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lina2011 [118]
3 years ago
12

Find the values of m and b that will create a system of equations with no solution. Check all that apply.

Mathematics
1 answer:
Dafna1 [17]3 years ago
7 0
Y = mx + b ; where m is the slope and b is the y-intercept.

A) y = -2x + 3/2 ; The slope is -2

For the system of equation to have no solution, they must form parallel lines. Parallel lines have the same slope. They don't intersect, thus, there is no solution.

<span>C.) m = –2 and b = -1/3
</span><span>F.) m = –2 and b = -2/3
</span>
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The volume of the sphere is 500 1 cubic units.
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The Value of X is 5 units

Step-by-step explanation:

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CAN SOMEONE HELP ME PLEASE
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Each morning, Joey leaves his house at 7:25 A.M. It takes Joey 10 minutes to walk to the bus stop. Joey rides the bus for 50 min
Akimi4 [234]

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8:30 AM

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3 years ago
In a psychology class, 44 students have a mean score of 96.6 on a test. Then 13 more students take the test and their mean score
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Hope this helped

5 0
2 years ago
What’s the differential equation of
worty [1.4K]

Answer:

y=\frac{1}{x^{2}-6x+13 }

Step-by-step explanation:

We have given,

                        \frac{dy}{dx}=y^{2}(6-2x)

and initial condition x=3,\  y=\frac{1}{4}

Now,

\frac{dy}{dx}=y^{2}(6-2x)

Rearranging the variables, we get

\frac{dy}{y^{2} }=(6-2x).dx

Applying integration both sides, we get

\int\ {\frac{dy}{y^{2} } } \,=\int\ {(6-2x).} \, dx

⇒\frac{-1}{y} =6x-\frac{2x^{2} }{2}

⇒ \frac{-1}{y}=6x-x^{2}  +C                  

Putting the initial condition (i.e., x=3,\  y=\frac{1}{4}), we get

⇒ -4=6\times3-(3)^{2}+C

⇒ -4=18-9+C

∴ C=-13

We have,  \frac{-1}{y}=6x-x^{2}  +C    

now putting the value of C in above equation, we get

⇒ \frac{-1}{y}=6x-x^{2}  -13

⇒  \frac{1}{y}=-6x+x^{2}  +13

y=\frac{1}{x^{2}-6x+13 }

5 0
3 years ago
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