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2 years ago

15

Harrison gets paid $22 per hour working at Izzy’s Ice Cream. If he works 40 hours per week and pays $145 per week in taxes, what

is his weekly net income?

2 answers:

AnnyKZ [126]2 years ago

7 0

Answer:

He makes $735.

Step-by-step explanation:

Multiply 22x40 and subtract 145.

kramer2 years ago

7 0

Answer:

$735

Step-by-step explanation:

To calculate the gross income, multiply the rate per hour ($22) by the number of hours (40) to get $880. Then you subtract the $145 in taxes, and you get a weekly net income of $735.

Which is really good for working at an ice cream store.

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−9,000÷3=−3,000 What does -3,000 tell me

soldier1979 [14.2K]

Answer:

Step-by-step explanation:

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3 years ago

The function h (d) = 2d + 4.3 relates the height (h) of the water in a fountain in feet to the diameter (d) of the pipe carrying

mezya [45]

Answer:

h(1.5) = 7.3 ft

h(10.3) = 24.9 ft

Step-by-step explanation:

Given the function h(d) = 2d + 4.3,

where:

h = height of the water in a fountain (in feet)

d = diameter of the pipe carrying the water (in inches)

<h3>h(1.5)</h3>

Substitute the input value of d = 1.5, into the function:

h(1.5) = 2(1.5) + 4.3

h(1.5) = 3 + 4.3

h(1.5) = 7 feet

The height of the water in a fountain is 7 feet when the diameter of the pipe is 1.5 inches.

<h3>h(10.3)</h3>

Substitute the input value of d = 10.3, into the function:

h(10.3) = 2(10.3) + 4.3

h(10.3) = 20.6 + 4.3

h(10.3) = 24.9 feet

The height of the water in a fountain is 24.9 feet when the diameter of the pipe is 10.3 inches.

<h3>Context of the solutions to h(1.5) and h(10.3):</h3>

The solutions to both functions show the relationship between the diameter of the pipe to the height of the water in a fountain. The height of the water in fountain increases relative to the diameter of the pipe. In other words, as the diameter or the size of the pipe increases or widens, the height of the water in a fountain also increases.

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2 years ago

Write the linear equation in slope intercept form 1,3 and -3,-5

oksian1 [2.3K]

For this case we have that by definition, the equation of the line of the slope-intersection form is given by:

$y = mx + b$

Where:

m: It's the slope

b: It is the cut-off point with the y axis

We have two points through which the line passes:

$(x_ {1}, y_ {1}) :( 1,3)\\(x_ {2}, y_ {2}): (- 3, -5)$

We found the slope:

$m = \frac{y_ {2} -y_ {1}} {x_ {2} -x_ {1}}$

Substituting we have:

$m = \frac {-5-3} {- 3-1} = \frac {-8} {- 4} = 2$

Thus, the equation is of the form:

$y = 2x + b$

We substitute one of the points and find the cut-off point:

$3 = 2 (1) + b\\3 = 2 + b\\3-2 = b\\b = 1$

Finally, the equation is:

$y = 2x + 1$

ANswer:

$y = 2x + 1$

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3 years ago

Which of the following sets are subspaces of R3 ?

Ratling [72]

Answer:

The following are the solution to the given points:

Step-by-step explanation:

for point A:

$\to A={(x,y,z)|3x+8y-5z=2} \\\\\to for(x_1, y_1, z_1),(x_2, y_2, z_2) \varepsilon A\\\\ a(x_1, y_1, z_1)+b(x_2, y_2, z_2) = (ax_1+bx_2,ay_1+by_2,az_1+bz_2)$

$=3(aX_l +bX_2) + 8(ay_1 + by_2) — 5(az_1+bz_2)\\\\=a(3X_l+8y_1- 5z_1)+b (3X_2+8y_2—5z_2)\\\\=2(a+b)$

The set A is not part of the subspace $R^3$

for point B:

$\to B={(x,y,z)|-4x-9y+7z=0}\\\\\to for(x_1,y_1,z_1),(x_2, y_2, z_2) \varepsilon B \\\\\to a(x_1, y_1, z_1)+b(x_2, y_2, z_2) = (ax_1+bx_2,ay_1+by_2,az_1+bz_2)$

$=-4(aX_l +bX_2) -9(ay_1 + by_2) +7(az_1+bz_2)\\\\=a(-4X_l-9y_1+7z_1)+b (-4X_2-9y_2+7z_2)\\\\=0$

$\to a(x_1,y_1,z_1)+b(x_2, y_2, z_2) \varepsilon B$

The set B is part of the subspace $R^3$

for point C: $\to C={(x,y,z)|x$

In this, the scalar multiplication can't behold

$\to for (-2,-1,2) \varepsilon C$

$\to -1(-2,-1,2)= (2,1,-1)$ ∉ C

this inequality is not hold

The set C is not a part of the subspace $R^3$

for point D:

$\to D={(-4,y,z)|\ y,\ z \ arbitrary \ numbers)$

The scalar multiplication s is not to hold

$\to for (-4, 1,2)\varepsilon D\\\\\to -1(-4,1,2) = (4,-1,-2)$ ∉ D

this is an inequality, which is not hold

The set D is not part of the subspace $R^3$

For point E:

$\to E= {(x,0,0)}|x \ is \ arbitrary) \\\\\to for (x_1,0 ,0) ,(x_{2},0 ,0) \varepsilon E \\\\\to a(x_1,0,0) +b(x_{2},0,0)= (ax_1+bx_2,0,0)\\$

The $x_1, x_2$ is the arbitrary, in which $ax_1+bx_2$ is arbitrary

$\to a(x_1,0,0)+b(x_2,0,0) \varepsilon E$

The set E is the part of the subspace $R^3$

For point F:

$\to F= {(-2x,-3x,-8x)}|x \ is \ arbitrary) \\\\\to for (-2x_1,-3x_1,-8x_1),(-2x_2,-3x_2,-8x_2)\varepsilon F \\\\\to a(-2x_1,-3x_1,-8x_1) +b(-2x_1,-3x_1,-8x_1)= (-2(ax_1+bx_2),-3(ax_1+bx_2),-8(ax_1+bx_2))$

The $x_1, x_2$ arbitrary so, they have $ax_1+bx_2$ as the arbitrary $\to a(-2x_1,-3x_1,-8x_1)+b(-2x_2,-3x_2,-8x_2) \varepsilon F$

The set F is the subspace of $R^3$

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3 years ago

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