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2 years ago

11

Evaluate the expression when x=10, y=−2, and z=−5.

1 answer:

Vlad [161]2 years ago

8 0

$\frac{ - {x}^{2} + 6z }{y}$

if x=10 y=-2 and z=-5 then...

$\frac{ { - 10}^{2} + 6 \times - 5}{ - 2}$

we solve this to get...

<em>65</em>

therefore the answer to your question would be 65

hope this helped-have a good day bro cya)

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The mean number of people per day visiting a museum in July was 120. If 20 more people each day visited the museum in August, wh

-BARSIC- [3]

The correct answer is 140 because the mean, average, number from July per day represents how many people would have visited each day if they were all even.

If there were 20 more people each day in August, the mean would be 140. This would represent how many people visited each day.

8 0

3 years ago

Read 2 more answers

Evaluate another function please

kari74 [83]

Answer:

70, I think

Step-by-step explanation:

f(10)= -10^2- 3(10)

f(10)= 100-30

f(10) = 70

6 0

2 years ago

kifflom [539]

$\\ \sf\longmapsto \dfrac{x^2-1}{x^2+5x+4}\leqslant 0$

Solving denominator

$\\ \sf\longmapsto x^2+5x+4>0$

$\\ \sf\longmapsto x^2+4x+x+4>0$

$\\ \sf\longmapsto x(x+4)+1(x+4)>0$

$\\ \sf\longmapsto (x+4)(x+1)>0$

$\\ \sf\longmapsto x>-4\:or x>-1$

Hence x $\neq$ -1
x can't be 0 as it makes function undefined

$\\ \sf\longmapsto -4$

5 0

2 years ago

Three cards are drawn from a standard deck of 52 cards without replacement. Find the probability that the first card is an ace,

MrRissso [65]

Answer:

$4.82\cdot 10^{-4}$

Step-by-step explanation:

In a deck of cart, we have:

a = 4 (aces)

t = 4 (three)

j = 4 (jacks)

And the total number of cards in the deck is

n = 52

So, the probability of drawing an ace as first cart is:

$p(a)=\frac{a}{n}=\frac{4}{52}=\frac{1}{13}=0.0769$

At the second drawing, the ace is not replaced within the deck. So the number of cards left in the deck is

$n-1=51$

Therefore, the probability of drawing a three at the 2nd draw is

$p(t)=\frac{t}{n-1}=\frac{4}{51}=0.0784$

Then, at the third draw, the previous 2 cards are not replaced, so there are now

$n-2=50$

cards in the deck. So, the probability of drawing a jack is

$p(j)=\frac{j}{n-2}=\frac{4}{50}=0.08$

Therefore, the total probability of drawing an ace, a three and then a jack is:

$p(atj)=p(a)\cdot p(j) \cdot p(t)=0.0769\cdot 0.0784 \cdot 0.08 =4.82\cdot 10^{-4}$

4 0

3 years ago

Please Help! The graph of y=f(x) is shown below. What are all of the real solutions of f(x)=0? (Picture of graph included)

qwelly [4]

Answer:

(7, 0), (-8, 0), (0, 0)

Step-by-step explanation:

y = f(x) = 0

the real solutions are the ones that cross the x-axis because when a point is on the x-axis, its y coordinate will be 0.

=> (7, 0), (-8, 0), (0, 0)

6 0

3 years ago