Help!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

A map of a rectangular park has a length of 4 inches and a width of 6 inches. It uses a scale of 1 inch for every 30 Miles. What

MakcuM [25]

Answer:

The actual area 21600 miles²

Step-by-step explanation:

* <em>Lets explain how to solve the problem</em>

- A drawing that shows a real object with accurate sizes reduced or

enlarged by a certain amount called the scale

- Drawing scale is a ratio between the drawing length and the

actual length

- To find the actual dimensions from the drawing dimensions use the

scale drawing

*<em> Lets solve the problem</em>

- A map of a rectangular park has a length of 4 inches and a width

of 6 inches

∴ The drawing dimensions are:

# length = 4 inches

# width = 6 inches

- The scale of the map is 1 inch foe every 30 miles

∴ The scale drawing is 1 : 30

- To find the actual area find the actual dimensions

∵ The scale drawing is 1 : 30

∵ The length = 4 inches and the width = 6 inches

- By using cross multiplication

∴ 1 : 30

4 : actual length

6 : actual width

∴ Actual length = (4 × 30)/1 = 120

∴ Actual length = 120 miles

∴ Actual width = (6 × 30)/1 = 180

∴ Actual Width = 180 miles

∴ The actual dimensions are 120 miles and 180 miles

∵ The area of any rectangle = length × width

∴ The actual area = 120 × 180 = 21600

∴ The actual area 21600 miles²

5 0

3 years ago

3. The following is raw data in respect to expenses incurred by a family on education from

dalvyx [7]

I don’t really understand what you have said. But I need help on my homework.

3 0

3 years ago

Find the derivative.

Aleksandr [31]

Answer:

Using either method, we obtain: $t^\frac{3}{8}$

Step-by-step explanation:

a) By evaluating the integral:

$\frac{d}{dt} \int\limits^t_0 {\sqrt[8]{u^3} } \, du$

The integral itself can be evaluated by writing the root and exponent of the variable u as: $\sqrt[8]{u^3} =u^{\frac{3}{8}$

Then, an antiderivative of this is: $\frac{8}{11} u^\frac{3+8}{8} =\frac{8}{11} u^\frac{11}{8}$

which evaluated between the limits of integration gives:

$\frac{8}{11} t^\frac{11}{8}-\frac{8}{11} 0^\frac{11}{8}=\frac{8}{11} t^\frac{11}{8}$

and now the derivative of this expression with respect to "t" is:

$\frac{d}{dt} (\frac{8}{11} t^\frac{11}{8})=\frac{8}{11}\,*\,\frac{11}{8}\,t^\frac{3}{8}=t^\frac{3}{8}$

b) by differentiating the integral directly: We use Part 1 of the Fundamental Theorem of Calculus which states:

"If f is continuous on [a,b] then

$g(x)=\int\limits^x_a {f(t)} \, dt$

is continuous on [a,b], differentiable on (a,b) and $g'(x)=f(x)$

Since this this function $u^{\frac{3}{8}$ is continuous starting at zero, and differentiable on values larger than zero, then we can apply the theorem. That means:

$\frac{d}{dt} \int\limits^t_0 {u^\frac{3}{8} } } \, du=t^\frac{3}{8}$

5 0

3 years ago

Cards are dealt, one at a time, from a standard 52-card deck. a if the first 2 cards are both spades, what is the probability th

ratelena [41]

So we already drew two cards and both of them are spades. So that means there are 50 cards and 11 spades left.

Then third card being spades would be 11 / 50

Now there are 10 spades and 49 cards left.
So the probability for fourth card would be 10 / 49

Fifth card would be 9 / 48.

So the probability of next three cards being spades would be
$\dfrac{11}{50}\cdot\dfrac{10}{49}\cdot\dfrac9{48}=\dfrac{33}{3920}\approx\boxed{0.00842}$

Hope this helps.

3 0

4 years ago

Jacob is twice as heavy as Ken Linda is half as heavy as Ken Giver 3 children is at least 35 kg. a By letting Linda's mass be x,

Usimov [2.4K]

==> she can buy 45 notebooks at most.

$\: \: \: \:$

6 0

2 years ago