All categories

4 years ago

How do you write 9,000,000+200,000+1,000+500+2 in standard form

Mathematics

2 answers:

Basile [38]4 years ago

8 0

9,201,502 I just added

maks197457 [2]4 years ago

4 0

You write it like this...9,201,502... hope this helps

You might be interested in

The volume of a gift box is 972 in. The height of the gift box is 12 inches

Serggg [28]

Answer:

The length of each side of the 9 in

Step-by-step explanation:

The volume of the gift box is 972 in. The height of the gift box is 12 inches

and the area of the base is 81 in².

We want to determine the length of each side of the square base if the base shape is a square.

Recall that, the area of a square is

${l}^{2}$

It was given that, the area of this base is 81 in²

${l}^{2} = 81$

Take positive square root:

$l = \sqrt{81} = 9in$

5 0

4 years ago

Use triangle ABC drawn below & only the sides labeled. Find the side of length AB in terms of side a, side b & angle C o

Brrunno [24]

Answer:

$AB = \sqrt{a^2 + b^2-2abCos\ C}$

Step-by-step explanation:

Given:

The above triangle

Required

Solve for AB in terms of a, b and angle C

Considering right angled triangle BOC where O is the point between b-x and x

From BOC, we have that:

$Sin\ C = \frac{h}{a}$

Make h the subject:

$h = aSin\ C$

Also, in BOC (Using Pythagoras)

$a^2 = h^2 + x^2$

Make $x^2$ the subject

$x^2 = a^2 - h^2$

Substitute $aSin\ C$ for h

$x^2 = a^2 - h^2$ becomes

$x^2 = a^2 - (aSin\ C)^2$

$x^2 = a^2 - a^2Sin^2\ C$

Factorize

$x^2 = a^2 (1 - Sin^2\ C)$

In trigonometry:

$Cos^2C = 1-Sin^2C$

So, we have that:

$x^2 = a^2 Cos^2\ C$

Take square roots of both sides

$x= aCos\ C$

In triangle BOA, applying Pythagoras theorem, we have that:

$AB^2 = h^2 + (b-x)^2$

Open bracket

$AB^2 = h^2 + b^2-2bx+x^2$

Substitute $x= aCos\ C$ and $h = aSin\ C$ in $AB^2 = h^2 + b^2-2bx+x^2$

$AB^2 = h^2 + b^2-2bx+x^2$

$AB^2 = (aSin\ C)^2 + b^2-2b(aCos\ C)+(aCos\ C)^2$

Open Bracket

$AB^2 = a^2Sin^2\ C + b^2-2abCos\ C+a^2Cos^2\ C$

Reorder

$AB^2 = a^2Sin^2\ C +a^2Cos^2\ C + b^2-2abCos\ C$

Factorize:

$AB^2 = a^2(Sin^2\ C +Cos^2\ C) + b^2-2abCos\ C$

In trigonometry:

$Sin^2C + Cos^2 = 1$

So, we have that:

$AB^2 = a^2 * 1 + b^2-2abCos\ C$

$AB^2 = a^2 + b^2-2abCos\ C$

Take square roots of both sides

$AB = \sqrt{a^2 + b^2-2abCos\ C}$

6 0

3 years ago

Determine the equation of the line that passes through the point (7, -23)and is perpendicular to the line y=1/3x-6Enter your ans

marishachu [46]

We have a line y = 1/3x -6

We want a line that is perpendicular to this line

Perpendicular lines have slopes that multiply to -1

1/3 * m = -1

3 * 1/3 *m = -1 * 3

m = -3

The slope of the perpendicular line is -3

y = mx+b where m is the slope and b is the y intercept

y = -3x+b

We have a point on the line ( 7 ,-23)

Substitute this point into the equation

-23 = -3(7)+b

-23 = -21+b

Add 21 to each side

-23+21 = -21+21+b

-2 = b

y = -3x-2

In slope intercept form, the line perpendicular passing through (7,-23) is

y = -3x-2

3 0

1 year ago

mai and priya were on scooters. mai traveled 15 meters in 6 seconds priya traveled 22 meters in 10 seconds. who was moving faste

Eduardwww [97]

Answer:

Its Mai

Step-by-step explanation:

I promise I know why

7 0

3 years ago

Solve using Quadratic Formula <br>x^2 + 3x - 3 = 0

katrin [286]

Answer:

a= 1 , b= 3 , c= –3

$x = \frac{ - b \frac{ + }{} \sqrt{ {b}^{2} - 4ac} }{2a} \\ x = \frac{ - 3 \frac{ + }{} {} \sqrt{ {3}^{2} - 4(1)( - 3)} }{2(1)} \\ = \frac{ - 3 \frac{ + }{} \sqrt{9 + 12} }{2} = \frac{ - 3 \frac{ + }{} \sqrt{21} }{2} \\ \\ x = \frac{ - 3 \frac{ + }{} \sqrt{21} }{2} \\ \\ x = 0.79 \\ x = - 3.79$

I hope I helped you^_^

4 0

3 years ago