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erastova [34]
3 years ago
9

Plzzz help me i hate math:(

Mathematics
1 answer:
krok68 [10]3 years ago
8 0

Answer:

D

Step-by-step explanation:

Combine the fractions by finding a common denominator (A denominator shared by two or more fractions.)

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What do you add to 2/9 to make 1?
marshall27 [118]

Answer:8

Step-by-step explanation:

8 0
4 years ago
The graph below displays the wrist circumferences and heights of six students in Alyssa’s classroom.
coldgirl [10]

Answer:

Option D(No because the wrist circumference of 16 cm is paired with two heights.


Step-by-step explanation:

In the given graph we have  set of points representing the wrist circumferences and heights of six students in Alyssa’s classroom.

x axis  represents the wrist circumference while  y axis represent the height.

Any relation is called a function if for one value of x we have only one value of y. If the graph passes the vertical line test that is if a vertical line drawn touches the graph at only one point then the graph is called a function.In the given graph for x value x=16 there are two y values 162 and 165.So the graph is not a function.

Among all the options Option D(No because the wrist circumference of 16 cm is paired with two heights.) is the right answer.

5 0
4 years ago
Read 2 more answers
Help me solve this please.​
poizon [28]
6, 7, 8, and 9. just plug the X into the equation
4 0
3 years ago
Solve: −3(5+8x)−20≤−11
Tatiana [17]
-3(5 + 8x) - 20 ≤ -11  |use distributive property: a(b + c) = ab + ac

-15 - 24x - 20 ≤ -11

-35 - 24x ≤ -11    |add 35 to both sides

-24x ≤ 24      |change signs

24x ≥ -24    |divide both sides by 24

x ≥ -1
3 0
3 years ago
<img src="https://tex.z-dn.net/?f=I%20%3D%20%5Cfrac%7BE%7D%7B%20%5Csqrt%7BR%20%7B%7D%5E%7B2%7D%20%2B%20W%20%7B%7D%5E%7B2%7D%20L%
AleksandrR [38]

Answer:

R = sqrt[(IWL)^2/(E^2 - I^2)] or R = -sqrt[(IWL)^2/(E^2 - I^2)]

Step-by-step explanation:

Squaring both sides of equation:

     I^2 = (ER)^2/(R^2 + (WL)^2)

<=>(ER)^2 = (I^2)*(R^2 + (WL)^2)

<=>(ER)^2 - (IR)^2 = (IWL)^2

<=> R^2(E^2 - I^2) = (IWL)^2

<=> R^2 = (IWL)^2/(E^2 - I^2)

<=> R = sqrt[(IWL)^2/(E^2 - I^2)] or R = -sqrt[(IWL)^2/(E^2 - I^2)]

Hope this helps!

7 0
3 years ago
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