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aleksandrvk [35]
3 years ago
10

Translate to an inequality. "A number m is at most 2"

Mathematics
1 answer:
Leni [432]3 years ago
8 0

Answer:

m \leq 2

Step-by-step explanation:

"at most" can translate to "less than or equal to," meaning m is less than or equal to 2.

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Answer:

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5 0
3 years ago
1. Write the word sentence as an equation. 2. Then solve the equation. 5/7a=25
seropon [69]

Answer:

a = 35

Step-by-step explanation:

5/7a = 25

multiply both sides by 7:

5a = 175

divide by 5:

a = 35

Hope this helps !!!

7 0
3 years ago
Two students in your class, Tucker and Karly, are disputing a function. Tucker says that for the function, between x = –3 and x
Effectus [21]
The difference between Tucker and Karly's take is that Tucker's solution is analytical while Karly's is graphical. But both are correct either way. 

For Tucker's solution, let's say at x=-3 the value for y is 4, and at x=3, the value of y is still 4, then the average rate of change or slope is 0. Note that the slope of the curve is Δy/Δx. Since there is no change for Δy, the slope is zero.

For Karly's solution, even if the curve travels high or low but would have the same elevation of x=-3 and x=3, the average rate of change is still zero. It is actually just same with Tucker's but Karly just verbalizes her solution that was observed visually.
7 0
3 years ago
X^2 -25 =0. Factor and use the zero product property to solve.
Gwar [14]
X^2 - 25 is a difference of squares which has a special factorization.

In general, a^2 - b^2 = (a + b)(a - b)

x^2 - 25 = 0

(x + 5)(x - 5) = 0

x + 5 = 0  or   x - 5 = 0

x = -5   or   x = 5
7 0
3 years ago
Read 2 more answers
f of x equals 4 x over quantity x squared minus 16. Show all work to identify the asymptotes and zero of the function.
yuradex [85]

Given:

The function is

f(x)=\dfrac{4x}{x^2-16}

To find:

The asymptotes and zero of the function.

Solution:

We have,

f(x)=\dfrac{4x}{x^2-16}

For zeroes, f(x)=0.

\dfrac{4x}{x^2-16}=0

4x=0

x=0

Therefore, zero of the function is 0.

For vertical asymptote equate the denominator of the function equal to 0.

x^2-16=0

x^2=16

Taking square root on both sides, we get

x=\pm \sqrt{16}

x=\pm 4

So, vertical asymptotes are x=-4 and x=4.

Since degree of denominator is greater than degree of numerator, therefore, the horizontal asymptote is y=0.

8 0
3 years ago
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