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3 years ago

Find the angles in this quadilateral

Mathematics

1 answer:

ch4aika [34]3 years ago

7 0

Answer:

Therefore a= 90°,b=54°, x=54°, y= 162°

Step-by-step explanation:

a=90°

a:b=5:3

5+3= 8

5/8 x A = 90

A is the sum of the angles of a and b divided in the ratio 5:3

5A/8 = 90

cross multiply

5A= 90 X8 = 720

5A=720

A= 720/5= 144°

b= 3/8 x 144 = 3x144/8 = 432/8 = 54

a= 90

b= 54

x:y is in the ratio of 1:3

the Sum of angles in a Quadrilateral is 360°

if the sum of a and b is 144°

then the reamining angles is 360-144= 216°

then x:y=1:3

1+3=4

x= 1/4 x 216= 54°

y= 3/4 x 216= 162°

Therefore a= 90°,b=54°, x=54°, y= 162°

we can see that b = x = 54°

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Using the distributive property to factorize the equation 3x^2+24x=0, you get

Natalka [10]

Answer:

Step-by-step explanation:

$3x^2+24x=0\\factor: 3x(x+8)=0\\x=0, -8$

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2 years ago

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What is the slope of a line that passes through the points (-5,-3) and (9,-6)

kipiarov [429]

To find the slope given two points, you need to subtract the two y values over the two x values
(-5,-3) (9,-6)
-5 will be x1, and -3 will be y1. 9 will be x2, and -6 will be y2. You then need to put the numbers in the correct place in the equation.
y2-y1 -6-(-3) -9
-------- = -------- = ----
x2-x1 9-(-5) 14
Now that we have the slope (-9/14) you can use one of the coordinates to find the value of b.
Let's use (9,-6).
-6=-9/14(9)+b
-6=-81/14+b
+81/14
b=-3/14
Your final equation is y=-9/15x-3/14

7 0

2 years ago

A 75-gallon tank is filled with brine (water nearly saturated with salt; used as a preservative) holding 11 pounds of salt in so

Debora [2.8K]

Let $A(t)$ = amount of salt (in pounds) in the tank at time $t$ (in minutes). Then $A(0) = 11$ .

Salt flows in at a rate

$\left(0.6\dfrac{\rm lb}{\rm gal}\right) \left(3\dfrac{\rm gal}{\rm min}\right) = \dfrac95 \dfrac{\rm lb}{\rm min}$

and flows out at a rate

$\left(\dfrac{A(t)\,\rm lb}{75\,\rm gal + \left(3\frac{\rm gal}{\rm min} - 3.25\frac{\rm gal}{\rm min}\right)t}\right) \left(3.25\dfrac{\rm gal}{\rm min}\right) = \dfrac{13A(t)}{300-t} \dfrac{\rm lb}{\rm min}$

where 4 quarts = 1 gallon so 13 quarts = 3.25 gallon.

Then the net rate of salt flow is given by the differential equation

$\dfrac{dA}{dt} = \dfrac95 - \dfrac{13A}{300-t}$

which I'll solve with the integrating factor method.

$\dfrac{dA}{dt} + \dfrac{13}{300-t} A = \dfrac95$

$-\dfrac1{(300-t)^{13}} \dfrac{dA}{dt} - \dfrac{13}{(300-t)^{14}} A = -\dfrac9{5(300-t)^{13}}$

$\dfrac d{dt} \left(-\dfrac1{(300-t)^{13}} A\right) = -\dfrac9{5(300-t)^{13}}$

Integrate both sides. By the fundamental theorem of calculus,

$\displaystyle -\dfrac1{(300-t)^{13}} A = -\dfrac1{(300-t)^{13}} A\bigg|_{t=0} - \frac95 \int_0^t \frac{du}{(300-u)^{13}}$

$\displaystyle -\dfrac1{(300-t)^{13}} A = -\dfrac{11}{300^{13}} - \frac95 \times \dfrac1{12} \left(\frac1{(300-t)^{12}} - \frac1{300^{12}}\right)$

$\displaystyle -\dfrac1{(300-t)^{13}} A = \dfrac{34}{300^{13}} - \frac3{20}\frac1{(300-t)^{12}}$

$\displaystyle A = \frac3{20} (300-t) - \dfrac{34}{300^{13}}(300-t)^{13}$

$\displaystyle A = 45 \left(1 - \frac t{300}\right) - 34 \left(1 - \frac t{300}\right)^{13}$

After 1 hour = 60 minutes, the tank will contain

$A(60) = 45 \left(1 - \dfrac {60}{300}\right) - 34 \left(1 - \dfrac {60}{300}\right)^{13} = 45\left(\dfrac45\right) - 34 \left(\dfrac45\right)^{13} \approx 34.131$

pounds of salt.

7 0

1 year ago

If it is 12:00 noon,then the sun is shining

LUCKY_DIMON [66]

Yes it is it's been that way for years

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3 years ago

What number am I?

Rudik [331]

Answer:

Step-by-step explanation:

The first digit is 9

the 2nd one from the right = 2

So far what you have is

911, 121

941,121

There seems to be a second 4 implied. What is it? I'll edit if you get it to me in the next hour.

6 0

2 years ago