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maria [59]
3 years ago
8

State the opposite value of the number: a. 1 ______ b. 2______ c. 15______

Mathematics
1 answer:
Sergio [31]3 years ago
5 0

A. -1

B. -2

C. -15

This is the anwser

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ZB is 3x-5 BA is x+3
Akimi4 [234]
If B is the midpoint of ZA, then ZB = <span>BA
</span>3x-5 = <span>x+3
3x - x = 3 + 5
2x = 8
x = 8/2
x = 4

</span>ZA = ZB + BA
      =  3x-5 + x+3
      = 3*4 - 5 + 4 + 3
      = 12 - 5 + 7 
      = 14

Answer: <span>D. 14</span>
3 0
3 years ago
8)<br> 8x + 6<br> 14x - 2<br> A) 11<br> C) 8<br> B) -9<br> D) -7
Travka [436]

Answer:

C) 8

Step-by-step explanation:

Consecutive angles are equal to 180. So 22x+4+180. Solve that to get x=8. Hope I helped and post more questions :)

6 0
3 years ago
5x +14+13x-2 i need help on these type of problems please
OleMash [197]

Answer:

18x+12

Step-by-step explanation:

5x+14+13x-2

combine Like terms:

5x+13x

14-2

18x+12

3 0
3 years ago
Read 2 more answers
A see-saw is 25 feet long with a fulcrum in the middle of the board. If a 60 lb. child sits three feet from the fulcrum, what is
My name is Ann [436]

Answer:

14.4 lb

Step-by-step explanation:

In a see-saw in equilibrium, the torque generated by one side needs to be the same generated in the other side. The torque is calculated by the product between the mass and the distance to the center of the see-saw.

The torque generated by the child is:

T1 = 60 * 3 = 180 lb*feet

So, the torque generated by the weight needs to be higher than T1 in order to lift the child.

The lowest mass is calculated when the mass is in the maximum distance, that is, 12.5 feet from the center.

So, we have that:

T2 = 180 = mass * 12.5

mass = 180/12.5 = 14.4 lb

So the lowest weight is 14.4 lb

4 0
3 years ago
Can someone please answer these questions to help me understand? Please and thank you! Will mark as brainliest!!
Nadya [2.5K]

QUESTION 1  

If a function is continuous at x=a, then \lim_{x \to a}f(x)=f(a)  

Let us find the limit first,  

\lim_{x \to 4} \frac{x-4}{x+5}  

As x \rightarrow 4, x-4 \rightarrow 0,x+5 \rightarrow 9 and f(x) \rightarrow \frac{0}{9}=0  

\therefore \lim_{x \to 4} \frac{x-4}{x+5}=0  

Let us now find the functional value at x=4  

f(4)=\frac{4-4}{4+5} =\frac{0}{9}=0  

Since  

\lim_{x \to 4} f(x)=\frac{x-4}{x+5}=f(4), the function is continuous at a=4.  

QUESTION 2  

The correct answer is table 2. See attachment.


In this table the values of x approaches zero from both sides.


This can help us determine if the one sided limits are approaching the same value.

As we are getting closer and closer to zero from both sides, the function is approaching 2.


The values are also very close to zero unlike those in table 4.


The correct answer is B


QUESTION 3


We want to evaluate;


\lim_{x \to 1} \frac{x^3+5x^2+3x-9}{x-1}


using the properties of limits.


A direct evaluation gives \frac{1^3+5(1)^2+3(1)-9}{1-1}=\frac{0}{0}.


This indeterminate form suggests that, we simplify the function first.


We factor to obtain,


\lim_{x \to 1} \frac{(x-1)(x+3)^2}{x-1}


We cancel common factors to get,


\lim_{x \to 1} (x+3)^2


=(1+3)^2=16


The correct answer is D



QUESTION 4

We can see from the table that as x approaches -2 from both sides, the function approaches -4


Hence the limit is -4.


See attachment


The correct answer is option A

3 0
3 years ago
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