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Ad libitum [116K]
2 years ago
12

Guys please solve it.

Mathematics
2 answers:
Rainbow [258]2 years ago
8 0

Answer:

1) a. y = 4x + 8

  b. It will take 9 days for it to grow 44 inches tall

hope this helps :)

ruslelena [56]2 years ago
4 0

Answer:

a. y= 4x+8

b. 23 days

Step-by-step explanation:

8 is the y-int.

Slope you go up 4 and 1 to the right

y = 4*44+8

y= 176+8

y=184/8

y= 23

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How do I work it out step by step y^2+5y+6
enyata [817]

Answer:

7y+6

Step-by-step explanation

You have to do 2 times y first. That is 2y. Then you have to add 2y plus 5y, which is 7y. 7y plus 6 equals 7y+6.

3 0
3 years ago
Read 2 more answers
Evaluate each expression for w=5, x=3, y=4 and z=8.<br> 3y+6/2x
Aliun [14]
Plug in your numbers so the equation will turn out to be 3(4)+6/2(3) , you need to simplify 6/2 which will be 3 and 3 times 3 = 9 so the equation will be 3(4)+9 and 3 times 4 = 12 so the equation will now be 12+9 = 21
Answer= 21
4 0
3 years ago
It is now 3:15 pm. Is it possible to drive 135 miles and arrive before 5:00 pm if ypu drove 55 mph? Explain your answer
Troyanec [42]
Well, how long will it take you to drive 135 miles at 55mph?

at 55 mph, you're doing 55 miles every hour, so we can simply get the quotient of 135/55 and that's how many hours it'll take you to drive 135 miles at that speed.

\bf \cfrac{135}{55}\implies \cfrac{27}{11}\implies 2\frac{5}{11}~hours\implies \textit{about 2hrs, 27 minutes}

so, it takes you that long, however, from 3:15 to 5:00pm there are only 45mins + 60mins or 1hr and 45 minutes, namely 1¾ hr.

so 2hrs and 27 minutes is much later than 1¾ hr, so, no dice, you can't arrive at 5pm, actually you'll arrive around 5:42pm.
6 0
3 years ago
Read 2 more answers
Find the product of (2x + 5y)(2x − 5y).
Nana76 [90]
The final answer for this is 4x^2-25y^2.
7 0
3 years ago
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Find the max and min values of f(x,y,z)=x+y-z on the sphere x^2+y^2+z^2=81
Anton [14]
Using Lagrange multipliers, we have the Lagrangian

L(x,y,z,\lambda)=x+y-z+\lambda(x^2+y^2+z^2-81)

with partial derivatives (set equal to 0)

L_x=1+2\lambda x=0\implies x=-\dfrac1{2\lambda}
L_y=1+2\lambda y=0\implies y=-\dfrac1{2\lambda}
L_z=-1+2\lambda z=0\implies z=\dfrac1{2\lambda}
L_\lambda=x^2+y^2+z^2-81=0\implies x^2+y^2+z^2=81

Substituting the first three equations into the fourth allows us to solve for \lambda:

x^2+y^2+z^2=\dfrac1{4\lambda^2}+\dfrac1{4\lambda^2}+\dfrac1{4\lambda^2}=81\implies\lambda=\pm\dfrac1{6\sqrt3}

For each possible value of \lambda, we get two corresponding critical points at (\mp3\sqrt3,\mp3\sqrt3,\pm3\sqrt3).

At these points, respectively, we get a maximum value of f(3\sqrt3,3\sqrt3,-3\sqrt3)=9\sqrt3 and a minimum value of f(-3\sqrt3,-3\sqrt3,3\sqrt3)=-9\sqrt3.
5 0
3 years ago
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