Question 3

How many square meters of brick pavements must be laid out for a 4 meter wide walk around a circular flower bed 22 meters wide i

Ronch [10]

Answer:

603.2 meters

Step-by-step explanation:

circle - the outside of the walk in this case. This circle has a

diameter of 22 + 8 m, therefore a radius of 15m. The inner circle

has a diameter of 22m - as given - a radius of 11m.

You subtract the area of the inner circle from that of the larger circle.

4 0

3 years ago

Read 2 more answers

Can someone help me please

Schach [20]

Answer:

Step-by-step explanation:

The system of equations is expressed as

y=-2x+7 - - - - - - - - 1

y=5x-7 - - - - - - - - -2

We would equate equation 1 and equation 2. It becomes

- 2x + 7 = 5x - 7

We would add 2x to the left hand side and the right hand side of the equation. Also add 7 to the left hand side and the right hand side of the equation. It becomes

- 2x + 2x + 7 + 7 = 5x + 2x - 7 + 7

14 = 7x

Dividing the left hand side and the right hand side of the equation by 7, it becomes

7x/7 = 14/2

x = 2

y= -2 × 2 + 7

y = - 4 + 7

y = 3

7 0

4 years ago

Eight adults and two children need to cross a river. A small boat is available that can hold one adult or one or two children. E

sleet_krkn [62]

Answer:

A = 8, C = 2 => T = 33

A=100, C = 2 => T = 401

A = any, C = 2 => T = 4A + 1

A = 8, C = 4 => T = 37

As long as the boat can carry one adult or one or two children, the following equation should solve the problem.

T = 4A + 2(C-2) + 1

Step-by-step explanation:

To transfer one adult, 4 one-way trips need to be made as follows:

1. Two children cross the river

2. One of them takes the boat back to the other side

3. One adult crosses the river

4. The other child takes the boat back

These four steps are repeated until all adults are crossed. At this point no children has crossed yet.

Then for every 2 children except the last 2, 2 one-way trips need to be made if the number of children is greater than 2:

1. Two children cross

2. One takes the boat back

For the last 2,only one trip needs to be made.

A = number of adults

C = number of children >= 2

T = total trips

T = 4A + 2(C-2) + 1

4 0

3 years ago

4. Solve the equation using the quadratic formula.

Vikentia [17]

Answer:

A. x = -2, x = 1.25

Step-by-step explanation:

Use the quadratic formula

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

Once in standard form, identify a, b, and c from the original equation and plug them into the quadratic formula.

4x² + 3x - 10 = 0

a = 4

b = 3

c = - 10

x = $\frac{-3+\sqrt{3^{2}-4x4(-10) } }{2x4}$

Simplify

Evaluate the exponent

x = $\frac{-3+\sqrt{9-4x4(-10)} }{2x4}$

Multiply the numbers

x = $\frac{-3+\sqrt{9+160} }{2x4}$

Add the numbers

x = $\frac{-3+\sqrt{169} }{2x4}$

Evaluate the square root

x = $\frac{-3+13}{2x4}$

Multiply the numbers

x = $\frac{-3+13}{8}$

Separate the equations

To solve for the unknown variable, separate into two equations: one with a plus and the other with a minus.

x = $\frac{-3+13}{8}$

x = $\frac{-3-13}{8}$

Solve

Rearrange and isolate the variable to find each solution

x = - 2

x = 1.25

6 0

3 years ago

Read 2 more answers

If S_1=1,S_2=8 and S_n=S_n-1+2S_n-2 whenever n≥2. Show that S_n=3⋅2n−1+2(−1)n for all n≥1.

Snezhnost [94]

You can try to show this by induction:

• According to the given closed form, we have $S_1=3\times2^{1-1}+2(-1)^1=3-2=1$ , which agrees with the initial value S₁ = 1.

• Assume the closed form is correct for all n up to n = k. In particular, we assume

$S_{k-1}=3\times2^{(k-1)-1}+2(-1)^{k-1}=3\times2^{k-2}+2(-1)^{k-1}$

and

$S_k=3\times2^{k-1}+2(-1)^k$

We want to then use this assumption to show the closed form is correct for n = k + 1, or

$S_{k+1}=3\times2^{(k+1)-1}+2(-1)^{k+1}=3\times2^k+2(-1)^{k+1}$

From the given recurrence, we know

$S_{k+1}=S_k+2S_{k-1}$

so that

$S_{k+1}=3\times2^{k-1}+2(-1)^k + 2\left(3\times2^{k-2}+2(-1)^{k-1}\right)$

$S_{k+1}=3\times2^{k-1}+2(-1)^k + 3\times2^{k-1}+4(-1)^{k-1}$

$S_{k+1}=2\times3\times2^{k-1}+(-1)^k\left(2+4(-1)^{-1}\right)$

$S_{k+1}=3\times2^k-2(-1)^k$

$S_{k+1}=3\times2^k+2(-1)(-1)^k$

$\boxed{S_{k+1}=3\times2^k+2(-1)^{k+1}}$

which is what we needed. QED

6 0

3 years ago