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3 years ago

1) Find the minimum and maximum values for the function with the given domain interval.

Mathematics

2 answers:

Julli [10]3 years ago

8 0

Answer:

"minimum value = 0; maximum value = 8"

Step-by-step explanation:

This is the absolute value function, which returns a positive value for any numbers (positive or negative).

For example,

| -9 | = 9

| 9 | = 9

| 0 | = 0

Now, the domain is from -8 to 7 and we want to find max and min value that we can get from this function.

If we look closely, putting 7 into x won't give us max value as putting -8 would do, because:

|7| = 7

|-8| = 8

So, putting -8 would give us max value of 8 for the function.

Now, we can't get any min values that are negative, because the function doesn't return any negative values. So the lowest value would definitely be 0!

|0| = 0

and

ex: |-2| = 2 (bigger), |-5| = 5 (even bigger).

So,

Min Value = 0

Max Value = 8

strojnjashka [21]3 years ago

6 0

Answer:

minimum value = 0; maximum value = 8

Step-by-step explanation:

The function $f(x)$ is an absolute value function, which means that for negative values in it's domain it gives positive values of $f(x)$ , and therefore it's minimum value is 0.

In the given domain interval the maximum value of the function is 8 because $f(-8)=8$ .

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Two times the sum of a number and 9 is half the number

Pavel [41]

Answer:

Equation:

2(a+9) = a/2

Solve:

The umber is:

-12

Step-by-step explanation:

Equation:

2(a+9) = a/2

Solve:

2*a + 2*9 = a/2

2a + 18 = a/2

2a - a/2 = -18

4a/2 - a/2 = -18

3a/2 = -18

a = -18*2/3

a = -12

Check:

2(-12+9) = -12/2

2(-3) = -12/2 = -6

4 0

3 years ago

You are leaving your parking spot and heading home at a velocity of 35 mph for 15 min. What is your acceleration?

tigry1 [53]

Answer:

When

solving distance problems we will use the relationship rt = d or rate (speed) times

time equals distance. For example, if a person were to travel 30 mph for 4 hours.

To find the total distance we would multiply rate times time or (30)(4) = 120.

Step-by-step explanation:

7 0

4 years ago

Read 2 more answers

In the expression 7³ - 4 · 3 +8, the first operation is? A. An Exponent B. Subtraction C. Multiplication D. Addition

Vikki [24]

Answer:

An exponent

Step-by-step explanation:

look at PEM/DA/S

Parenthesis

EXPONENTS

and then you can stop the first operation is 7^3

this is an exponent

(also brainliest if this helped please!)

6 0

4 years ago

PLEASE HELP

sweet [91]

To model and solve our situation we are going to use the equation: $s= \frac{d}{t}$
where
$s$ is speed
$d$ is distance
$t$ is time

1. We know that the distance between the cities is 2400 miles, so $d=2400$ . We also know that the speed of the plane is 450 mi/h. Since we don't know the speed of the air, $S_{a}=?$ . We don't know how much the westward trip takes, so $t_{w}=?$ , and we also don't know how much the eastward trip takes, so $t_{e}=?$ .

Going westward. Here the plane is flying against the air, so we need to subtract the speed of the air from the speed of the plane:
$450-S_{a}= \frac{2400}{t_{w} }$
Going eastward. Here the plane is flying with the the air, so we need to add the speed of the air to the speed of the plane:
$450+S_{a}= \frac{2400}{t_{e} }$

2. We know for our problem that the round trip takes 11 hours; so the total time of the trip is 11, $t_{t}=11$ . Notice that we also know that the total time of the trip equals time of the tip going westward plus time of the trip going eastward, so $t_{t}=t_{w}+t_{e}$ . Since we know that the total trip takes 11 hours, we can replace that value in our total time equation and solve for $t_{w}$ :
$11=t_{w}+t_{e}$
$t_{w}=11-t_{e}$

Now we can replace $t_{w}$ in our going westward equation to model our round trip with a system of equations:
$450-S_{a}= \frac{2400}{t_{w}}$
$450-S_{a}= \frac{2400}{11-t_{e} }$ equation (1)
$450+S_{a}= \frac{2400}{t_{e}}$ equation (2)

3. To solve our system of equations, we are going to solve for $t_{e}$ in equations (1) (2):

From equation (1)
$450-S_{a}= \frac{2400}{11-t_{e} }$
$11-t_{e}= \frac{2400}{450-S_{a} }$
$-t_{e}= \frac{2400}{450-S_{a} } -11$
$t_{e}=11- \frac{2400}{450-S_{a} }$
$t_{e}= \frac{4950-11S_{a} -2400}{450-S_{a} }$
$t_{e}= \frac{2550-11S_{a} }{450-S_{a} }$ equation (3)

From equation (2):
$450+S_{a}= \frac{2400}{t_{e} }$
$t_{e}= \frac{2400}{450+S_{a} }$ equation (4)

Replacing (4) in (3)
$\frac{2400}{450+S_{a}} = \frac{2550-11S_{a}}{450-S_{a} }$
Now, we can solve for $S_{a}$ to find the speed of the wind:
$2400(450-S_{a})=(450+S_{a})(2550-11S_{a})$
$1080000-2400S_{a}=1147500-4950S_{a}+2550S_{a}-11(S_{a})^{2}$
$11(S_{a})^{2}-67500=0$
$11(S_{a})^{2}=67500$
$(S_{a})^{2}= \frac{67500}{11}$
$S_{a}=+/- \sqrt{ \frac{67500}{11} }$
Since speed cannot be negative, the solution of our equation is:
$S_{a}= \sqrt{ \frac{67500}{11} }$
$S_{a}=78.33$

We can conclude that the speed of the wind is 78 mph.

3 0

4 years ago

SOLVING EACH SYSTEM BY ELIMINATION x+3y=5 x=-6y+14 HELP

nata0808 [166]

Answer:

$\mathrm{The\:solutions\:to\:the\:system\:of\:equations\:are:}$