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alukav5142 [94]
2 years ago
14

One half the square of a number is less than a second number. The sum of 3 and the opposite of the second number is greater than

the square of the first number. Which graph represents the solution set?​
Mathematics
2 answers:
Ivanshal [37]2 years ago
7 0

Answer: D

Step-by-step explanation:

melomori [17]2 years ago
3 0

Answer:

d

Step-by-step explanation:

You might be interested in
If x is equal to -6 which equality is true​
Papessa [141]

Answer:

-5 - 3x > 10

Step-by-step explanation:

x = -6

Plug in the x value for each inequality:

-5 - 3(-6) > 10

-5 + 18 > 10

13 > 10

True

-3 - 5(-6) < -14

-3 + 30 < -14

27 < -14

False

1 - 2(-6) > 13

1 + 12 > 13

13 > 13

False

2 - (-6) < -3

2 + 6 < -3

8 < -3

False

Only one of these is true; that is the answer.

8 0
2 years ago
Jimmy threw a baseball in the air from the roof of his house. The path followed by the baseball can be modeled by the function f
erastovalidia [21]

Answer:

Step-by-step explanation:

The first part of A is easy. Look at the quadratic function, and the constant, the very last number with no t stuck to it represents the height from which the object in question was originally launched. Our constant is 40, so the height of the roof from which the baseball was thrown is 40 feet. Part 2 of A is not quite as simple because it requires factoring using the quadratic formula.Before we do that, let's make our numbers a bit more manageable, shall we? Let's factor out a -8 to get

f(t) = -(t^2-6t-5) and a = 1, b = -6, c = -5.

Filling in the quadratic formula now looks like this:

t=\frac{6+-\sqrt{6^2-4(1)(-5)} }{2(1)} and

t=\frac{6+-\sqrt{36+40} }{2} and

t=\frac{6+-\sqrt{56} }{2} so the 2 solutions are

t=\frac{6+\sqrt{56} }{2}=6.74sec and

t=\frac{6-\sqrt{56} }{2}=-.742sec and since we know time can NEVER be negative, the time it takes for the baseball to hit the ground from a height of 40 feet is 6.74 seconds. Onto part B.

In order to determine exactly how high the baseball did go, we have to find the vertex of the function. We do this by completing the square and getting the function into vertex, or work, form. Begin by setting the quadratic equal to 0, moving over the constant, and then factoring out the leading coefficient. The rule for completing the square are kinda picky in that you have to have a 1 as the leading coefficient, and righ now ours is a -8. So following the rules I stated above:

-8(t^2-6t)=-40 Next is the take half the linear term, square it, and then add it to both sides. Our linear term is a -6. Half of -6 is -3, and -3 squared is 9, so we add 9 into the parenthesis first:

-8(t^2-6t+9)=-40+??

Because this is an equation, we can't add 9 to one side without adding the equivalent to the other side. But, we cannot forget about that -8 sitting out front there, refusing to be ignored. We didn't just add in a 9, we actually added in a -8 times 9 which is -72. That's what goes on the right side in place of the ??.

-8(t^2-6t+9)=-40-72

The reason we complete the square is found on the left side of the equals sign. We have, in the process of completing the square, formed a perfect square binomial that will serve as the h in our vertex (h, k) where h is the number of seconds it takes for the baseball to reach its max height of k, whatever k is. That's what we have to find out. Putting the left side into its simplified perfect square binomial and adding the numbers on the right gives us:

-8(t-3)^2=-112

For the last step, add over the -112 and set it back equal to f(t):

-8(t-3)^2+112=f(t) From that we determine that the vertex is (3, 112). The max height of this baseball was 112 feet...so no, it did not make it up to the height of 120 feet that Jimmy wanted for the baseball.

6 0
2 years ago
Did the scale factor of 0.75 enlarge reduce or stay the same?
elixir [45]

Answer:

\huge\boxed{\sf Reduces}

Step-by-step explanation:

Since the scale factor of 0.75 is less than 1, So the dilated image is a reduced image.

Thus the scale factor of 0.75 reduces the dilated image.

\rule[225]{225}{2}

Hope this helped!

<h3>~AnonymousHelper1807</h3>
6 0
2 years ago
Read 2 more answers
Point b has coordinates (-8,15) and lies on the circle whose equation is x^2+y^2=289. If an angles is drawn in standard position
Alenkinab [10]

Answer:

\cos \theta=-\dfrac{8}{17}

Step-by-step explanation:

Coordinates of Point b=(-8,15)

b lies on the circle whose equation is x^2+y^2=289

x^2+y^2=17^2

Comparing with the general form a circle with center at the origin: x^2+y^2=r^2

The radius of the circle =17 which is the length of the hypotenuse of the terminal ray through point b.

For an angle drawn in standard position through point b,

x=-8 which is negative

y=15 which is positive

Therefore, the angle is in Quadrant II.

\cos \theta=\dfrac{Adjacent}{Hypotenuse} \\$Adjacent=-8\\Hypotenuse=17\\\cos \theta=\dfrac{-8}{17} \\\cos \theta=-\dfrac{8}{17}

8 0
2 years ago
Freee pointsss iggg :)
OlgaM077 [116]
Yayyy thanks a lottt can i also get brainliest
8 0
3 years ago
Read 2 more answers
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