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Ket [755]
3 years ago
15

Which of the following is not a scatter plot below?

Mathematics
1 answer:
Lubov Fominskaja [6]3 years ago
7 0

Answer:

the answer is an outlier

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3x-7&gt;5<br> Find the solution, please show your work. Thank You!1
miskamm [114]
To find x just change the <span>> to a =
</span><span>3x-7=5
add 7
3x=12
then divide by 3
x=4

</span>

7 0
3 years ago
Read 2 more answers
What is the answer?
mezya [45]

Answer:

18.1 cm

Step-by-step explanation:

AD segment breakup: 11=4+7

Pythagorean Theorem:

7^2+b^2=16^2

49+b^2=256

x^2=207

b=sqrt(207)

b=3sqrt(23)

Pythagorean Theorem:

11^2+3sqrt(23)^2=c^2

121+207=c^2

328=c^2

18.11=c

So the length of AC is around 18.1 cm

5 0
3 years ago
Evaluate the function at the indicated values. f(x)= 6x^2+ 5x -12
seropon [69]

Step-by-step explanation:

We need to find the value of the function at the indicated value.

f(x)= 6x^2+ 5x -12

Indicated values are : f(-7), f(0), f(6), f(7)

To find f(-7), put x = -7 in the given function

f(-7)= 6(-7)^2+ 5(-7) -12=247

To find f(0), put x = 0 in the given function

f(0)= 6(0)^2+ 5(0) -12=-12

To find f(6), put x = 6 in the given function

f(6)= 6(6)^2+ 5(6) -12=234

To find f(7), put x = 7 in the given function

f(7)= 6(7)^2+ 5(7) -12=317

Hence, this is the required solution.

8 0
4 years ago
which three numbers make a right triangle the square root of 4 the square root of 5 the square root of 9 + square root of 6 the
statuscvo [17]
Please, group your sets of numbers, using {   }  notation or at least semicolons ( ; ).  Thanks.

Looking at what I think is your first set:   { sqrt(4), sqrt(5), sqrt(16) }

Square each of these and then subst. the results into the Pythagorean Theorem:

  { 4, 5, 16 }        Do 4 and 5 when added together result in 16?  NO.
Therefore,   { sqrt(4), sqrt(5), sqrt(16) }  does not produce a right triangle.

Your turn.  Pick out the next 3 numbers and test them using the Pyth. Thm.               
8 0
3 years ago
Simplify the rational expression. State any restrictions on the variable n^4-11n^2+30/ n^4-7n^2+10
djyliett [7]
To factor both numerator and denominator in this rational expression we are going to substitute n^{2} with x; so n^{2} =x and n ^{4} =  x^{2}. This way we can rewrite the expression as follows:
\frac{n^{4}-11n^{2} +30 }{n^{2}-7n^{2} +10 } =  \frac{ x^{2} -11x+30}{ x^{2} -7x+10}
Now we have two much easier to factor expressions of the form a x^{2} +bx+c. For the numerator we need to find two numbers whose product is c (30) and its sum b (-11); those numbers are -5 and -6. (-5)(-6)=30 and -5-6=-11.
Similarly, for the denominator those numbers are -2 and -5. (-2)(-5)=10 and -2-5=-7. Now we can factor both numerator and denominator:
\frac{ x^{2} -11x+30}{ x^{2} -7x+10} = \frac{(x-6)(x-5)}{(x-2)(x-5)}
Notice that we have (x-5) in both numerator and denominator, so we can cancel those out:
\frac{x-6}{x-2}
But remember than x= n^{2}, so lets replace that to get back to our original variable:
\frac{n^{2}-6 }{n^{2}-2 }
Last but not least, the denominator of rational expression can't be zero, so the only restriction in the variable is n^{2} -2 \neq 0
n^{2}  \neq 2
n \neq +or- \sqrt{2}
5 0
4 years ago
Read 2 more answers
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