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3 years ago

Please help me Find YZ

Mathematics

1 answer:

Ad libitum [116K]3 years ago

4 0

Answer:

blah

Step-by-step explanation:

blah

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-6x-8<4 plz help me this is sixty percent of my grade

masya89 [10]

Answer:

x>-2

Step-by-step explanation:

5 0

3 years ago

If there are five numbers in a data set, how many modes can there be? (Check all that apply.)

STatiana [176]

(a) 1

Explanation:

The mode of the data set is the number that occurs most frequently in the set. To easily find the mode. put the numbers in order from least to greatest and count how many times each number occurs. The number that occurs the most is the mode.

If 5 different numbers are there in a data set then, the mode would be 1 as a number is present minimum one number of times.

4 0

3 years ago

Which one of the following statements is true of perpendicular lines?

miv72 [106K]

A. Thay intersect at one point.

5 0

3 years ago

karen amad an hans have a total of 113 in their wallets? if hans has 4 times the amount of karen an karen has 7 dollars more tha

zvonat [6]

Amount in Han wallet = $ 80

Amount in amad wallet = $ 13

Amount in karen wallet = $ 20

Solution:

Let "h" be the amount in Han's wallet

Let "k" be the amount in karen wallet

Let "a" be the amount in amad wallet

They have a total of 113 in their wallet

Therefore,

h + k +a = 113 ------- eqn 1

If hans has 4 times the amount of karen

Amount with Han = 4 times the amount with Karen

h = 4k ------- eqn 2

And karen has 7 dollars more than amad

k = 7 + a

a = k - 7 ---------- eqn 3

Substitute eqn 2 and eqn 3 in eqn 1

4k + k + k - 7 = 113

6k - 7 = 113

6k = 113 + 7

6k = 120

k = 20

Substitute k = 20 in eqn 3

a = 20 - 7

a = 13

Substitute k = 20 in eqn 2

h = 4k = 4(20) = 80

h = 80

Summarizing the results:

Amount in Han wallet = $ 80

Amount in amad wallet = $ 13

Amount in karen wallet = $ 20

6 0

3 years ago

The value of -1 is a lower bound for the zeros of the function show below.

dedylja [7]

To perform this check, you must use the following theorem: $a$ is a lower bound for the zeroes of $f(x)$ if, when divide $f(x)$ by $(x-a)$ , the quotient and the remainter alternate signs.

Since the long division yields the result

$\dfrac{x^4+x^3-11x^2-9x+18}{x+1}=x^3-11x+2+\dfrac{16}{x+1}$

You can see that the signs don't alternate (the last two terms are positive). So, -1 is not a lower bound.

In fact, the actual roots of the polynomial are -3, -2, 1, 3, so as you can see there are roots smaller than -1.

7 0

3 years ago