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1 year ago

WZ = 32, YZ = 6, and X is the midpoint of WY. Find WX.

Mathematics

1 answer:

GrogVix [38]1 year ago

3 0

We are given the length of two segments:

WZ = 32

YZ = 6

and we are told that x is the midpoint of the segment WY

We are asked to find the length of the segment WX

Notice that the total length of the segment WZ is 32. from the point Y to the point Z we have 6 units. therefore, between W and Y there is 32 - 6 = 26 units.

SInce X is the midpoint of the distance between W and Y, then it has to cut the segment WY (26 units long) in two equal parts, each of length 13 units (half of 26).

Therefore, WX must be of length 13 units.

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Need the solution to this: 34x + 85 = 3(13x + 7)

yanalaym [24]

Answer:

Isolate the variable by dividing each side by factors that don't contain the variable.

Exact Form:

x=64/5

Decimal Form:

x=12.8

Mixed Number Form:

x=12 4/5

6 0

3 years ago

A large tank is filled to capacity with 300 gallons of pure water. Brine containing 5 pounds of salt per gallon is pumped into t

Gwar [14]

Answer:

(a) $A = 1500(1 - e^{-0.01t})$

(b) $t = 0$

Step-by-step explanation:

Given

$V= 300gal$ --- Volume of tank

$B = 5lb/gal$ --- Brine solution

$R_1 = 3gal/min$ --- Rate in, in gallon/min

$R_2 = 6gal/min$ --- Rate out, in gallon/min

Required

Determine A(t)

First, calculate the rate in (R in) and (R out) in lb/min

$R_{in} = B * R_1$

$R_{in} = 5lb/gal * 3gal/min$

$R_{in} = 15lb/min$

R(out) is calculated by multiply the rate at which brine leaves the tank (lb/gal) * R2

So, we have:

$R_{out} = \frac{A(t)}{300} lb/gal * 3gal/min$

$R_{out} = \frac{3*A(t)}{300} lb/min$

$R_{out} = \frac{A(t)}{100} lb/min$

The change in the amount of brin e in the tank at time t is given as:

$A'(t) = R_{in} - R_{out}$

$A'(t) = 15 - \frac{A(t)}{100}$

Rewrite:

$\frac{dA}{dt} = 15 - \frac{A}{100}$

Multiply through by dt

$dA = [15 - \frac{A}{100}]* dt$

Make dt stand alone

$\frac{dA}{15 - \frac{A}{100}} = dt$

$ln|15 - \frac{A}{100}| = -0.01t + ln\ c$

Express as exponents

$15 - \frac{A}{100} = ce^{-\frac{t}{100}}$

Multiply through by 100

$1500 - A = 100ce^{-\frac{t}{100}}$

$A = 1500 - 100ce^{-\frac{t}{100}}$

$A = 1500 - 100ce^{-0.01t}$

Apply initial conditions

$A(0) = 0$

Substitute 0 for t in: $A = 1500 - 100ce^{-0.01t}$

$0 = 1500 - 100ce^{-0.01*0}$

$0 = 1500 - 100ce^{0}$

$0 = 1500 - 100c$

$100c = 1500$

$c = 15$

Substitute 15 for c in $A = 1500 - 100ce^{-0.01t}$

$A = 1500 - 100*15e^{-0.01t}$

$A = 1500 - 1500e^{-0.01t}$

Factorize:

$A = 1500(1 - e^{-0.01t})$

How long to empty the tank

Set A to 0

$1500(1 - e^{-0.01t}) = 0$

Divide through by 1500

$1 - e^{-0.01t} = 0$

Collect Like Terms

$e^{-0.01t} = 1$

Take ln of both sides

$ln\ (e^{-0.01t}) = ln\ 1$

$-0.0t = 0$

$t = 0$

8 0

3 years ago

In the problem 4 times 12 equals 48 which numbers are the factors

stich3 [128]

4 and 12 are the factors

4 0

4 years ago

Use the figure to find the measures. Apothem=

ad-work [718]

See the attached picture with letters to better understand the problem

we know that
the figure is a regular hexagon

A regular hexagon has:

Interior Angles of 120°
so
in the triangle ABC
BC=2 units
∠ABC=30°
∠BAC=60°

tan 30°=AC/BC
solve for AC
AC=BC*tan 30°-------> AC=2*(√3)/3
Apothem is equal to AC
Apothem=(2/3)√3

the answer is
(2/3)√3

3 0

3 years ago

What is the value of 3 in 305

goldfiish [28.3K]

Answer:

Hundred

Step-by-step explanation:

We are asked to find the value of 3 in 305

3- hundred

0- tens

5- units

Hence the value of 3 in 305 is hundred

7 0

3 years ago