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3 years ago

If the equation of the line is y=1/2x-3, find the x-coordinate of the Lundy whose y coordinate is 5.

Mathematics

1 answer:

Anna71 [15]3 years ago

5 0

Y=1/2x -3 find the x coordinate of the point whose y

coordinate is 5

y = 5. Write the equation as:

1/2F2x - 3 = 5

multiply both sides by 2, and you have

x - 6 = 10

x = 10 + 6

x - 16

Check solution in original equation, replace x with 16

y = 1/2F2(16) - 3

y = 8 - 3

y = 5

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Darlene wants to buy a sweater that originally cost 32$ it is on sale for 20% off. Sales tax is 6.75%. What is the final price o

AURORKA [14]

Answer:

Step-by-step explanation:

32 x 0.20 = $6.40

32 - 6.4 = $25.60

25.6 x .0675 ≈ 1.738

25.6 + 1.738 = 27.338 ≈$27.34

6 0

3 years ago

Choose ALL possible answers

muminat

Select all the correct answers:
1) Yes

2) No
x=8→h(8)=2(8)^2+5(8)+2=2(64)+40+2=128+40+2→h(8)=170
x=8→f(8)=3^8+2=6,561+2→f(8)=5,563>170=h(8)

3) Yes

4) No

5) Yes
rg=[g(3)-g(2)]/(3-2)=[g(3)-g(2)]/1→rg=g(3)-g(2)
g(3)=20(3)+4=60+4→g(3)=64
g(2)=20(2)+4=40+4→g(2)=44
rg=64-44→rg=20

rf=f(3)-f(2)
f(3)=3^3+2=27+2→f(3)=29
f(2)=3^2+2=9+2→f(2)=11
rf=29-11→rf=18

rh=h(3)-h(2)
h(3)=2(3)^2+5(3)+2=2(9)+15+2=18+15+2→h(3)=35
h(2)=2(2)^2+5(2)+2=2(4)+10+2=8+10+2→h(2)=20
rh=35-20→rh=15

rg=20>18=rf
rg=20>15=rh

6) No
x=4→g(4)=20(4)+4=80+4→g(4)=84
x=4→h(4)=2(4)^2+5(4)+2=2(16)+20+2=32+20+2→h(4)=54
x=4→f(4)=3^4+2=81+2→f(4)=83>54=h(4)
f(4)=83<84=g(4)

7 0

3 years ago

a radioactive substance has a half-life of 2 seconds. A scientist begins with a sample of 20 grams. How much of the substance re

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Answer:

1.25 g

Step-by-step explanation:

you start with 20g at 0 seconds.

divide by half after 2 seconds and you have 10g.

at 4 seconds you have 5g.

at 6 seconds you have 2.5g.

at 8 seconds you have 1.25g.

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2 years ago

What is the value of ⅚÷¾

Harman [31]

Answer:

10/9

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

Solve using the box method

Lena [83]

$\huge\text{Hey there!}$

$\large\text{Just SIMPLIFY the given EQUATION or find the DIFFERENCE}\\\large\text{OF the SQUARES... Here is the formula: }\mathsf{\bf a^2 - b^2 =(a+b)(a-b)}$

$\large\text{Equation: }\mathsf{\dfrac{(4x^2-9)}{(2x + 3)}}$

$\large\text{Rewrite }\mathsf{ 4x^9 - 9}\large\text{ in the formation of }\mathsf{a^2 - b^2}\large\text{ whereas}\mathsf{a = 2x \ \&\ b = 3.}$

$\large\text{Equation: }\mathsf{\dfrac{(2x)^2-3^2}{2x + 3}}$

$\large\text{This is where you try to do the DIFFERENCE OF its SQUARES}$

$\mathsf{\dfrac{(2x + 3)(2x - 3)}{2x + 3}}$

$\large\text{CANCEL out: }\mathsf{(2x + 3)\ - (2x +3)}\large\text{ because it gives you 0}$

$\large\text{This leaves us with }\mathsf{\bf 2x - 3}\large\text{ as your POSSIBLE ANSWER}$

$\boxed{\boxed{\large\text{Answer: \huge \bf 2x - 3}}}\huge\checkmark$

$\text{Good luck on your assignment and enjoy your day!}$

~ $\frak{Amphitrite1040:)}$

$\large\text{Note: There is/are many ways to solve for equations like this.... this was just}\\\large\text{the quickest and easiest way to understand it!}$

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3 years ago