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Drupady [299]
2 years ago
11

When solving a system by substitution, one should

Mathematics
1 answer:
neonofarm [45]2 years ago
6 0
B, because to substitute for a variable you have to have an equation to substitute first.
You might be interested in
Sketch the area under the standard normal curve over the indicated interval and find the specified area. (Round your answer to f
Angelina_Jolie [31]

Answer:

P(z>1.51) =1-P(Z

And i fwe look into the normalstandar dtable we can find the desired probability:

P(z>1.51) =1-P(Z

And then the The area to the right of z = 1.51 is 0.0655

Step-by-step explanation:

For this case we want to find the following probability:

P(z>1.51)

And for this case we can use the normal standard distribution and the complement rule and we have this:

P(z>1.51) =1-P(Z

And i fwe look into the normalstandar dtable we can find the desired probability:

P(z>1.51) =1-P(Z

And then the The area to the right of z = 1.51 is 0.0655

7 0
3 years ago
Which of the following statements is true about the division expression 686.54 ÷ a? A. If a is a number greater than 686.54, the
borishaifa [10]

Answer:

D-- if a is a number between 0 and 1, the quotient will be greater

Step-by-step explanation:

5 0
3 years ago
Calculate the length of x.
Naya [18.7K]

Answer:

B

Step-by-step explanation:

Use the pythagorean theorem to find the hypotenuse. You can do this by squaring 12 and 11 (separately) and then adding those together. You'll get 265. However, since this is c squared and not c, you have to find the square root for the correct answer. The square root of 265 is approximately 16.28. Hope this helps!

6 0
3 years ago
Solve: (⅓) x – 13 &gt; 14<br> A) x &lt; 81 <br> B) x &gt; 81<br> C) x &gt; 9 <br> D) x &lt; 9
frozen [14]

\dfrac{1}{3}x-13>14\qquad\text{add 13 to both sides}\\\\\dfrac{1}{3}x>27\qquad\text{multiply both sides by 3}\\\\\boxed{x>81}\to\boxed{B)}

4 0
3 years ago
Consider the quadratic equation ax^2+bx+5=0,where a, b and c are rational numbers and the quadratic has two distinct zeros.
FrozenT [24]
You have shared the situation (problem), except for the directions:  What are you supposed to do here?  I can only make a educated guesses.  See below:

Note that if    <span>ax^2+bx+5=0    then it appears that c = 5 (a rational number).

Note that for simplicity's sake, we need to assume that the "two distinct zeros" are real numbers, not imaginary or complex numbers.  If this is the case, then the discriminant,    b^2 - 4(a)(c), must be positive.  Since c = 5, 

b^2 - 4(a)(5) > 0, or b^2 - 20a > 0.

Note that if the quadratic has two distinct zeros, which we'll call "d" and "e," then 

(x-d) and (x-e) are factors of ax^2 + bx + 5 = 0, and that because of this fact,

         - b plus sqrt( b^2 - 20a )
d =  ------------------------------------
                      2a

and

 </span>         - b minus sqrt( b^2 - 20a )
e =  ------------------------------------
                      2a

Some (or perhaps all) of these facts may help us find the values of "a" and "b."  Before going into that, however, I'm asking you to share the rest of the problem statement.  What, specificallyi, were you asked to do here?

7 0
3 years ago
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