All categories

3 years ago

Describe how you know whether an equation will be true for all values of x or true for no values of x

Mathematics

1 answer:

love history [14]3 years ago

5 0

Answer:

<h2>brainiest me</h2>

Step-by-step explanation:

The value returned is TRUE if all of the values in x are TRUE (including if there are no values), and FALSE if at least one of the values in x is FALSE. Otherwise the value is NA (which can only occur if na.rm = FALSE and ... contains no FALSE values and at least one NA value).

You might be interested in

CALCULATOR COLOR THEME 0 ZOOM 2. Danielle wants to translate triangle XYZ so that vertex Yis mov (-4,2) to (0,-3). Which of the

pogonyaev

Answer:The answer is (x,y)->(+4,y-5)

Step-by-step explanation:

since -4 to 0 is + 4 and 2 to -3 is -5

6 0

3 years ago

Find the outlier of the set of data: 24,18, 33, 20, 18, 24, 8, 24

sertanlavr [38]

8, because it's the has the greatest difference from the number average.

3 0

3 years ago

Read 2 more answers

which number comes first when The numbers are listed from least to greatest a. 1/6 b. -1 c. 5 squared d. -7/2 e. 4.6

Marina86 [1]

the answer is C. -1

Step-by-step explanation:

-1 is the smallest number

8 0

3 years ago

Kate sets a goal to walk 3 miles every day for a year. How many miles more or less than 1000 miles will she walk in the year

maw [93]

Answer:

She would walk 95 miles more because

Step-by-step explanation:

3 miles per day multiplied by 365 days = 1095

You have to subtract the number of miles currently by 1095 so that would be

1095-1000=95

Therefore 95 Is your answer

7 0

3 years ago

Which shows the following expression after the negative exponents have been eliminated?

Mazyrski [523]

Answer:

$a^2b^2$

Step-by-step explanation:

Given the algebraic expression:

$\dfrac{a^3b^{-2}}{ab^{-4}} , a\neq 0, b\neq 0$

We are required to eliminate the negative exponents and find the resulting expression.

Using the Negative Exponent Law of Indices: $x^{-y}=\dfrac{1}{x^y}$

Using the Division Law of Indices: $\dfrac{t^x}{t^y}=t^{x-y}$

Therefore:

$\dfrac{a^3b^{-2}}{ab^{-4}} = a^{3-1}\cdot b^{-2-(-4)}\\$Simplifying\\a^{3-1}\cdot b^{-2-(-4)}=a^2b^{-2+4}\\=a^2b^2\\Therefore:\\\dfrac{a^3b^{-2}}{ab^{-4}}=a^2b^2$

7 0

3 years ago

Read 2 more answers