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3 years ago

#1 Find the slope given the points (3,-20) and (5,8) *

Mathematics

1 answer:

Artyom0805 [142]3 years ago

6 0

Hey there :)

To find the slope with two points we will need to use the equation $\frac{y^2-y^1}{x^2-x^1}$

$x^1$ $x^2$ $y^1$ $y^2$

( 3 , -20 ) ( 5 , 8 )

Here is what the equation should look like when replaced with the numbers (it will be a fraction) :

$\frac{8 - (-20)}{5 - 3}$

We know that two negatives cancel out and make a positive, so the numerator would actually be 8 + 20.

Solve the fraction and we get $\frac{28}{2}$

28 divided by 2 is 14. The slope is 14.

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each side of triangle xyz has length 9 .Find the area of the region inside the circumcircle of the triangle but outside the tria

mote1985 [20]

Answer:

The area of the region inside the circumcircle of the triangle but outside the triangle is

$A=\frac{27}{4}[\pi-3\sqrt{3}]\ units^2$

Step-by-step explanation:

see the attached figure to better understand the problem

step 1

Find the area of triangle

we have an equilateral triangle

Applying the law of sines

$A_t=\frac{1}{2}(b^2)sin(60^o)$

where b is the length side of the equilateral triangle

we have

$b=9\ units$

$A_t=\frac{1}{2}(81)sin(60^o)$

$A_t=\frac{1}{2}(81)\frac{\sqrt{3}}{2}$

$A_t=81\frac{\sqrt{3}}{4}\ units^2$

step 2

Find the area of circle

The area of the circle is equal to

$A_c=\pi r^{2}$

The formula to calculate the radius of the circumcircle of the triangle equilateral is equal to

$r=b\frac{\sqrt{3}}{6}$

where b is the length side of the equilateral triangle

we have

$b=9\ units$

substitute

$r=(9)\frac{\sqrt{3}}{6}$

$r=3\frac{\sqrt{3}}{2}\ units$

Find the area

$A_c=\pi (3\frac{\sqrt{3}}{2})^{2}$

$A_c=\frac{27}{4} \pi\ units^2$

step 3

Find the area of the shaded region

we know that

The area of the region inside the circumcircle of the triangle but outside the triangle is equal to the area pf the circle minus the area of triangle

$A=(\frac{27}{4} \pi-81\frac{\sqrt{3}}{4})\ units^2$

Simplify

$A=\frac{27}{4}[\pi-3\sqrt{3}]\ units^2$

6 0

3 years ago

Pls help me for this

Firlakuza [10]

The answer for 50 is (61-7 divides by 7)times 1/2+17

3 0

3 years ago

Please help solve algebra question

Ksenya-84 [330]

Let be:Speed of the wind: WSpeed of the plane in still air: P
Against the wind the plane flew:Distance: d=175 milesTime: ta=1 hour 10 minutesta=1 hour (10 minutes)*(1 hour/60 minutes)ta=1 hour + 1/6 hourta=(6+1)/6 hourta=7/6 hourSpeed against the wind: Sa=d/taSa=(175 miles) / (7/6 hour)Sa=175*(6/7) miles/hourSa=1,050/7 miles per hourSa=150 mph
(1) P-W=Sa(1) P-W=150
The return trip only took 50 minutesDistance: d=175 milesTime: tr=50 minutestr=(50 minutes)*(1 hour/60 minutes)tr=5/6 hour
Speed retur trip: Sr=d/trSr=(175 miles) / (5/6 hour)Sr=175*(6/5) miles/hourSr=1,050/5 miles per hourSr=210 mph
(2) P+W=Sr(2) P+W=210
We have a system of 2 equations and 2 unknows:(1) P-W=150(2) P+W=210
Adding the equations:P-W+P+W=150+2102P=360Solving for P:2P/2=360/2P=180
Replacing P by 180 in equation (2):(2) P+W=210180+W=210
Solving for W:180+W-180=210-180W=30
Answers:The speed of the plane in still air was 180 mphThe speed of the wind was 30 mph

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3 years ago

-4x4+1.5-2.5-2x4-6.5

nekit [7.7K]

Answer:

0.5

please mark me as Branliest

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3 years ago

Read 2 more answers

The length of chord AB in D is 9mm. If the measure of angle AB is 32 then find the length of AB

Taya2010 [7]

The length of arc AB is 9.12 mm:

We first calculate for the radius r of the circle using the equation
r = c/(2 sin[theta/2])

where c is the length of chord AB that is given as 9 millimeters
angle given is 32 degrees

To convert theta 32 degrees into radians:
32 degrees * (pi/180) = 32 degrees * (3.14/180) = 0.5583 radians

We now substitute the values into the equation to find the radius r:
r = 9/(2 sin[0.5583/2])
r = 16.33 mm
.
We can finally solve for the length s of arc:
s = r theta = 16.33 * 0.5583 = 9.12 mm

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4 years ago