All categories

3 years ago

Distance = 40km. Time = 2h 40 min. Calculate the speed.

Mathematics

1 answer:

PolarNik [594]3 years ago

8 0

Answer: 15 km/h

Step-by-step explanation:

You might be interested in

Solve for n. 7 + 2n || = submit

Lunna [17]

Answer:

Step-by-step explanation:

7+2n = 11

2n = 11-7

2n = 4

n = 4/2

therefore n = 2

8 0

2 years ago

Can someone help me with this?

Mandarinka [93]

Answer:

Step-by-step explanation:

Area of rectangle = length * width

= (x + 4) (5x)

= x *5x + 4 *5x

= 5x² + 20x

Perimeter of rectangle = 2*(length + width)

= 2*(x + 4 + 5x)

= 2*(6x + 4)

= 2*6x + 2 *4

= 12x + 8

4 0

3 years ago

Help please! I need the area of this given figure. The formula is A=1/2(b1+b2)h

Pie

Answer:

108 ft

Correct me, if I am wrong :)

have a great day!

Thanks!

8 0

3 years ago

Read 2 more answers

Find the 2th term of the expansion of (a-b)^4.

vladimir1956 [14]

The second term of the expansion is $-4a^3b$ .

Solution:

Given expression:

$(a-b)^4$

To find the second term of the expansion.

$(a-b)^4$

Using Binomial theorem,

$(a+b)^{n}=\sum_{i=0}^{n}\left(\begin{array}{l}n \\i\end{array}\right) a^{(n-i)} b^{i}$

Here, a = a and b = –b

$$(a-b)^4=\sum_{i=0}^{4}\left(\begin{array}{l}4 \\i\end{array}\right) a^{(4-i)}(-b)^{i}$

Substitute i = 0, we get

$$\frac{4 !}{0 !(4-0) !} a^{4}(-b)^{0}=1 \cdot \frac{4 !}{0 !(4-0) !} a^{4}=a^4$

Substitute i = 1, we get

$$\frac{4 !}{1 !(4-1) !} a^{3}(-b)^{1}=\frac{4 !}{3!} a^{3}(-b)=-4 a^{3} b$

Substitute i = 2, we get

$$\frac{4 !}{2 !(4-2) !} a^{2}(-b)^{2}=\frac{12}{2 !} a^{2}(-b)^{2}=6 a^{2} b^{2}$

Substitute i = 3, we get

$$\frac{4 !}{3 !(4-3) !} a^{1}(-b)^{3}=\frac{4}{1 !} a(-b)^{3}=-4 a b^{3}$

Substitute i = 4, we get

$$\frac{4 !}{4 !(4-4) !} a^{0}(-b)^{4}=1 \cdot \frac{(-b)^{4}}{(4-4) !}=b^{4}$

Therefore,

$$(a-b)^4=\sum_{i=0}^{4}\left(\begin{array}{l}4 \\i\end{array}\right) a^{(4-i)}(-b)^{i}$

$=\frac{4 !}{0 !(4-0) !} a^{4}(-b)^{0}+\frac{4 !}{1 !(4-1) !} a^{3}(-b)^{1}+\frac{4 !}{2 !(4-2) !} a^{2}(-b)^{2}+\frac{4 !}{3 !(4-3) !} a^{1}(-b)^{3}+\frac{4 !}{4 !(4-4) !} a^{0}(-b)^{4}$ $=a^{4}-4 a^{3} b+6 a^{2} b^{2}-4 a b^{3}+b^{4}$

Hence the second term of the expansion is $-4a^3b$ .

3 0

3 years ago

X+y+z=3 x-y+z=2 x+y-z=1

Dima020 [189]

Answer:

Step-by-step explanation:

I'm assuming you are solving this system for x, y, and z. Do it this way:

Combine the first 2 equations and the second 2 equations and eliminate the y's. The first 2 equations:

x + y + z = 3

x - y + z = 2

which solves to

2x + 2z = 5

The second 2 equations:

x - y + z = 2

x + y - z = 1

which solves to

2x = 3 so

x = 3/2 or 1.5

Now sub that x value into the bold equation above to get:

2(1.5) + 2z = 5 and

3 + 2z = 5 and

2z = 2 so

z = 1

Now we plug back in to solve for y:

1.5 + y + 1 = 3 and

2.5 + y = 3 so

y = .5

The solution set then is (1.5, .5, 1) or in fraction form:

$(\frac{3}{2},\frac{1}{2},1)$

6 0

3 years ago

Distance = 40km. Time = 2h 40 min. Calculate the speed.​

Distance = 40km. Time = 2h 40 min. Calculate the speed.