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Gelneren [198K]
2 years ago
15

Wich is the missing number in the prime factorization will rate branleist

Mathematics
1 answer:
gogolik [260]2 years ago
5 0
The answer is a. 5. Good luck
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Find the slope, Y-intercept, and equation (make sure there is no decimals if there is a decimal make it into a fraction)!
adelina 88 [10]

Answer:

Slope is 5/3

y-intercept = -3

y = 5/3x - 3

Step-by-step explanation:

To find the slope, take to coordinates and use the equation below

m = \frac{\text{rise}}{\text{run}} = \frac{y_2 - y_1}{x_2 - x_1}

7-2 = 5

6-3= 3

Now you have your slope which is 5/3

Then take the slope and input it into point slope form to get your y-intercept.

y-y₁=m(x-x₁)

y-2 = 5/3(x-3)

y-2=5/3x-5

add two to both sides to get Y by itself

y = 5/3x - 3

now you have your y-intercept, which is -3

Then check your answer by inputing two different coordinates for X and Y

7 = 5/3(6)-3

7=7

12= 5/3(9)-3

12=12

6 0
3 years ago
Kari is flying a kite. She releases 50 feet of string. What is the approximate difference in the height of the kite when the str
nirvana33 [79]
Sinα=h/L  where h=height, L=string length...

h=Lsinα  so

h(25°)=50sin25≈21.1ft

h(45°)=50sin45≈35.4ft
5 0
3 years ago
Read 2 more answers
How are vectors used in real life?
aleksandr82 [10.1K]

Answer:

Vectors can be used in sports to define how a baseball player moves to catch a high fly baseball they can also be used with aircrafts.


8 0
3 years ago
Read 2 more answers
True or False: The following sequence of numbers is an arithmetic sequence.
slava [35]

Answer:

False

Step-by-step explanation:

3 0
3 years ago
wo point charges lie on the $x$ axis. A charge of +6.24 $\mu C$ is at the origin, and a charge of -9.55 $\mu C$ is at $x$ = 12.0
Andreyy89

Answer:

E_n=34,467,075.42\ N/C

Step-by-step explanation:

<u>Electric Field</u>

The electric field produced by a point charge Q at a distance d is given by

\displaystyle E=K\cdot \frac{Q}{d^2}

Where

K = 9\cdot 10^9\ Nw.m^2/c^2

The net electric field is the vector addition of the individual electric fields produced by each charge. The direction is given by the rule: If the charge is positive, the electric field points outward, if negative, it points inward.

Let's calculate the electric fields of each charge at the given point. The first charge q_1=+6.24\mu C=6.24\cdot 10^{-6}C is at the origin. We'll calculate its electric field at the point x=-3.85 cm. The distance between the charge and the point is d=3.85 cm = 0.0385 m, and the electric field points to the left:

\displaystyle E_1=9\cdot 10^9\cdot \frac{6.24\cdot 10^{-6}}{0.0385^2}

E_1=37,888,345.42\ N/C

Similarly, for q_2=-9.55\mu C=-9.55\cdot 10^{-6}C, the distance to the point is 12 cm + 3.85 cm = 15.85 cm = 0.1585 m. The electric field points to the right:

\displaystyle E_2=9\cdot 10^9\cdot \frac{9.55\cdot 10^{-6}}{0.1585^2}

E_2=3,421,270\ N/C

Since E1 and E2 are opposite, the net field is the subtraction of both

E_n=37,888,345.42\ N/C-3,421,270\ N/C

\boxed{E_n=34,467,075.42\ N/C}

6 0
3 years ago
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