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2 years ago

Evaluate the function g(x) = –2x2 + 3x – 5 for the input values –2, 0, and 3

Mathematics

2 answers:

Verizon [17]2 years ago

8 0

Answer: g(-2) = -15, g(0) = -9,

g(3) = -18

Step-by-step explanation:

g(-2) = -2*2 + 3(-2) - 5

g(-2) = -4 - 6 - 5

g(-2) = -15

g(0) = -4 - 0 - 5

g(0) = -9

g(3) = -4 - 3(3) - 5

g(3) = -4 - 9 - 5

g(3) = -18

Sunny_sXe [5.5K]2 years ago

4 0

Answer:

g(x)= -19

g(x)= -5

g(x)= -14

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Ben and Josh went to the roof of their 40-foot tall high school to throw their math books offthe edge.The initial velocity of Be

Taya2010 [7]

Answer

Josh's textbook reached the ground first

Josh's textbook reached the ground first by a difference of $t=0.6482$

Step-by-step explanation:

Before we proceed let us re write correctly the height equation which in correct form reads:

$h(t)=-16t^2 +v_{o}t+s$ Eqn(1).

Where:

$h(t)$ : is the height range as a function of time

$v_{o}$ : is the initial velocity

$s$ : is the initial heightin feet and is given as 40 feet, thus Eqn(1). becomes:

$h(t)=-16t^2 + v_{o}t + 40$ Eqn(2).

Now let us use the given information and set up our equations for Ben and Josh.

Ben:

We know that $v_{o}=60ft/s$

Thus Eqn. (2) becomes:

$h(t)=-16t^2+60t+40$ Eqn.(3)

Josh:

We know that $v_{o}=48ft/s$

Thus Eqn. (2) becomes:

$h(t)=-16t^2+48t+40$ Eqn. (4).

Now since we want to find whose textbook reaches the ground first and by how many seconds we need to solve each equation (i.e. Eqns. (3) and (4)) at $h(t)=0$ . Now since both are quadratic equations we will solve one showing the full method which can be repeated for the other one.

Thus we have for Ben, Eqn. (3) gives:

$h(t)=0-16t^2+60t+40=0$

Using the quadratic expression to find the roots of the quadratic we have:

$t_{1,2}=\frac{-b+/-\sqrt{b^2-4ac} }{2a} \\t_{1,2}=\frac{-60+/-\sqrt{60^2-4(-16)(40)} }{2(-16)} \\t_{1,2}=\frac{-60+/-\sqrt{6160} }{-32} \\t_{1,2}=\frac{15+/-\sqrt{385} }{8}\\\\t_{1}=4.3276 sec\\t_{2}=-0.5776 sec$

Since time can only be positive we reject the $t_{2}$ solution and we keep that Ben's book took $t=4.3276$ seconds to reach the ground.

Similarly solving for Josh we obtain

$t_{1}=3.6794sec\\t_{2}=-0.6794sec$

Thus again we reject the negative and keep the positive solution, so Josh's book took $t=3.6794$ seconds to reach the ground.

Then we can find the difference between Ben and Josh times as

$t_{Ben}-t_{Josh}= 4.3276 - 3.6794 = 0.6482$

So to answer the original question:

Whose textbook reaches the ground first and by how many seconds?

Josh's textbook reached the ground first
Josh's textbook reached the ground first by a difference of $t=0.6482$

3 0

1 year ago

∠E and ∠F are vertical angles with m∠E=5x+10 and m∠F=7x−12 . What is the value of x?

In-s [12.5K]

Vertical angles are equal...so set ur angles equal to each other and solve for x

5x + 10 = 7x - 12
12 + 10 = 7x - 5x
22 = 2x
22/2 = x
11 = x <==

6 0

2 years ago

The perimeter of the base of a regular quadrilateral prism is 60 cm, the area of one of the lateral faces is 105 cm2.

yawa3891 [41]

For a better understanding of the solution, please follow the diagram in the attached file.

A regular quadrilateral is basically a square.

So, if the base of the prism has a perimeter of 60 cm, then the length of the side of the square will be $\frac{60}{4}=15 cm$ . It is shown of the diagram.

Now, from the diagram, it is clear that the lateral face area, which is given as 105 cm^2, is the product of the side of the square, which is known, and the unknown height, let us call it h. Thus, we will get the following equation:

$15\times h=105$

$\therefore h=\frac{105}{15}=7 cm$

This is depicted on the diagram.

Now, all our required parameters are in place. Thus, let us find what has been asked.

SURFACE AREA

Surface Area (SA) will be the sum of the areas of the two bases (squares) and the areas of the four lateral faces.

Since the side of one square base is 15 cm, therefore, the area of one square base will be $15^2$ .

Likewise, the area of one lateral surface is actually the area of a rectangle with length 15 cm and height 7 cm. Thus, it's area will be given as: $15\times 7$ .

Thus, our equation will be:

$SA=2\times 15^2+4\times 15\times 7=870 cm^2$

Therefore, Surface Area=870 cm^2

VOLUME OF THE PRISM

The volume of the prism will simply be the area of the base times the height of the prism.

Thus, the volume is:

$Volume=15^2\times 7=1575 cm^3$

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2 years ago

Bob has 40 cents in his pocket. If Bob has no pennies, how many different combinations of quarters, dimes, and/or nickles does h

Anton [14]

Answer:

He could have 4 dimes, or 8 nickles, or 1 quater and 3 nickels, or 1 quater with one dime and 1 nickel, or 3 dimes and 2 nickles, or 1 quater with obe nickel and one dime.

Step-by-step explanation:

hope this helps.

3 0

2 years ago

A bar graph titled Album Type Sold per Year has year on the x-axis and number of albums sold on the y-axis. For C D s, in 2008 t

stira [4]

Answer:

True, True, False, True, False, True

Step-by-step explanation:

CDs have a higher mean than digital

Let's check. CD mean is (1000 + 800 + 800 + 600 + 400 + 200)/6 = 633.33

Digital mean: (100 + 300 + 300 + 500 + 700 + 900)/6 = 466.67

CD's mean is higher. Since you are dividing byt he same number you also could have just added the amount and found which was a larger number. But yes, this is true.

The range of digital is 800

The range is the highest number inus the lowest number. For digital the highest is 900 and lowest is 100 so the ange is 900-100 = 800. So this is true.

The median of CDs is 400.

The median is the middle value for odd numbers of values, or the average of the two middle values. 6 total values means you have to take the third and fourth and average them. in CDs the middle values are 800 and 600, the average is 700, so that is what the median is. this is false.

Both have the same interquartile range.

Interquartile range is to find the middle number or numbers like in median, then take the two parts it is split into and find the "median" of those. Then subtract the larger one fromt he smaller one.

IQR of CD the first half is 200, 400, 600 so the middle here is 400, second half has a middle number of 800 so IQR here is 800-400=400

IQR of digital is 700-300 = 400 so yes both are the same.

Both have the same median

We know the medan of CDs is 700 then findign the median of digital is (300+500)/2 = 400. So no, the medians are not the same.

Digital’s mean is around 467.

We founf the mean for digital to be 466.67 which rounds up to 467, So I would say it is true.

8 0

2 years ago

Read 2 more answers