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2 years ago

Square root of 16r raised to the 6

1 answer:

7 0

If you mean: (√(16r)^6 that is the same as:

(16r)^(1/2)^6

relevant rule: (a^b)^c=a^(b*c), in this case

(16r)^(6*1/2)

(16r)^3

4096r^3

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A triangle has one side that is 5 units long and an adjacent angle

Alexxx [7]

Answer:

<u>you can get help with this link given</u>

<u>you can get help with this link givenhttps://www.sbps.net/cms/lib/NE01912194/Centricity/Domain/603/Unit%2007%20Practice%20Problems.docx</u>

Step-by-step explanation:

<h3><u>hope it's helpful to </u><u>you</u></h3>

<h3><u>Mark as brain list</u></h3>

5 0

2 years ago

Read 2 more answers

Which exponential equation is equivalent to the logarithmic equation below? <br><br> 4= ln x

ra1l [238]

The exponential function equivalent to 4=ln x will be evaluated as follows:
from
4=ln x
introducing the exponential function
e^4=e^(lnx)
but
e^(lnx)=x
thus
e^4=x
answer:
e^4=x

5 0

3 years ago

Ivan draws PQR on the coordinate plane.

Ivahew [28]

Answer:

Perimeter of PQR = 37 units (Approx.)

Step-by-step explanation:

Using graph;

Coordinate of P = (-2 , -4)

Coordinate of Q = (16 , -4)

Coordinate of R = (7 , -7)

Find:

Perimeter of PQR

Computation:

Distance between two point = √(x1 - x2)² + (y1 - y2)²

Distance between PQ = √(-2 - 16)² + (-4 - 4)²

Distance between PQ = 18 unit

Distance between QR = √(16 - 7)² + (-4 + 7)²

Distance between QR = √81 + 9

Distance between QR = 9.48 unit (Approx.)

Distance between RP = √(7 + 2)² + (-7 + 4)²

Distance between RP = √81 + 9

Distance between RP = 9.48 unit (Approx.)

Perimeter of PQR = PQ + QR + RP

Perimeter of PQR = 18 + 9.48 + 9.48

Perimeter of PQR = 36.96

Perimeter of PQR = 37 units (Approx.)

8 0

2 years ago

Barbara drives between Miami, Florida and West Palm Beach, Florida. she drives 45 miles in clear weather and then encounters a t

WINSTONCH [101]

Answer:

Barbara's speed in clear weather is $60mph$ and in the thunderstorm is $38mph$ .

Step-by-step explanation:

Let $v_1$ be the speed and $t_1$ be the time Barbara drives in clear weather, and let $v_2$ be the speed and $t_2$ be the time she drives in the thunderstorm.

Barbara drives 22 mph lower in the thunderstorm than in the clear weather; therefore,

(1). $v_2 = v_1 -22$

Also,

(2). $v_1t_1 = 45miles$

(3). $v_2 t_2 = 57$ ,

and

(4). $t_1+t_2 = 2.25hr$

From equations (2) and (3) we get:

$t_1 = \dfrac{45}{v_1}$

$t_2 =\dfrac{57}{v_2}$

putting these in equation (4) we get:

$\dfrac{45}{v_1}+\dfrac{57}{v_2}=2.25$

and substituting for $v_2$ from equation (1) we get:

$\dfrac{45}{v_1}+\dfrac{57}{v_1-22}=2.25$

This equation can be rewritten as

$2.25v_1^2-151.5v_1+990=0$

which has solutions

$v_1 = 60$

$v_1 = 7.33$

We take the first solution $v_1 =60$ because it gives a positive value for $v_2:$

$v_2 = v_1 -22$

$v_2 = 60 -22\\$

$v_2 = 38$ .

Thus, Barbara's speed in clear weather is $60mph$ and in the thunderstorm is $38mph$ .

4 0

3 years ago

PLEASE ANSWER picture shown

Alexxx [7]

$\frac{x}{2} + \frac{7}{9x}$
$= \frac{x*9x}{2*9x} + \frac{7*2}{9x*2}$
$= \frac{9x^2}{18x} + \frac{14}{18x}$
$= \frac{9x^2+14}{18x}$

Therefore the answer is B.

Hope this helps !

Photon

5 0

2 years ago