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3 years ago

Help I need these quick 8/5*(4/9)^-3

Mathematics

1 answer:

otez555 [7]3 years ago

8 0

Answer:

18.225

Step-by-step explanation:

You do the -3 power to 4/9 first then multiply by 8/5 to get your answer.

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Find the area of the composite figure.

Roman55 [17]

Answer:

The Area of the composite figure would be 76.26 in^2

Step-by-step explanation:

According to the Figure Given:

Total Horizontal Distance = 14 in

Length = 6 in

To Find :

The Area of the composite figure

Solution:

Firstly we need to find the area of Rectangular part.

So We know that,

$\boxed{ \rm \: Area \: of \: Rectangle = Length×Breadth}$

Here, Length is 6 in but the breadth is unknown.

To Find out the breadth, we’ll use this formula:

$\boxed{\rm \: Breadth = total \: distance - Radius}$

According to the Figure, we can see one side of a rectangle and radius of the circle are common, hence,

$\longrightarrow\rm \: Length \: of \: the \: circle = Radius$

Since Length = 6 in ;

$\longrightarrow \rm \: 6 \: in = radius$

Hence Radius is 6 in.

So Substitute the value of Total distance and Radius:

Total Horizontal Distance= 14
Radius = 6

$\longrightarrow\rm \: Breadth = 14-6$

$\longrightarrow\rm \: Breadth = 8 \: in$

Hence, the Breadth is 8 in.

Then, Substitute the values of Length and Breadth in the formula of Rectangle :

Length = 6
Breadth = 8

$\longrightarrow\rm \: Area \: of \: Rectangle = 6 \times 8$

$\longrightarrow \rm \: Area \: of \: Rectangle = 48 \: in {}^{2}$

Then, We need to find the area of Quarter circle :

We know that,

$\boxed{\rm Area_{(Quarter \; Circle) } = \cfrac{\pi{r} {}^{2} }{4}}$

Now Substitute their values:

r = radius = 6
π = 3.14

$\longrightarrow\rm Area_{(Quarter \; Circle) } = \cfrac{3.14 \times 6 {}^{2} }{4}$

Solve it.

$\longrightarrow\rm Area_{(Quarter \; Circle) } = \cfrac{3.14 \times 36}{4}$

$\longrightarrow\rm Area_{(Quarter \; Circle) } = \cfrac{3.14 \times \cancel{{36} } \: ^{9} }{ \cancel4}$

$\longrightarrow\rm Area_{(Quarter \; Circle)} =3.14 \times 9$

$\longrightarrow\rm Area_{(Quarter \; Circle) } = 28.26 \: {in}^{2}$

Now we can Find out the total Area of composite figure:

We know that,

$\boxed{ \rm \: Area_{(Composite Figure)} =Area_{(rectangle)}+ Area_{ (Quarter Circle)}}$

So Substitute their values:

$\rm Area_{(rectangle)}$ = 48
$\rm Area_{(Quarter Circle)}$ = 28.26

$\longrightarrow \rm \: Area_{(Composite Figure)} =48 + 28 .26$

Solve it.

$\longrightarrow \rm \: Area_{(Composite Figure)} =\boxed{\tt 76.26 \:\rm in {}^{2}}$

Hence, the area of the composite figure would be 76.26 in² or 76.26 sq. in.

$\rule{225pt}{2pt}$

I hope this helps!

3 0

2 years ago

Find each equation with the correct solution

stepladder [879]

Answer:

15;9;-8

Step-by-step explanation:

1. 5/3x-3=2/3x+12

x=15 once you combine like terms

2. 5x+35-3x+12=7x+2

2x+47=7x+2

5x=45

x=9

3. 12x+20-3=9x-7

12x+17=9x-7

3x=-24

x=-8

3 0

2 years ago

The cost of sending an overnight package from Kansas City to Miami is $23.50 for a package weighing up to but not including 1

Dimas [21]

Answer:

For every pound that the package weight increases, the price of sending it increases by $3.80.

Step-by-step explanation:

The coefficient 3.80 multiplies the variable x (package weight), so if x is 1 pound the cost increase $3.80, if x is 2 pound the cost increase 2*3.80 = $7.6, and so on. Generally speaking, for every pound that the package weight increases the cost increases by $3.80.

4 0

3 years ago

Read 2 more answers

Based on information from Harper’s Index, 37% of adult Americans believe in Extraterrestrials. Out of a random sample of 100 adu

san4es73 [151]

Answer:

z(s) is in the rejection zone , therefore we reject H₀

We have enough evidence to claim the proportion of individuals who attended college and believe in extraterrestrials is bigger than 37%

Step-by-step explanation:

We have a prortion test.

P₀ = 37 % P₀ = 0,37

sample size = n = 100

P sample proportion = P = 47 % P = 0,47

confidence interval 95 %

α = 0,05

One tail-test (right tail) our case is to show if sample give enough information to determine if proportion of individual who attended college is higher than the proportion found by Harper´s index.

1.-Hypothesis:

H₀ null hypothesis P₀ = 0,37

Hₐ alternative hypothesis P₀ > 0,37

2.-Confidence interval 95 %

α = 0,05 and z(c) = 1.64

3.-Compute of z(s)

z(s) = [ P - P₀ ] /√(P₀Q₀/n) ]

z(s) = [ ( 0,47 - 0,37 ) / √0.37*0,63/100

z(s) = 0,1 /√0,2331/100 ⇒ z(s) = 0,1 /0,048

z(s) = 2.08

4.-Compare z(s) and z(c)

z(s) > z(c) 2.08 > 1.64

5.-Decision:

z(s) is in the rejection zone , therefore we reject H₀

We have enough evidence to claim the proportion of individuals who attended college and believe in extraterrestrials is bigger than 37%

5 0

3 years ago

Price-club sells a 12-ounce box of crackers for $2.59 and Shop Mart sells 1-pound box of crackers for $4.68. The unit price at P

Vesna [10]

Answer:

The unit price at Price-Club is $0.2158 per ounce.

The unit price at Shop Mart is $0.2925 per ounce.

Step-by-step explanation:

Price Club:

12-ounce box of crackers for $2.59

12 ounces - $2.59

1 ounce - x

$12x = 2.59$

$x = \frac{2.59}{12}$

$x = 0.2158$

The unit price at Price-Club is $0.2158 per ounce.

Shop Mart:

1-pound box of crackers for $4.68.

1 pound is 16 ounces. So

16 ounces - $4.68

1 ounce - x

$16x = 4.68$

$x = \frac{4.68}{16}$

$x = 0.2925$

The unit price at Shop Mart is $0.2925 per ounce.

8 0

3 years ago