All categories

3 years ago

What is the circle's circumference? The diameter of a circle is 6 millimetres.

Mathematics

1 answer:

dolphi86 [110]3 years ago

6 0

Answer:

C = 6pi or approximately 18.84 mm

Step-by-step explanation:

The circumference of a circle is

C = pi *d

C = pi *6

We can approximate pi by 3.14

C = 3.14*6

C =18.84

You might be interested in

Find the range of f(x) = –x + 4 for the domain {–3, –2, –1, 1}.

marishachu [46]

Answer:

$\boxed{ \{7, 6, 5, 3 \} }$

Step-by-step explanation:

The domain is all possible values for x.

The range is all possible values for f(x) or y.

The domain given is {-3, -2, -1, 1}.

Plug x as {-3, -2, -1, 1} and find the f(x) or y values.

$f(-3)=-(-3)+4=7\\f(-2)=-(-2)+4=6\\f(-1)=-(-1)+4=5\\f(1)=-(1)+4=3$

The range is {7, 6, 5, 3}, when the domain is {-3, -2, -1, 1}.

5 0

2 years ago

What is the area of this trapezoid?

Ivanshal [37]

Answer: 357 cm²

Step-by-step explanation: In this problem, were asked to find the area of the trapezoid shown. Remember that a trapezoid is a quadrilateral with one pair of parallel sides. The formula for the area of a trapezoid is shown below

The b's are the bases or the parallel sides and the h represents the height.

So in the trapezoid shown, the bases are 15 cm and 27 cm and the height is 17 cm.

Plugging this information into the formula, we have .

Next, the order of operations tells us that we must simplify inside the parentheses first. 15 cm + 27 cm is 42 cm and we have

1/2 × 42 cm is 21 cm and we have (21 cm)(17 cm) which is 357 cm².

So the area of the trapezoid shown is 357 cm².

5 0

3 years ago

Carys is checking her tax bill for the last year .the tax rates were as follows: no tax on the first 11000 of earnings .earnings

faust18 [17]

Answer:

$7,440.

Step-by-step explanation:

We have been given the tax rates as follows:

$0-$11,000 = No tax.

$11,000-$43,000 = 20%.

$43,000-$150,000 = 40%.

Earnings over $150,000 = 45%.

Since last year Carys earned $45,600, so we will have to find the tax paid by her for 11,000-43,000 at rate of 20%, then the tax for amount over 43,000 at the rate of 40%.

$\text{The amount of tax paid by Carys for 11000-43000}=\frac{20}{100}\times (43,000-11,000)$

$\text{The amount of tax paid by Carys for 11000-43000}=0.20\times (32,000)$

$\text{The amount of tax paid by Carys for 11000-43000}=6400$

$\text{The amount of tax paid by Carys for the amount over 43000}=\frac{40}{100}\times (45,600-43,000)$

$\text{The amount of tax paid by Carys for the amount over 43000}=0.40\times (2600)$

$\text{The amount of tax paid by Carys for the amount over 43000}=1040$

$\text{The total amount of tax paid by Carys}=6400+1040$

$\text{The total amount of tax paid by Carys}=7440$

Therefore, Carys paid an amount of $7,440 in tax.

6 0

2 years ago

Read 2 more answers

Ben and Josh went to the roof of their 40-foot tall high school to throw their math books offthe edge.The initial velocity of Be

Taya2010 [7]

Answer

Josh's textbook reached the ground first

Josh's textbook reached the ground first by a difference of $t=0.6482$

Step-by-step explanation:

Before we proceed let us re write correctly the height equation which in correct form reads:

$h(t)=-16t^2 +v_{o}t+s$ Eqn(1).

Where:

$h(t)$ : is the height range as a function of time

$v_{o}$ : is the initial velocity

$s$ : is the initial heightin feet and is given as 40 feet, thus Eqn(1). becomes:

$h(t)=-16t^2 + v_{o}t + 40$ Eqn(2).

Now let us use the given information and set up our equations for Ben and Josh.

Ben:

We know that $v_{o}=60ft/s$

Thus Eqn. (2) becomes:

$h(t)=-16t^2+60t+40$ Eqn.(3)

Josh:

We know that $v_{o}=48ft/s$

Thus Eqn. (2) becomes:

$h(t)=-16t^2+48t+40$ Eqn. (4).

Now since we want to find whose textbook reaches the ground first and by how many seconds we need to solve each equation (i.e. Eqns. (3) and (4)) at $h(t)=0$ . Now since both are quadratic equations we will solve one showing the full method which can be repeated for the other one.

Thus we have for Ben, Eqn. (3) gives:

$h(t)=0-16t^2+60t+40=0$

Using the quadratic expression to find the roots of the quadratic we have:

$t_{1,2}=\frac{-b+/-\sqrt{b^2-4ac} }{2a} \\t_{1,2}=\frac{-60+/-\sqrt{60^2-4(-16)(40)} }{2(-16)} \\t_{1,2}=\frac{-60+/-\sqrt{6160} }{-32} \\t_{1,2}=\frac{15+/-\sqrt{385} }{8}\\\\t_{1}=4.3276 sec\\t_{2}=-0.5776 sec$

Since time can only be positive we reject the $t_{2}$ solution and we keep that Ben's book took $t=4.3276$ seconds to reach the ground.

Similarly solving for Josh we obtain

$t_{1}=3.6794sec\\t_{2}=-0.6794sec$

Thus again we reject the negative and keep the positive solution, so Josh's book took $t=3.6794$ seconds to reach the ground.

Then we can find the difference between Ben and Josh times as

$t_{Ben}-t_{Josh}= 4.3276 - 3.6794 = 0.6482$

So to answer the original question:

Whose textbook reaches the ground first and by how many seconds?

Josh's textbook reached the ground first
Josh's textbook reached the ground first by a difference of $t=0.6482$

3 0

2 years ago

What is the area, in square centimeters, of the figure shown?

lisabon 2012 [21]

For this one, instead of breaking it down into 3 shapes we can simply subtract the missing square area from the total enclosing area...

15*8-4*4

120-16

104 cm^2

4 0

3 years ago

Read 2 more answers