All categories

3 years ago

Find the weekly wages for 43 hours of work at $11 per hour if the employee receives time and a half for all the hours over 40

Mathematics

1 answer:

guapka [62]3 years ago

3 0

40hours x $11 = $440
3hours x $16.50 = $49.50
$440 + $49.50 = $489.50

You might be interested in

I need help what is 1+1=

irina [24]

Answer:

Step-by-step explanation:

i hope it will help u hmm

4 0

3 years ago

Read 2 more answers

Help pls fast

Andrews [41]

Answer:

32.40cm²

Step-by-step explanation:

Given the parallelogram ABCD ;

side AB is always Parallel to CD, hence ;

AB = CD = 7.2cm

Perpendicular from C on AB is the height of the parallelogram = 4.5cm

Area of a parallelogram : base * height

Area of a parallelogram = length base * height

Area of a parallelogram = 7.2 cm * 4.5cm

Area of parallelogram = 32.40cm²

6 0

3 years ago

(a) Find parametric equations for the line through (1, 4, 4) that is perpendicular to the plane x − y + 2z = 7. (Use the paramet

Alex777 [14]

Answer:

$x=1+t\\y=4-t\\z=4+2t$

Step-by-step explanation:

A vector perpendicular to the plane ax+by+cz+d=0 is of the form $(a,b,c)$ .

So, a vector perpendicular to the plane x − y + 2z = 7 is $(1,-1,2)$ .

The parametric equations of a line through the point $(x_0,y_0,z_0)$ and parallel to the vector $(a,b,c)$ are as follows:

$x=x_0+at\\y=y_0+bt\\z=z_0+ct$

Put $(x_0,y_0,z_0)=(1,4,4)$ and $(a,b,c)=(1,-1,2)$

Therefore,

$x=1+t\\y=4-t\\z=4+2t$

xy-plane:

Put z = 0 ⇒ t = -2 ⇒x = - 1 , y = 6

So, at point (-1,6,0)

yz-plane:

Put x = 0 ⇒ t = -1 ⇒ y = 5, z =2

So, at point (0,5,2)

xz-plane:

Put y = 0 ⇒ t = 4 ⇒ x = 5, z = 12

So, at point (5,0,12)

5 0

3 years ago

Determine if the following equations are equal by evaluating both sides

padilas [110]

Answer:

Yes, the both sides of the given equation are equal.

Step-by-step explanation:

The given equation is

$\log((1-i)^3)=3\log(1-i)$

Taking LHS,

$LHS=\log((1-i)^3)$

Using the power property of logarithm, we get

$LHS=3\log(1-i)$ $[\because log_ax^n=nlog_ax]$

$LHS=RHS$ $[\because RHS=3\log(1-i)]$

Both sides of the given equation are equal.

5 0

3 years ago

Runner A is initially 6.0 km west of a flagpole and is running with a constant velocity of 3.0 km/h due east. Runner B is initia

ArbitrLikvidat [17]

Answer:

The distance of the two runners from the flagpole is 0.0882 km

Step-by-step explanation:

We are going to use the minus sign when the runner is on the west and the plus sign when the runner is on the east.

Then, Taking into account the runner A is 6 km west and is running with a constant velocity of 3 Km/h east, the distance of the runner A from the flagpole is given by the following equation:

Xa = -6 Km + (3 Km/h)* t

Where Xa is the position of the runner A from the flagpole and t is the time in hours.

At the same way the distance of the runner B, Xb, from the flagpole is given by the following equation:

Xb = 7.4 Km - (3.8 Km/h)*t

Then, the two runners cross their path when Xa is equal to Xb, so if we solve this equation for t, we get:

Xa = Xb

-6 + (3*t) = 7.4 - (3.8*t)

(3.8*t) + (3*t) = 7.4 + 6

(6.8*t) = 13.4

t = 13.4/6.8

t = 1.9706

Therefore, at time t equal to 1.9706 hours, both runners cross their path. The distance of the two runners from the flagpole can be calculated replacing the value of t in equation for Xa or in equation for Xb as:

Xa = -6 Km + (3 Km/h)* t

Xa = -6 Km + (3 Km/h)*(1.9706 h)

Xa = -0.0882 Km

That means that both runners are 0.0882 Km west of a flagpole.

4 0

3 years ago