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AfilCa [17]
3 years ago
12

Which statement is not true? X squared equals 100 if and only if X equals 10 orX equals -10. If X squared equal 100 then X equal

s 10
Mathematics
1 answer:
mamaluj [8]3 years ago
7 0

Answer:

the second statement is not true

Step-by-step explanation:

If x squared equals 100, it doesn't mean necessarily that x is positive ten, since you can get 100 also by squaring the value (-10):

(-10)(-10)^{2} = (-10) x (-10) = 100

The second statement is ignoring the possibility of x being a negative ten which also gives 100 as answered when squared.

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Y=1/3x-4<br> Click<br> slope = here<br> Click<br> y-intercept = here
dmitriy555 [2]
Slope=1/3
y intercept=-4
8 0
3 years ago
Pls help! I'm confused. Can you explain this if you can.
alexdok [17]

Answer:

From a <u>table</u>, for an ordered pair (0, y), <em>y</em> will not be <u>zero</u>. From a <u>graph</u>, the y-intercept will not be <u>zero</u>.  From an equation, it will have the form, y = mx + b where b is <u>≠ 0</u>.

Step-by-step explanation:

  • From a <u>table</u>, for an ordered pair (0, y), <em>y</em> will not be <u>zero</u>. If there is not a constant rate of change in the data displayed in a table, then the table represents a nonlinear nonproportional relationship.
  • From a <u>graph</u>, the y-intercept will not be <u>zero</u>. This means that it doesn't contain or go through the origin.  
  • From an equation, it will have the form, y = mx + b where b is <u>≠ 0.</u> (not equal to zero). If an equation is not a linear equation, it represents a nonproportional relationship. A <u>linear equation</u> of the form y = mx + b may represent either a <em>proportional</em> (b = 0) or <em>nonproportional</em> (b ≠ 0) relationship.  Therefore, when b ≠ 0, the relationship between <em>x</em> and <em>y</em> is <u>nonproportional</u>.
7 0
3 years ago
Estimate the integral ∫6,0 x^2dx by the midpoint estimate, n = 6
Anettt [7]
Splitting up the interval [0, 6] into 6 subintervals means we have

[0,1]\cup[1,2]\cup[2,3]\cup\cdots\cup[5,6]

and the respective midpoints are \dfrac12,\dfrac32,\dfrac52,\ldots,\dfrac{11}2. We can write these sequentially as {x_i}^*=\dfrac{2i+1}2 where 0\le i\le5.

So the integral is approximately

\displaystyle\int_0^6x^2\,\mathrm dx\approx\sum_{i=0}^5({x_i}^*)^2\Delta x_i=\frac{6-0}6\sum_{i=0}^5({x_i}^*)^2=\sum_{i=0}^5\left(\frac{2i+1}2\right)^2

Recall that

\displaystyle\sum_{i=1}^ni^2=\frac{n(n+1)(2n+1)}6
\displaystyle\sum_{i=1}^ni=\frac{n(n+1)}2
\displaystyle\sum_{i=1}^n1=n

so our sum becomes

\displaystyle\sum_{i=0}^5\left(\frac{2i+1}2\right)^2=\sum_{i=0}^5\left(i^2+i+\frac14\right)
=\displaystyle\frac{5(6)(11)}6+\frac{5(6)}2+\frac54=\frac{143}2

8 0
3 years ago
How can you tell that the inequality 3t+1&gt; 3t + 2
lutik1710 [3]

Answer:

see below

Step-by-step explanation:

3t+1> 3t + 2

Subtract 3t from each side

1>2

This is never true so there is no solution

Since the variable terms are the same, we only have to look at the constants

6 0
3 years ago
What is the simplified answer to 5+3w+3-w​
DaniilM [7]

Answer:

<h2>5 + 3w + 3 - w = 2w + 8</h2>

Step-by-step explanation:

5+3w+3-w\qquad\text{combine like terms}\\\\=(3w-w)+(5+3)\\\\=2w+8

6 0
3 years ago
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