.e 3 m

=> π6²/2 - π3²/2= 27π/2= 13.5π square m

Therefore, the area of the archway = (13.5 π + 24) square meters

4 0

3 years ago

M=0 passing through p(-2,-2)

valentinak56 [21]

Answer:

The equation of line is y+2=0

Step-by-step explanation:

The equation of line passing through the point $(x_{1},y_{1} )$ and having slope m is

$y-y_{1} = m (x-x_{1}$

Given slope m=0 and given point is is (-2,-2)

$y-(-2)=0(x-(-2))$

$y+2=0$

therefore equation of the straight line is $y+2=0$

Therefore the straight is parallel to negative x-axis(y=-2)

3 0

3 years ago

I beg of you to help me I need to be able to graduate

Ne4ueva [31]

Answer:

this i not high school work

6 0

2 years ago

Let g(x) = 2x and h(x)= x^2 -4. find (g o h)(0)

finlep [7]

The answer is -8

====================================================

Explanation:

There are two ways to get this answer

Method 1 will have us plug x = 0 into h(x) to get
h(x) = x^2 - 4
h(0) = 0^2 - 4
h(0) = 0 - 4
h(0) = -4
Then this output is plugged into g(x) to get
g(x) = 2x
g(-4) = 2*(-4)
g(-4) = -8 which is the answer
This works because (g o h)(0) is the same as g(h(0)). Note how h(0) is replaced with -4
So effectively g(h(0)) = -8 which is the same as (g o h)(0) = -8

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

The second method involves a bit algebra first
Start with the outer function g(x). Then replace every x with h(x). On the right side, we will replace h(x) with x^2-4 because h(x) = x^2-4

g(x) = 2x
g(x) = 2( x )
g(h(x)) = 2( h(x) ) ... replace every x with h(x)
g(h(x)) = 2( x^2-4 ) ... replace h(x) on the right side with x^2-4
g(h(x)) = 2x^2-8
(g o h)(x) = 2x^2-8

Now plug in x = 0
(g o h)(x) = 2x^2-8
(g o h)(0) = 2(0)^2-8
(g o h)(0) = 2(0)-8
(g o h)(0) = 0-8
(g o h)(0) = -8

Regardless of which method you use, the answer is -8

5 0

3 years ago

Write the product as a sum: 14cos(39x)sin(19x)

charle [14.2K]

Recall that

sin(a + b) = sin(a) cos(b) + cos(a) sin(b)

sin(a - b) = sin(a) cos(b) - cos(a) sin(b)

Adding these together gives

sin(a + b) + sin(a - b) = 2 sin(a) cos(b)

To get 14 cos(39x) sin(19x) on the right side, multiply both sides by 7 and replace a = 19x and b = 39x :

7 (sin(19x + 39x) + sin(19x - 39x)) = 14 cos(39x) sin(19x)

7 (sin(58x) + sin(-20x)) = 14 cos(39x) sin(19x)

7 (sin(58x) - sin(20x)) = 14 cos(39x) sin(19x)

4 0

2 years ago