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How to write 100,203 in word form

pantera1 [17]

One hundred thousand, two hundred and three.

3 0

3 years ago

Read 2 more answers

Really easy math question

Eddi Din [679]

Number of sodas sold = 130

Number of hot dogs sold = 65

Step-by-step explanation:

Given,

Combined number of sodas and hot dogs sold = 195

Let,

x represent the number of sodas sold

y represent the number of hot dogs sold

According to given statement;

x+y=195 Eqn 1

x = 2y Eqn 2

Putting value of x from Eqn 2 in Eqn 1

$2y+y=195\\3y=195$

Dividing both sides by 3

$\frac{3y}{3}=\frac{195}{3}\\y=65$

Putting y=65 in Eqn 2

$x=2(65)\\x=130$

Number of sodas sold = 130

Number of hot dogs sold = 65

Keywords: linear equation, substitution method

Learn more about substitution method at:

#LearnwithBrainly

4 0

3 years ago

Read 2 more answers

In the equation 16x=128,is the value 8 a solution to this equation? Explain your reasoning by filling in the blanks.

GrogVix [38]

Answer:

yes

Step-by-step explanation:

The value 8 is the solution to the equation because plugging 8 into the equation in place of the unknown variable x like 16(8) will equal 128.

6 0

2 years ago

The monthly sales (in thousands of units) of a seasonal product are approximated by

Elodia [21]

Don't forget S is measured in thousands of units so you are solving for :

100 < 74.5 + 43.75Sin(πt/6)
25.5 < 43.75Sin(πt/6)
Sin(πt/6) >25.5/43.75 = 0.582857
ASrcSin(πt/6) > 0.62224 radians
πt/6 > 0.62224
t > 6 x 0.62224/π = 1.1884 (4dp)

This initial value occurs when the sine value is increasing and it will reach its maximum value of 1 when Sin(πt/6) = Sinπ/2, that is when t = 3.
Consequently, monthly sales exceed 100,000 during the period between t = 1.1884 and 4.8116
[3 - 1.1884 = 1.8116 so the other extreme occurs at 3 + 1.8116]

Note : on the basis of these calculations, January is 0 ≤ t < 1 : February is 1 ≤ t < 2 :....May is 4 ≤ t < 5
So the period when sales exceed 100,000 occurs between Feb 6 and May 25 and annually thereafter.

3 0

3 years ago

Help me on this please

zalisa [80]

Answer:

1. (x, y) → (x + 3, y - 2)

Vertices of the image

a) (-2, - 3)

b) (-2, 3)

c) (2, 2)

2. (x, y) → (x - 3, y + 5)

Vertices of the image

a) (-3, 2)

b) (0, 2)

c) (0, 4)

d) (2, 4)

3. (x, y) → (x + 4, y)

Vertices of the image

a) (-1, -2)

b) (1, -2)

c) (3, -2)

4. (x, y) → (x + 6, y + 1)

Vertices of the image

a) (1, -1)

b) (1, -2)

c) (2, -2)

d) (2, -4)

e) (3, -1)

f) (3, -3)

g) (4, -3)

h) (1, -4)

5. (x, y) → (x, y - 4)

Vertices of the image

a) (0, -2)

b) (0, -3)

c) (2, -2)

d) (2, -4)

6. (x, y) → (x - 1, y + 4)

Vertices of the image

a) (-5, 3)

b) (-5, -1)

c) (-3, 0)

d) (-3, -1)

Explanation:

To identify each IMAGE you should perform the following steps:

List the vertex points of the preimage (the original figure) as ordered pairs.

Apply the transformation rule to every point of the preimage

List the image of each vertex after applying each transformation, also as ordered pairs.

1. (x, y) → (x + 3, y - 2)

The rule means that every point of the preimage is translated three units to the right and 2 units down.

Vertices of the preimage Vertices of the image

a) (-5,2) (-5 + 3, -1 - 2) = (-2, - 3)

b) (-5, 5) (-5 + 3, 5 - 2) = (-2, 3)

c) (-1, 4) (-1 + 3, 4 - 2) = (2, 2)

2. (x,y) → (x - 3, y + 5)

The rule means that every point of the preimage is translated three units to the left and five units down.

Vertices of the preimage Vertices of the image

a) (0, -3) (0 - 3, -3 + 5) = (-3, 2)

b) (3, -3) (3 - 3, -3 + 5) = (0, 2)

c) (3, -1) (3 - 3, -1 + 5) = (0, 4)

d) (5, -1) (5 - 3, -1 + 5) = (2, 4)

3. (x, y) → (x + 4, y)

The rule represents a translation 4 units to the right.

Vertices of the preimage Vertices of the image

a) (-5, -2) (-5 + 4, -2) = (-1, -2)

b) (-3, -5) (-3 + 4, -2) = (1, -2)

c) (-1, -2) (-1 + 4, -2) = (3, -2)

4. (x, y) → (x + 6, y + 1)

Vertices of the preimage Vertices of the image

a) (-5, -2) (-5 + 6, -2 + 1) = (1, -1)

b) (-5, -3) (-5 + 6, -3 + 1) = (1, -2)

c) (-4, -3) (-4 + 6, -3 + 1) = (2, -2)

d) (-4, -5) (-4 + 6, -5 + 1) = (2, -4)

e) (-3, -2) (-3 + 6, -2 + 1) = (3, -1)

f) (-3, -4) (-3 + 6, -4 + 1) = (3, -3)

g) (-2, -4) (-2 + 6, -4 + 1) = (4, -3)

h) (-2, -5) (-2 + 3, -5 + 1) = (1, -4)

5. (x, y) → (x, y - 4)

This is a translation four units down

Vertices of the preimage Vertices of the image

a) (0, 2) (0, 2 - 4) = (0, -2)

b) (0,1) (0, 1 - 4) = (0, -3)

c) (2, 2) (2, 2 - 4) = (2, -2)

d) (2,0) (2, 0 - 4) = (2, -4)

6. (x, y) → (x - 1, y + 4)

This is a translation one unit to the left and four units up.

Vertices of the pre-image Vertices of the image

a) (-4, -1) (-4 - 1, -1 + 4) = (-5, 3)

b) (-4 - 5) (-4 - 1, -5 + 4) = (-5, -1)

c) (-2, -4) (- 2 - 1, -4 + 4) = (-3, 0)

d) (-2, -5) (-2 - 1, -5 + 4) = (-3, -1)

8 0

3 years ago