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Whitepunk [10]
3 years ago
5

Positive numbers have negative square roots. A. True O B. False

Mathematics
2 answers:
qwelly [4]3 years ago
7 0
The Answer is A. True
DanielleElmas [232]3 years ago
3 0

Answer:

A.  True

Step-by-step explanation:

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How to solve 22 – 3y > 7
nikklg [1K]

Answer:

y < 5

Step-by-step explanation:

22 – 3y > 7

=> 22 – 7 > 3y

=> 15 > 3y

=> y < 5

Therefore, y < 5 is the solution.

Hoped this helped.

3 0
2 years ago
The graph of a function f is shown below.<br> Find f (2) and find one value of x for which f(x) = -4
nadya68 [22]

Answer: a: -2, b: 0

Step-by-step explanation:

f(2) means that x=2, and we need to solve for y.

On the graph, it has a point of (2, -2) and since it's a function that is the only possible output. Therefore, f(2)=-2.

f(x)=-4 means that y=-4, and we need to solve for x.

On the graph, the point (0,-4) exists, and since the function contains only one output, it must be correct. Therefore, x=0.

5 0
2 years ago
Dhruv is at a car dealership. He is going to randomly select a vehicle to test drive. There are
andrew-mc [135]

Answer:

P(compact cars) = 4/25

Step-by-step explanation:

We know that there are 13 trucks, 8 vans, and 4 compact cars.

Let P denotes the probability of an event

Now we are asked to find the probability of compact cars

Number of favorable events = number of compact cars = 4

total number of outcomes = 13+8+4 = 25

hence, P(compact cars) = number of favorable events/total outcomes

                                       =4/25

8 0
1 year ago
Last year, 350 people come to see a spring concert at REL. The cost per ticket was $5.00. The orchestra committee wants to incre
Anit [1.1K]
Let 0.25x is the price increase. Therefore, the number of attendees will be 350-10x and the revenue R=(350-10x)*(5+0.25x)=1750+37.5x-2.5x^2. This parabola has maximum at x=10, meaning the price=7.5
6 0
3 years ago
Read 2 more answers
Solve the following equation by completing the square. 3x^2-3x-5=13
mr Goodwill [35]

we'll start off by grouping some

\bf 3x^2-3x-5=13\implies (3x^2-3x)-5=13\implies 3(x^2-x)-5=13 \\\\\\ 3(x^2-x)=18\implies (x^2-x)=\cfrac{18}{3}\implies (x^2-x)=6\implies (x^2-x+~?^2)=6

so we have a missing guy at the end in order to get the a perfect square trinomial from that group, hmmm, what is it anyway?

well, let's recall that a perfect square trinomial is

\bf \qquad \textit{perfect square trinomial} \\\\ (a\pm b)^2\implies a^2\pm \stackrel{\stackrel{\text{\small 2}\cdot \sqrt{\textit{\small a}^2}\cdot \sqrt{\textit{\small b}^2}}{\downarrow }}{2ab} + b^2

so we know that the middle term in the trinomial, is really 2 times the other two without the exponent, well, in our case, the middle term is just "x", well is really -x, but we'll add the minus later, we only use the positive coefficient and variable, so we'll use "x" to find the last term.

\bf \stackrel{\textit{middle term}}{2(x)(?)}=\stackrel{\textit{middle term}}{x}\implies ?=\cfrac{x}{2x}\implies ?=\cfrac{1}{2}

so, there's our fellow, however, let's recall that all we're doing is borrowing from our very good friend Mr Zero, 0, so if we add (1/2)², we also have to subtract (1/2)²

\bf \left( x^2 -x +\left[ \cfrac{1}{2} \right]^2-\left[ \cfrac{1}{2} \right]^2 \right)=6\implies \left( x^2 -x +\left[ \cfrac{1}{2} \right]^2 \right)-\left[ \cfrac{1}{2} \right]^2=6 \\\\\\ \left(x-\cfrac{1}{2} \right)^2=6+\cfrac{1}{4}\implies \left(x-\cfrac{1}{2} \right)^2=\cfrac{25}{4}\implies x-\cfrac{1}{2}=\sqrt{\cfrac{25}{4}} \\\\\\ x-\cfrac{1}{2}=\cfrac{\sqrt{25}}{\sqrt{4}}\implies x-\cfrac{1}{2}=\cfrac{5}{2}\implies x=\cfrac{5}{2}+\cfrac{1}{2}\implies x=\cfrac{6}{2}\implies \boxed{x=3}

6 0
3 years ago
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