Use the drop-down menus to complete each equation so the statement about its solution is true. 5−4+7x+1=

Find the product. ( 5/6 )( 5/8 )( 3/2 )

AVprozaik [17]

Answer:

The answer is C

Step-by-step explanation:

(5/6)(5/8) = 25/48

(25/48)(3/2) = 75/96

simplify 75/96 = 25/32

7 0

3 years ago

Find the range of the following piecewise function.

mash [69]

Answer:

$\huge\boxed{[-5;\ -2)\ \cup\ [3;\ 13)}$

Step-by-step explanation:

The following piecewise functions are linear functions. The graph of any of them is a line segment.

We just need to calculate the value of the function at each end specified in the brace.

$y=3x-2\ \text{if}\ -1\leq x$

Substitute x =-1 and x = 0:

$x=-1\\y=3(-1)-2=-3-2=-5\\\\x=0\\y=3(0)-2=0-2=-2$

Range of this piece is [-5; -2)

$y=2x+3\ \text{if}\ 0\leq x$

Substitute x =0and x = 5:

$x=0\\y=2(0)+3=0+3=3\\\\x=5\\y=2(5)+3=10+3=13$

Range of this piece is [3; 13)

Therefore the range of the following piecewise function is:

$[-5;\ -2)\ \cup\ [3;\ 13)$

Look at the picture.

6 0

3 years ago

A 9-pound bag of dog food costs $40, and a 20-pound bag of dog food costs $85. Which bag is the better buy?

xz_007 [3.2K]

The 20-pound bag for $85 is a better deal.

40/9= $4.44

85/20= $4.25

7 0

4 years ago

In △ABC, m∠ABC=40°, BL (L∈AC ) is the angle bisector of ∠

ira [324]

check the picture below.

M is perpendicular to AB and ∈ AB, and stemming from point L.

N is perpendicular to BC and ∈ BC, and stemming from point L as well.

BL is the bisector, that means the angle at verte B, gets cut into two equal halves.

now, we know what ∠CLN =3∠ALM, namely that ∠CLN is 3 times greater than ∠ALM.

so, once the bisector kicks in, you get two angles of 20° each, the angles at M are 90° each and the angles at N are 90° each as well, that pretty much narrows down what the missing angle is in triangles MBL and NBL, so is 70°.

now, the little sliver angles at CLN and ALM are on a 3:1 ratio, so, the flat-line of AC affords us 180°, subtract the 140°, so CLN and ALM will have to share only the remaining 40°, and they have to do so on a 3:1 ratio, that leaves us with, notice the blue angles.

now, from the triangles ALM and CLN, you can pretty much tell what the missing angle is, and those are the values for angles A and C.

Read more on Brainly.com - brainly.com/question/6781441#readmore

8 0

3 years ago

You throw a baseball into the air at an initial velocity of 30 feet per second. When the ball leaves your hand it is 5 feet abov

ValentinkaMS [17]

Assuming this little game of catch took place on planet earth, negative acceleration due to gravity is -9.8 m/s .
Converting 30ft/s to m/s, initial velocity was 30ft/s x 0.305ft/meter = 9.14m/s

Let's find how long it took for the velocity to equal zero, meaning when the ball reached it's highest point and, for a split second, stopped in mid-air before falling back down.

V(t) = Vi + a*t , where V(t) is velocity as a function of time, a is acceleration due to gravity, and t is time. Set V(t) = 0
0 = 9.14 + (-9.8)* t Add -9.8t to both sides
9.8t = 9.14 Divide both sides by 9.8
t = 0.93 seconds

Let's say your hand is the base point, or where h=zero. We want to find how high above your hand the ball went before it started coming down. Using the distance, or in this case height, formula:
h = Vi*t + (1/2)at² Plug in Vi, a, and our t value, 0.93
h= 9.14 * 0.93 + (1/2)(9.8)(0.93²)
h= 8.5 + 4.9 (0.865)
h = 8.5+ 4.27
h = 12.74 meters

The ball made it 12.74 meters above your hand. Your friends had was one foot above yours, so let's subtract .305 meters to see how far it dropped from the peak height to his hand.
12.74-.305 = 12.43 meters

Let's use the distance formula again to see how long it took to come down. Remember that this time, initial velocity is zero, since the ball starts off suspended in the air.
-12.43 = 0*t + (1/2)(-9.8)(t²) Divide both sides by -9.8/2, or -4.9
2.5374 = t²
t = 1.59

The ball took .93 seconds to go up, and 1.59 seconds to come down to your friend's glove. The total time the ball was in the air:
.93 + 1.59 = 2.52 seconds

7 0

4 years ago