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3 years ago

I need help with someone explaining how to find the range of this! This course didn’t explain anything!

Mathematics

1 answer:

zaharov [31]3 years ago

4 0

Answer:

$(-infinity,3)$ , D

Step-by-step explanation:

3 if x is greater than or equal to 1 is nothing. That leaves us with $h(x)=2x+1$ if x<1. If you substitute in 1 for x, you get 3, but of course that isn't possible, so the range is $(-infinity,3)$ , which is D.

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I need a step by step explanation on how you solved the equation.

Ghella [55]

Answer: x=3 and y=−2

Step-by-step explanation:

x=−2y−1;4x−4y=20

Step: Solve x=−2y−1 for x:

Step: Substitute−2y−1 for x in 4x−4y=20:

4x−4y=20

4(−2y−1)−4y=20

−12y−4=20(Simplify both sides of the equation)

−12y−4+4=20+4(Add 4 to both sides)

−12y=24

−12y

−12

(Divide both sides by -12)

y=−2

Step: Substitute−2 for y x=−2y−1:

x=−2y−1

x=(−2)(−2)−1

x=3(Simplify both sides of the equation)

7 0

3 years ago

Add the following polynomials -4x^2+3xy-y^2 with -7x^2-Xy+6y^2 and x^2+4xy-2y^2

Anestetic [448]

Answer:

-10x2 + 6xy + 3y2

Step-by-step explanation:

3 0

3 years ago

Please please help and explain!

artcher [175]

Answer:

440 Meters

Step-by-step explanation:

The question asks what his altitude was at the beginning, the beginning is the y-intercept and the y-intercept is 440 according to this graph so that's why the answer is 440.

8 0

3 years ago

How to write 9 and 15 thousandths in decimals

poizon [28]

Answer:

0.915

Step-by-step explanation:

8 0

2 years ago

Read 2 more answers

Someone help me with this

puteri [66]

Answer:

A) t = 1 second to reach the highest point

B) h = 96 feet

C) t = 3.449 seconds to hit the ground

Step-by-step explanation:

We notice that the equation that describes the position of the object in terms of the time, is a parabola (quadratic) with negative leading coefficient. therefore this is a parabola with arms pointing down, and with vertex at the top. It is at that top vertex that the maximum altitude of the object is reached.

We need therefore to find the position of the vertex (time "t" is the horizontal coordinate and height "h" is the vertical coordinate).

We know that the vertex (maximum of a parabola of this type - normally described by $y=ax^2+bx+c$ ) is given by $x_{vertex} = -\frac{b}{2*a}$ . Therefore in our case, understanding that time "t" is equivalent to the variable "x", and eight "h" is equivalent to the variable "y", that "-16" is $a$ , and "32" is $b$ , we have that the maximum occurs for:

$t_{max}=-\frac{32}{2*(-16)} = \frac{-32}{-32} = 1$

which means at one second from the launching.

We then can find the answer to part B by replacing t with "1" second in the formula for the height "h":

$h(1)=-16*(1)^2+32(1)+80=-16+32+80=96$

That is : a height of 96 feet.

To find the answer for part C: the time to hit the ground, we solve for the variable "t" in the given expression for the height, by setting the height to zero (object touching the ground) and using the quadratic formula:

$0=-16t^2+32t+80\\t=\frac{-32+/-\sqrt{32^2-4*(-16)*80} }{2*(-16)} \\ t=\frac{-32+/-\sqrt{6144} }{-32}$

which gives us two solutions: t = -1.449 seconds, and t = 3.449 seconds

Since the negative time does not have physical meaning in our case (time before the object was launched), we adopt the second answer: t = 3.449 seconds

5 0

3 years ago