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3 years ago

Can someone help me?

Mathematics

1 answer:

geniusboy [140]3 years ago

7 0

Step-by-step explanation:

If you have any questions about the way I solved it, don't hesitate to ask me in the comments below ;)

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What is greater 27% or 0.48

mylen [45]

Well since 27% in decimal form is .27, that mean .48 is greater.

27% < 0.48

3 0

3 years ago

Help With These Special Right Triangles!!!

NikAS [45]

Answer:

I hope it will be useful.

Hint:

Since its an isosceles right angled triangle! (Apply the properties)

Or, you can use Pythagoras theorem.

(2x^2=9 v.i.z., on rationalizing 1.5*√2)

4 0

3 years ago

Which expression is equivalent to 5^10•5^5?

Tju [1.3M]

When we have exponents multiplied with the same base we add them so 5^5*5^10=5^15

4 0

3 years ago

Read 2 more answers

Directions: Calculate the percent increase or decrease between the starting and ending

Sveta_85 [38]

I'll do the first two to get you started

===============================================

Problem 1

A = 3 = starting value

B = 10 = ending value

C = percent change

C = [ (B - A)/A ] * 100%

C = [ (10-3)/3 ] * 100%

C = (7/3) * 100%

C = 2.3333333 * 100%

C = 233.33333%

C = 233.3%

The positive C value means we have a percent increase. If C was negative, then we'd have a percent decrease.

<h3>Answer: 233.3% increase</h3>

===============================================

Problem 2

A = 9 = start value

B = 20 = end value

C = percent change

C = [ (B - A)/A ] * 100%

C = [ (20-9)/9 ] * 100%

C = (11/9)*100%

C = 1.2222222222*100%

C = 122.22222222%

C = 122.2%

<h3>Answer: 122.2% increase</h3>

7 0

3 years ago

Read 2 more answers

The perimeter of a rectangle is 30.8 km and it’s diagonal length is 11 km. Find it’s length and width

blsea [12.9K]

Answer:

Length of the rectangle is 15.0325 km and width is 0.3765 km.

Explanation:

Given:

Perimeter of a rectangle = 30.8 km

Length of diagonal of rectangle = 11 km

To find:

The length and width of rectangle=?

Solution:

Lets assume length of the rectangle = x km

And assume width of the rectangle = y km

Lets first create equation using given perimeter

perimeter of rectangle = 2 ( length + width )

=> 30.8 km = 2 ( x + y )

=> $x + y = \frac{30.8}{2}$

=> y = 15.4 – x ------(1)

As diagonal and two sides of rectangle forms right angle triangle whose hypoteneus is diagonal ,

=> $length^2 + width^2 = diagonal^2$

=> $x^2 + y^2 = 11^2$

=> $x^2 + y^2 = 121$

On substituting value of y from (1) in above equation we get

=> $x^2 + (15.4-x)^2 = 121$

=> $x^2 + (15.4)^2 + x^2 – 2 x 15.4 \times x = 121$

=> $2x^2-30.8x + 237.16 -121 = 0$

=> $2x^2-30.8x + 116.16 = 0$

Solving above quadratic equation using quadratic formula

General form of quadratic equation is

$ax^2 +bx +c = 0$

And quadratic formula for getting roots of quadratic equation is

$x= \frac{ -b\pm\sqrt{(b^2-4ac)}}{2a}$

As equation is $2x^2-30.8x + 116.16 = 0$ , in our case

a = 2 , b = -30.8 and c = 116.16

Calculating roots of the equation we get

$x=\frac{ -(-30.8)\pm\sqrt{(-30.8)^2-4(2)( 11)} } {(2\times2)}$

$x=\frac{30.8\pm\sqrt{(948.64-88)}}{4}$

$x=\frac{30.8\pm\sqrt{860.64}}{4}$

$x=\frac{(30.8\pm29.33)}4$

$x=\frac{(30.8+29.33)}{4}$

$x=\frac{(30.8-29.33)}{4}$

=> x = 15.0325 or x = 0.3675

As generally length is longer one ,

So x = 1.0325

From equation (1) y = 15.4 – x = 0.3765

Hence length of the rectangle is 15.0325 km and width is 0.3765 km.

6 0

3 years ago