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Mrrafil [7]
3 years ago
12

10 + Brainiest Please Help!

Mathematics
1 answer:
Agata [3.3K]3 years ago
3 0

no. mike added the two fractions incorrectly by not changing them to common denominators.

the correct answer:

4/7 + 1/4 = ?

4/7 = 16/28 (multiply both the numerator and denominator by 4 to get the common denominator fraction)

1/4 = 7/28 (multiply both the numerator and denominator by 7 to get the common denominator fraction)

16/28 + 7/28 = 23/28 (only add the numerators!!)

mike covered <u>23/28 </u>of the distance from his home to the library. (always check to see if your answer can be simplified)

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Pleas help, Enter an inequality that represents the graph in the box.
ddd [48]

Answer:

the answer is -20

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
2 times a number is the same as 12 minus 3 times the number. What
Ksju [112]

Answer:

a. 2n = 3n − 12, 12

Step-by-step explanation:

Let the number be represented by n.

Then 2 times the number is 2*n = 2n

3 times the number = 3*n = 3n

12 minus 3 times the number is 3n-12

It is given that 2 times the number is the same as 12 minus 3 times the number.

Expressing this equation form:

\[2n=3n-12\]

Solving the equation,

\[3n-2n=12\]

\[=> n=12\]

Hence the required number is 12.

3 0
3 years ago
Read 2 more answers
D=R(100-x)/100 for r '<br><br> i need help thank you
balandron [24]
Times oth sides by 100
100d=r(100-x)
divide both sides by (100-x)
\frac{100D}{100-x}=r
x= \frac{100D}{100-x}

6 0
3 years ago
I’ll put you as brainliest
kari74 [83]

The law of an object moving with constant acceleration is

s(t)=s_0+v_0t+\dfrac{1}{2}at^2

Where s is space, t is time, s_0 is the initial position, v_0 is the initial velocity and a is the acceleration.

In this case, if we choose a reference grid with the vertical axis pointing upwards, the acceleration of gravity will point downwards (and thus be negative). The initial position is zero, because the rocket is on the ground, and the initial velocity is 100 (positive because pointing upwards).

So, its law is

h(t)=100t-\dfrac{1}{2}gt^2

(I changed s for h since the rocket is moving vertically, so its position is actually its height. Also, g is the acceleration due to gravity).

The rocket hits the ground if its height is zero, so if we set h=0 we have

0=100t-\dfrac{1}{2}gt^2 \iff t(100-\dfrac{1}{2}gt)=0

Solving for t, we have either t=0, or

100-\dfrac{1}{2}gt=0 \iff 100=\dfrac{1}{2}gt \iff 200=gt \iff t=\dfrac{200}{g}

The solution t=0 means that at the beginning the rocket is on the ground. So, we're interested in the other solution. Considering that g is about 32.2 feet/s^2, we have

t=\dfrac{200}{g}\approx \dfrac{200}{32.2}\approx 6.21

3 0
3 years ago
Graph x+y=9<br> And y=x squared
Akimi4 [234]
Just type it into the graphing calculator, but reorder and solve for y first. 
x + y = 9
y = 9 - x

y = x^2

In the graphing calculator, 
Y1 = 9 - x, 
Y2 = x^2

If you don't have a graphing calculator try desmos.
4 0
4 years ago
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