Need help. Find the value of x

Ok so

t=number of text messsages
p=number of picture messages

total number is 33
t+p=33

if text is 0.18 per pic and picture is 0.45 per text
total cost is 8.3
0.18t+0.45p=8.3
times 100 both sides for ease
18t+45p=830

we now have 2 equations
t+p=33
18t+45p=830

multiply first equation by -18 and add to second equation

18t+45p=830
-18t-18p=-594 +
0t+27p=236

27p=236
divide both sides by 27
p=8.740740740740744
this is impossible becasue you can't send 0.740740740740744 of a message so you did a mistype somewhere

7 0

3 years ago

The measure of central angle QRS is radians. What is the area of the shaded sector? 36 72 144 324

SpyIntel [72]

The picture in the attached figure

we know that

the angle measurement of central angle QRS is less than pi radians

area of the circle=pi*r²
for r=18 units
area of the circle=pi*18²------> 324*pi units²
area of the half of a circle=324*pi/2-----> 162*pi units²
area of a quarter of a circle=324*pi/4----> 81*pi units²

analyzing the graph we can say that the shaded area is smaller than the area of the middle of a circle but it is larger than the area of a quarter of a circle
so
the area of the shaded sector < 162*pi units²
and
the area of the shaded sector > 81*pi units²

therefore
the answer must be 144*pi units²

8 0

4 years ago

Read 2 more answers

In an attempt to escape a desert island, a castaway builds a raft and sets out to sea. The wind shifts a great deal during the d

nikdorinn [45]

Answer:

R = 7,34 km

Ф = 63,43° north of east

Step-by-step explanation:

Let´s analyze each movement and find its perpendicular components. Then, we are going to add the components along the x and y axes to get the resultant and finally to calculate its magnitude and direction.

To determine the perpendicular components, we will use right triangles trigonometrics ratios.

Please see images for each displacement in attached file.

Displacement A: 2.50 km and 45° north of west

Ax= AcosФ = 2.50cos45° = 1.77 km (neg. direction)

Ay= AsenФ = 2.50sen45° = 1.77 km (pos. direction)

Displacement B: 4.70 km and 60° south of east

Bx= BcosФ = 4.70cos60° = 2,35 km (pos. direction)

By= BsenФ = 4.70sen60° = 4,07 km (neg. direction)

Displacement C: 1.30 km and 25° south of west

Cx= CcosФ = 1.30cos25° = 1,18 km (neg. direction)

Cy= CsenФ = 1.30sen25° = 0.55 km (neg. direction)

Displacement D: 5.10 km due east (0°)

Dx= DcosФ = 5.10 km (pos. direction)

Dy= DsenФ = 0 km

Displacement E: 1.70 km and 5° east of north

Ex= EsenФ = 1.70sen5° = 0,15 km (pos. direction)

Ey= EcosФ = 1.70cos5° = 1.69 km (pos. direction)

Displacement F: 7.20 km and 55° south of west

Fx= FcosФ = 7.20cos55° = 4.13 km (neg. direction)

Fy= FsenФ = 7.20sen55° = 5.90 km (neg. direction)

Displacement G: 2.80 km and 10° north of east

Gx= GcosФ = 2.80cos10° = 2.76 km (pos. direction)

Gy= GsenФ = 2.80sen10° = 0.49 km (pos. direction)

Now, we add the components along the x- and y- axis to find the components of the resultant (R):

Rx = Ax + Bx + Cx + Dx + Ex + Fx + Gx

Ry = Ay + By + Cy + Dy + Ey + Fy + Gy

Therefore,

Rx= (-1.77)+(2.35)+(-1.18)+(5.10)+(0.15)+(-4.13)+(2.76)

Ry= (1.77)+(-4.07)+(-0.55)+(0)+(1.69)+(-5.90)+(0.49)

Note: minus (-) symbol to negative directions

Rx = 3.28 km

Ry = - 6.57 km

Let´s use the Theorem of Pythagoras to find the magnitud ot the resultant:

$R = \sqrt{Rx^{2} +Ry^{2} }$

$R =\sqrt{(3.28)^{2}+(-6.57)^{2} }$

R = 7,34 km

To find the direction of the resultant:

tanФ = $\frac{Ry}{Rx}$

tanФ = $\frac{6.57}{3.28}$

tanФ = 2.00

Ф = tan-1 (2.00)

Ф = 63,43° north of east

Download pdf

3 0

4 years ago