Ask question

Ask question

All categories

2 years ago

15

Given that $x < 5$, rewrite $5x - |x - 5|$ without using absolute value signs. WILL GIVE BRAINLIEST

1 answer:

Olegator [25]2 years ago

3 0

Answer:

6x-5 if x<5

Step-by-step explanation:

If x<5, then x-5<0.

If x-5 is negative, then |x-5)=-(x-5) or -x+5.

5x - |x - 5|

5x-(-x+5) provided x<5

5x+x-5 by distributive property

6x-5 combined liked terms

You might be interested in

The gross domestic product (GDP) of a certain country, which measures the overall size of the economy in billions of dollars, ca

DiKsa [7]

Answer:

The estimated GDP in 2007 is 12782 billion dollars.

The estimated GDP in 2011 is 15010 billion dollars.

The estimated GDP in 2012 is 15567 billion dollars.

Step-by-step explanation:

Consider the provided function.

$g(y) = 557y +8883$

Where y = 10 corresponds to the year 2010.

Part (b) Estimate the GDP (to the nearest billion dollars) in 2007.

If y = 10 corresponds to the year 2010 then y = 7 corresponds to the year 2007.

Substitute the value of y = 7 in the provided function.

$g(y) = 557(7) +8883$

$g(y) = 3899+8883$

$g(y) = 12782$

Hence, the estimated GDP in 2007 is 12782 billion dollars.

Part (b) Estimate the GDP (to the nearest billion dollars) in 2011.

If y = 10 corresponds to the year 2010 then y = 11 corresponds to the year 2011.

Substitute the value of y = 11 in the provided function.

$g(y) = 557(11) +8883$

$g(y) = 6127+8883$

$g(y) = 15010$

Hence, the estimated GDP in 2011 is 15010 billion dollars.

Part (C) Estimate the GDP (to the nearest billion dollars) in 2012.

If y = 10 corresponds to the year 2010 then y = 12 corresponds to the year 2012.

Substitute the value of y = 12 in the provided function.

$g(y) = 557(12) +8883$

$g(y) = 6684+8883$

$g(y) = 15567$

Hence, the estimated GDP in 2012 is 15567 billion dollars.

4 0

3 years ago

An ice cube is melting, and the lengths of its sides are decreasing at a rate of 0.8 millimeters per minute At what rate is the

julia-pushkina [17]

Answer:

The rate of decrease is: $43.2mm^3/min$

Step-by-step explanation:

Given

$l = 18mm$

$\frac{dl}{dt} = -0.8mm/min$ ---- We used minus because the rate is decreasing

Required

Rate of decrease when: $l = 18mm$

The volume of the cube is:

$V = l^3$

Differentiate

$\frac{dV}{dl} = 3l^2$

Make dV the subject

$dV = 3l^2 \cdot dl$

Divide both sides by dt

$\frac{dV}{dt} = 3l^2 \cdot \frac{dl}{dt}$

Given that: $l = 18mm$ and $\frac{dl}{dt} = -0.8mm/min$

$\frac{dV}{dt} = 3 * (18mm)^2 * (-0.8mm/min)$

$\frac{dV}{dt} = 3 * 18 *-0.8mm^3/min$

$\frac{dV}{dt} = -43.2mm^3/min$

<em>Hence, the rate of decrease is: 43.2mm^3/min</em>

8 0

2 years ago

How many variable terms are in the expression 3x3y + 5xz − 4y + x − z + 9?

Dimas [21]

Answer:

3

Step-by-step explanation:

there are only x y and z used in the expression no matter how many times they multiply.

8 0

2 years ago

of the pets in the pet show 5/6 are cats.4/5 of the calico cats what fraction of the pets are calico cats

Bond [772]

The answer is 0.664 because you have to divide 4/5 and 5/6

3 0

2 years ago

Y=-x^2+2x+10<br> y=x+2<br><br> Substitution <br> Please show your work<br> Need ASAP

erik [133]

Answer:

The solutions of the system of equations are the points

$(\frac{1-\sqrt{33}} {2},\frac{5-\sqrt{33}} {2})$

$(\frac{1+\sqrt{33}} {2},\frac{5+\sqrt{33}} {2})$

Step-by-step explanation:

we have

$y=-x^{2} +2x+10$ ----> equation A

$y=x+2$ ----> equation B

Solve the system by substitution

substitute equation B in equation A

$x+2=-x^{2} +2x+10$

solve for x

$-x^{2} +2x+10-x-2=0$

$-x^{2} +x+8=0$

The formula to solve a quadratic equation of the form

$ax^{2} +bx+c=0$

is equal to

$x=\frac{-b\pm\sqrt{b^{2}-4ac}} {2a}$

in this problem we have

$-x^{2} +x+8=0$

so

$a=-1\\b=1\\c=8$

substitute in the formula

$x=\frac{-1\pm\sqrt{1^{2}-4(-1)(8)}} {2(-1)}$

$x=\frac{-1\pm\sqrt{33}} {-2}$

$x=\frac{-1+\sqrt{33}} {-2}$ -----> $x=\frac{1-\sqrt{33}} {2}$

$x=\frac{-1-\sqrt{33}} {-2}$ -----> $x=\frac{1+\sqrt{33}} {2}$

<em>Find the values of y</em>

For $x=\frac{1-\sqrt{33}} {2}$

$y=x+2$

$y=\frac{1-\sqrt{33}} {2}+2$ ----> $y=\frac{5-\sqrt{33}} {2}$

For $x=\frac{1+\sqrt{33}} {2}$

$y=x+2$

$y=\frac{1+\sqrt{33}} {2}+2$ ----> $y=\frac{5+\sqrt{33}} {2}$

therefore

The solutions of the system of equations are the points

$(\frac{1-\sqrt{33}} {2},\frac{5-\sqrt{33}} {2})$

$(\frac{1+\sqrt{33}} {2},\frac{5+\sqrt{33}} {2})$

5 0

2 years ago