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sergeinik [125]
2 years ago
12

Please help me with these it’s due TODAY!!!!! ( This is Simplifying expressions in 7th grade)

Mathematics
2 answers:
son4ous [18]2 years ago
5 0
Picture won’t open for me... if you reply to this with the questions I can definitely help you.
ANTONII [103]2 years ago
5 0

Answer:

left side up : -16x+5

left side down: 8x+3

right side up: -x

right side down:-5x

Step-by-step explanation:

hope i helped <3 :P

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What are the length(s) of the sides labeled a and b in the triangle below?
neonofarm [45]

Answer:

14 and 14

Step-by-step explanation:

this is because it is a isosceles triangle and it has a specific rule.

8 0
2 years ago
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Which of the following points lies in Quadrant II?
RUDIKE [14]
The answer is d. (-3,8.)

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Solve by factorization<br> 2x^2 + 3x -20 = 0
saul85 [17]

Answer:

x=2.5 or x=-4

Step-by-step explanation:

factorisation= factors:1, 2, 4, 5, 10, 20 (only the positive factors)

We will use the factors 4 and 5.

=(2x-5)(x+4)=0

2x-5=0   || x+4=0

2x=+5      || x=-4

x=2.5  or x=-4

4 0
2 years ago
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The force of gravity on Mars is different than on Earth. The function of the same situation on Mars would be represented by the
sweet-ann [11.9K]

Answer:

If thrown up with the same speed, the ball will go highest in Mars, and also it would take the ball longest to reach the maximum and as well to return to the ground.

Step-by-step explanation:

Keep in mind that the gravity on Mars; surface is less (about just 38%) of the acceleration of gravity on Earth's surface. Then when we use the kinematic formulas:

v=v_0+a\,*\,t\\y-y_0=v_0\,* t + \frac{1}{2} a\,\,t^2

the acceleration (which by the way is a negative number since acts opposite the initial velocity and displacement when we throw an object up on either planet.

Therefore, throwing the ball straight up makes the time for when the object stops going up and starts coming down (at the maximum height the object gets) the following:

v=v_0+a\,*\,t\\0=v_0-g\,*\,t\\t=\frac{v_0}{t}

When we use this to replace the 't" in the displacement formula, we et:

y-y_0=v_0\,* t + \frac{1}{2} a\,\,t^2\\y-y_0=v_0\,(\frac{v_0}{g} )-\frac{g}{2} \,(\frac{v_0}{g} )^2\\y-y_0=\frac{1}{2} \frac{v_0^2}{g}

This tells us that the smaller the value of "g", the highest the ball will go (g is in the denominator so a small value makes the quotient larger)

And we can also answer the question about time, since given the same initial velocity v_0 , the smaller the value of "g", the larger the value for the time to reach the maximum, and similarly to reach the ground when coming back down, since the acceleration is smaller (will take longer in Mars to cover the same distance)

3 0
2 years ago
What is y=3x^2-2 in vertex form
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\bf \qquad \qquad \textit{parabola vertex form}&#10;\\\\\\&#10;y=a(x-{{ h}})^2+{{ k}}\\&#10;x=a(y-{{ k}})^2+{{ h}}\qquad\qquad  vertex\ ({{ h}},{{ k}})\\\\&#10;-----------------------------\\\\&#10;y=3x^2-2\iff y = 3(x-0)^2-2
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3 years ago
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