Joshua plans to buy x pencils and one notebook at the schoolstore. A pencil costs $0.10 and a notebook costs $1.25. Joshua

Please help 30 points and brainly answer :(

Tasya [4]

The graphs insect at 3 and 13

4 0

3 years ago

Gina shows the steps she took to find the solution of the inequality below.

Leviafan [203]

Answer:

hello,2 th made from the wrong place.

because She should have distributed the -2 inside the parentheses.

the process should be like this;

19-2(1-x)<13

17(1-x)<13

17-17x<13

-17x<13-17 (17 goes to the opposite side as -17)

-17x<-4 (we divide each side by 17)

x= -4/-17

I hope you understand. There are errors in the first 2 parts of the question.

6 0

3 years ago

Point A is the midpoint of side XZ and point B is the midpoint of side YZ. Triangle X Y Z is cut by line segment A B. Point A is

AveGali [126]

Answer:

<h3>B. 4units</h3>

Step-by-step explanation:

Find the diagram attached. The diagram is a similar triangle.

From the diagram, XZ/XY = CA/AB

Given XZ = 2x-2+2x-2 = 4x-4

XY =5x-7

CA = 2x-2

AB= x+1

On substituting this parameters into the formula to get x first

4x-4/5x-7 = 2x-2/x+1

cross multiply

(4x-4)(x+1) = 5x-7(2x-2)

open the parenthesis

4x²+4x-4x-4 = 10x²-10x-14x+14

4x²-4 = 10x²-24x+14

10x²-4x²-24x+14+4 = 0

6x²-24x+18 = 0

x²-4x+3 = 0

x²-3x-x+3 = 0

x(x-3)-1(x-3) =0

(x-3)(x-1) = 0

x = 3 and 1

Next is to get length AX.

Given AX = 2x-2

Substitute x = 3 into the expression

AX = 2(3)-2

AX = 6-2

AX = 4 units

Hence the measure of length AX is 4 units

4 0

3 years ago

Bryan decided to ride his bike along a new route home from school to avoid some construction. Before leaving school, he estimate

docker41 [41]

<h3>Answer: 20 minutes.</h3>

Work Shown:

Estimate = 1.25*(actual time)

Estimate = 1.25*(16)

Estimate = 20

Bryan estimated it would take 20 minutes.

Note: The multiplier 1.25 represents an increase of 25%

8 0

2 years ago

The indicated function y1(x is a solution of the given differential equation. use reduction of order or formula (5 in section 4.

Taya2010 [7]

Given a solution $y_1(x)=\ln x$ , we can attempt to find a solution of the form $y_2(x)=v(x)y_1(x)$ . We have derivatives

$y_2=v\ln x$
${y_2}'=v'\ln x+\dfrac vx$
${y_2}''=v''\ln x+\dfrac{v'}x+\dfrac{v'x-v}{x^2}=v''\ln x+\dfrac{2v'}x-\dfrac v{x^2}$

Substituting into the ODE, we get

$v''x\ln x+2v'-\dfrac vx+v'\ln x+\dfrac vx=0$
$v''x\ln x+(2+\ln x)v'=0$

Setting $w=v'$ , we end up with the linear ODE

$w'x\ln x+(2+\ln x)w=0$

Multiplying both sides by $\ln x$ , we have

$w' x(\ln x)^2+(2\ln x+(\ln x)^2)w=0$

and noting that

$\dfrac{\mathrm d}{\mathrm dx}\left[x(\ln x)^2\right]=(\ln x)^2+\dfrac{2x\ln x}x=(\ln x)^2+2\ln x$

we can write the ODE as

$\dfrac{\mathrm d}{\mathrm dx}\left[wx(\ln x)^2\right]=0$

Integrating both sides with respect to $x$ , we get

$wx(\ln x)^2=C_1$
$w=\dfrac{C_1}{x(\ln x)^2}$

Now solve for $v$ :

$v'=\dfrac{C_1}{x(\ln x)^2}$
$v=-\dfrac{C_1}{\ln x}+C_2$

So you have

$y_2=v\ln x=-C_1+C_2\ln x$

and given that $y_1=\ln x$ , the second term in $y_2$ is already taken into account in the solution set, which means that $y_2=1$ , i.e. any constant solution is in the solution set.

4 0

3 years ago