Ask question

Ask question

All categories

3 years ago

10

Patrick works a 40-hour week at $10.70 an hour with time and a half for overtime. Last week he worked 45 hours. What were his to

tal earnings for the week?
A. $444.05 B. $508.25 C. $510.00 D. $481.15

1 answer:

MA_775_DIABLO [31]3 years ago

3 0

D. $
481.15 there you go

You might be interested in

The arch of a bridge forms the upper half of an ellipse. The arch is 6 meters above the 20-meter-wide river. Write an equation f

aksik [14]

Answer:

$\frac{x^2}{36} + \frac{y^2}{100} = 1$

Step-by-step explanation:

Required

The equation of the ellipse

This is calculated as:

$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$

Where:

$b =6$ ---- height

$a = 0.5 * 20 =10$ --- width

So:

$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$

$\frac{x^2}{6^2} + \frac{y^2}{10^2} = 1$

$\frac{x^2}{36} + \frac{y^2}{100} = 1$

3 0

3 years ago

a larger gear wheel has 40 teeth and a small gear wheel has 25 teeth. if the larger one makes 100 revolutions in a minute, how m

nevsk [136]

Lets calculate gear ration T2/T1
so the gear ratio will be 1.6 it simply means dat if u rotate large gears one full revolution it tends to turn another short wheel to 1.6 revs, so in ur qtion the large whell revs 100 den u must multiply wid 1.6*100= 160 revs

5 0

3 years ago

miracle did 1/6 of her homework in class. She did 1/6 on the bus. She did 3/6 at home. How much of her homework does she have le

sergij07 [2.7K]

It is not reasonable that she has 1/2 of her homework to do. She did 3/6 (1/2) at home. She actually has 1/6 of her homework left to do.

8 0

3 years ago

QUICKK!! GIVING BRAINLIEST TO CORRECT ANSWER

Liono4ka [1.6K]

Step-by-step explanation:

Insufficient information!

however, volume will remain same in both of these figures!

7 0

3 years ago

The polynomial P(x) = 2x^3 + mx^2-5 leaves the same remainder when divided by (x-1) or (2x + 3). Find the value of m and the rem

Zigmanuir [339]

Answer:

m=7

Remainder =4

If q=1 then r=3 or r=-1.

If q=2 then r=3.

They are probably looking for q=1 and r=3 because the other combinations were used earlier in the problem.

Step-by-step explanation:

Let's assume the remainders left when doing P divided by (x-1) and P divided by (2x+3) is R.

By remainder theorem we have that:

P(1)=R

P(-3/2)=R

$P(1)=2(1)^3+m(1)^2-5$

$=2+m-5=m-3$

$P(\frac{-3}{2})=2(\frac{-3}{2})^3+m(\frac{-3}{2})^2-5$

$=2(\frac{-27}{8})+m(\frac{9}{4})-5$

$=-\frac{27}{4}+\frac{9m}{4}-5$

$=\frac{-27+9m-20}{4}$

$=\frac{9m-47}{4}$

Both of these are equal to R.

$m-3=R$

$\frac{9m-47}{4}=R$

I'm going to substitute second R which is (9m-47)/4 in place of first R.

$m-3=\frac{9m-47}{4}$

Multiply both sides by 4:

$4(m-3)=9m-47$

Distribute:

$4m-12=9m-47$

Subtract 4m on both sides:

$-12=5m-47$

Add 47 on both sides:

$-12+47=5m$

Simplify left hand side:

$35=5m$

Divide both sides by 5:

$\frac{35}{5}=m$

$7=m$

So the value for m is 7.

$P(x)=2x^3+7x^2-5$

What is the remainder when dividing P by (x-1) or (2x+3)?

Well recall that we said m-3=R which means r=m-3=7-3=4.

So the remainder is 4 when dividing P by (x-1) or (2x+3).

Now P divided by (qx+r) will also give the same remainder R=4.

So by remainder theorem we have that P(-r/q)=4.

Let's plug this in:

$P(\frac{-r}{q})=2(\frac{-r}{q})^3+m(\frac{-r}{q})^2-5$

Let x=-r/q

This is equal to 4 so we have this equation:

$2u^3+7u^2-5=4$

Subtract 4 on both sides:

$2u^3+7u^2-9=0$

I see one obvious solution of 1.

I seen this because I see 2+7-9 is 0.

u=1 would do that.

Let's see if we can find any other real solutions.

Dividing:

1 | 2 7 0 -9

| 2 9 9

-----------------------

2 9 9 0

This gives us the quadratic equation to solve:

$2x^2+9x+9=0$

Compare this to $ax^2+bx+c=0$

$a=2$

$b=9$

$c=9$

Since the coefficient of $x^2$ is not 1, we have to find two numbers that multiply to be $ac$ and add up to be $b$ .

Those numbers are 6 and 3 because $6(3)=18=ac$ while $6+3=9=b$ .

So we are going to replace $bx$ or $9x$ with $6x+3x$ then factor by grouping:

$2x^2+6x+3x+9=0$

$(2x^2+6x)+(3x+9)=0$

$2x(x+3)+3(x+3)=0$

$(x+3)(2x+3)=0$

This means x+3=0 or 2x+3=0.

We need to solve both of these:

x+3=0

Subtract 3 on both sides:

x=-3

----

2x+3=0

Subtract 3 on both sides:

2x=-3

Divide both sides by 2:

x=-3/2

So the solutions to P(x)=4:

$x \in \{-3,\frac{-3}{2},1\}$

If x=-3 is a solution then (x+3) is a factor that you can divide P by to get remainder 4.

If x=-3/2 is a solution then (2x+3) is a factor that you can divide P by to get remainder 4.

If x=1 is a solution then (x-1) is a factor that you can divide P by to get remainder 4.

Compare (qx+r) to (x+3); we see one possibility for (q,r)=(1,3).

Compare (qx+r) to (2x+3); we see another possibility is (q,r)=(2,3).

Compare (qx+r) to (x-1); we see another possibility is (q,r)=(1,-1).

6 0

3 years ago