Can someone help me Solve 6

Whats the value of x????

Serjik [45]

Answer:

X = 23

Step-by-step explanation:

oppsosite angles are equal so the brackets equal 90 degrees like the right angle opposite.

90-21 = 69

69 / 3 = 23

x = 23

8 0

3 years ago

Read 2 more answers

Explain why should record a 1 in the tens column when you regroup in an addition problem

daser333 [38]

Well you only do that when you are in the ones place and you have to regroup a 1in the tens place if you go past 19 then you are going to have to regroup a 2

8 0

4 years ago

A store buys 17 sweaters for $204 and sells them for $646. how much profit does the store make per sweater?

uranmaximum [27]

Total profit: 646 - 204 = 442

profit per shirt = 442 / 17 = 26

Answer: $26

6 0

2 years ago

A 100 gallon tank initially contains 100 gallons of sugar water at a concentration of 0.25 pounds of sugar per gallon suppose th

Vsevolod [243]

At the start, the tank contains

(0.25 lb/gal) * (100 gal) = 25 lb

of sugar. Let $S(t)$ be the amount of sugar in the tank at time $t$ . Then $S(0)=25$ .

Sugar is added to the tank at a rate of P lb/min, and removed at a rate of

$\left(1\frac{\rm gal}{\rm min}\right)\left(\dfrac{S(t)}{100}\dfrac{\rm lb}{\rm gal}\right)=\dfrac{S(t)}{100}\dfrac{\rm lb}{\rm min}$

and so the amount of sugar in the tank changes at a net rate according to the separable differential equation,

$\dfrac{\mathrm dS}{\mathrm dt}=P-\dfrac S{100}$

Separate variables, integrate, and solve for S.

$\dfrac{\mathrm dS}{P-\frac S{100}}=\mathrm dt$

$\displaystyle\int\dfrac{\mathrm dS}{P-\frac S{100}}=\int\mathrm dt$

$-100\ln\left|P-\dfrac S{100}\right|=t+C$

$\ln\left|P-\dfrac S{100}\right|=-100t-100C=C-100t$

$P-\dfrac S{100}=e^{C-100t}=e^Ce^{-100t}=Ce^{-100t}$

$\dfrac S{100}=P-Ce^{-100t}$

$S(t)=100P-100Ce^{-100t}=100P-Ce^{-100t}$

Use the initial value to solve for C :

$S(0)=25\implies 25=100P-C\implies C=100P-25$

$\implies S(t)=100P-(100P-25)e^{-100t}$

The solution is being drained at a constant rate of 1 gal/min; there will be 5 gal of solution remaining after time

$1000\,\mathrm{gal}+\left(-1\dfrac{\rm gal}{\rm min}\right)t=5\,\mathrm{gal}\implies t=995\,\mathrm{min}$

has passed. At this time, we want the tank to contain

(0.5 lb/gal) * (5 gal) = 2.5 lb

of sugar, so we pick P such that

$S(995)=100P-(100P-25)e^{-99,500}=2.5\implies\boxed{P\approx0.025}$

5 0

3 years ago

The taxi fare in a city is as follows: For the first kilometer, the

Ksju [112]

Answer:

The linear equation is y = 6 x + 9

The graph is attached down

Step-by-step explanation:

The given information:

The cost of 1st km is 9 Rs

The cost of every kilometer after the first is 6 Rs

x represents the number of km

y represents the total cost of x km

The form of the linear equation is

y = m x + b, where

m is the slope of the line (rate of change)
b is the y-intercept (initial value)

From the given information above

m = 6 Rs/km

b = 9 Rs

Substitute them in the form of the equation

y = 6 x + 9

The graph:

3 0

3 years ago