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Rina8888 [55]
3 years ago
14

Give an example of an open equation? my answer is, x + 4 = 7 Until the value of x is found, it is unknown whether the equation i

s true or false. am i right?
Mathematics
1 answer:
Sphinxa [80]3 years ago
8 0
Yes you’re right you cannot know if the equation is true or false
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Use the discriminant to determine the number of real solutions to the quadratic equation.
MAVERICK [17]

Answer:

12x2+12x+1=0

Step-by-step explanation:

well first you need to add all the x's and add your additional numbers, keep all +'s signals and then minuse the addition numbers which makes your anserw negitive and then divide your ansew by your x.

3 0
2 years ago
Solve for u.<br> U/-1 +-4=-1<br> U =
kherson [118]

Answer:

U = -3

Step-by-step explanation:

U/-1 +-4=-1

Add 4 to each side

U/-1 +-4+4=-1+4

U / -1 = 3

Multiply each side by -1

U / -1  * -1 = 3* -1

U = -3

3 0
3 years ago
Oomygawd plz help me with this math hw
aev [14]

Hi There!

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

Slope Intercept Form: y = mx + b

Where: m = slope and b = y-intercept

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

Question 11: The slope being 0.5 is the pay for the amount of miles he pays daily.

Question 12: y = 2x + 20

Question 13: y = 25x + 50

Question 14: y = 1/2x + 6

Question 15: y = 2x + 15

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Hope This Helps :)

6 0
3 years ago
Help pls!
4vir4ik [10]

Answer:

the probability that the next gumball that comes out will be neither yellow nor blue = 6/17

Step-by-step explanation:

neither yellow nor blue gumball = pink gumball = 6

3 yellow + 8 blue + 6 pink = total =17

3 0
3 years ago
Find the directional derivative of f(x,y,z)=2z2x+y3f(x,y,z)=2z2x+y3 at the point (−1,4,3)(−1,4,3) in the direction of the vector
Feliz [49]

f(x,y,z)=2z^2x+y^3

f has gradient

\nabla f(x,y,z)=2z^2\,\vec\imath+3y^2\,\vec\jmath+4xz\,\vec k

which at the point (-1, 4, 3) has a value of

\nabla f(-1,4,3)=18\,\vec\imath+48\,\vec\jmath-12\,\vec k

I'm not sure what the given direction vector is supposed to be, but my best guess is that it's intended to say \vec u=15\,\vec\imath+25\,\vec\jmath, in which case we have

\|\vec u\|=\sqrt{15^2+25^2}=5\sqrt{34}

Then the derivative of f at (-1, 4, 3) in the direction of \vec u is

D_{\vec u}f(-1,4,3)=\nabla f(-1,4,3)\cdot\dfrac{\vec u}{\|\vec u\|}=\boxed{\dfrac{294}{\sqrt{34}}}

4 0
3 years ago
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