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3 years ago

FIRST ONE TO ANSER WILL BE MARKED BRAINLIEST!!

Mathematics

2 answers:

FromTheMoon [43]3 years ago

7 0

Ok so
-10, -6
4,-6
-3,-13
-3,1
should be it

Mars2501 [29]3 years ago

3 0

Answer:

(-6,3)(4,3)(-6,-1)(4,-1)

Step-by-step explanation:

The example to the right shows a

rectangle with side lengths and units. The

coordinates of the rectangle’s vertices are (−,),

(, ), (, −), and (−, −).

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How do you graph a direct variation

Artyom0805 [142]

I confirm with the answer that k is the slope of the graph<span>. If the variables x and y vary directly when x = 3 and y = 15, then: a. Write an equation that relates x and y.</span>

3 0

3 years ago

Read 2 more answers

5 1/3 - 2 4/9= What is the answer?

Andrews [41]

Answer:

5 1/3 - 2 4/9 = 26/9

26/9 simplified equals 2 8/9

Step-by-step explanation:

I hope that my answer has helped you to understand you questions asked. If you have any further questions, please put them below.

Have a great rest of your day/night!

8 0

2 years ago

Read 2 more answers

PLEASE HELP! NO ONE HAS BEEN HELPING WITH MY QUESTIONS SO IT WOULD MEAN A LOT IF YOU DID!

kvv77 [185]

Answer:

80 m²

Step-by-step explanation:

The area (A) of a trapezoid is calculated as

A = $\frac{1}{2}$ h (a + b)

where h is the perpendicular height between bases and a, b are the bases

Here h = 8, a = 14 and b = 6, thus

A = $\frac{1}{2}$ × 8 × (14 + 6) = 4 × 20 = 80 m²

3 0

3 years ago

The area of rectangular field is 1440 sq meters and perimeter is 152 metre. Out of length or breadth which one is to be decrease

Arturiano [62]

Answer:

If Length is decreased by 10% then it would become a square.

Step-by-step explanation:

Given:

Area of Rectangular field = 1440 sq. m

Perimeter = 152 m.

We need to find which one needs to be decreased by what percentage to make it a square.

Solution:

Let length be denoted by 'l'.

Also Let width be denoted by 'w'.

Now we know that Area of rectangular field is given by Length times width.

framing in equation form we get;

$l\times w = 1440 \ \ \ \ equation\ 1$

Also we know that Perimeter of rectangular field is given by 2 times sum of Length and width.

framing in equation form we get;

$2(l+w) =152\\\\l+w =\frac{152}{2}\\\\l+w=76\\\\l=76-w \ \ \ \ equation \ 2$

Now Substituting equation 2 in equation 1 we get;

$l\times w =1440\\\\(76-w)w=1440\\\\76w-w^2=1440\\\\w^2-76w+1440 =0$

Now we will find the root by factorizing the same we get;

$w^2-36w-40w+1440=0\\\\w(w-36)-40(w-36)=0\\\\(w-36)(w-40) =0 \\$

Now solving for each to find the value of 'w' we get;

$w-36 =0\\\\w=36\\\\also \\\\w-40=0\\\\w=40$

So We can say that;

Width = 36 m and Length = 40 m

Now we need to find out of length or breadth which one is to be decreased by what percentage to make it a square.

We know that square has all sides same.

So from the data we found we can say that if length is decreased by 4 then length and breadth(width) both will be equal.

So to find percentage decreased we will divide length to be decreased by total length and then multiply by 100.

framing in equation form we get;

percentage decreased = $\frac{4}{40}\times 100 = 10\%$

Hence If Length is decreased by 10% then it would become a square.

6 0

3 years ago

Two circles, one of radius 5 inches, the other of radius 2 inches, are tangent at point P. Two bugs start crawling at the same t

lawyer [7]

Answer:

40 minutes

Step-by-step explanation:

The circumference of the larger circle is ...

C = 2πr = 2π(5 in) = 10π in

The bug navigates the circumference at 3π in/min, so will take

time = distance/speed = (10π in)/(3π in/min) = 10/3 min

to travel once around.

The circumference of the smaller circle is ...

C = 2πr = 2π(2 in) = 4π in

The bug navigates this circumference at 2.5π in/min, so will take

(4π in)/(2.5π in/min) = 8/5 min

to travel once around.

The bugs will meet at a time that is the least common multiple of these times. Both can be expressed in 15ths of a minute as ...

{50/15, 24/15}

Then the LCM of these will be ...

(1/15)LCM(50, 24) = (1/15)(50×24)/GCD(50, 24) = 1200/30 = 40

It will be 40 minutes before the bugs next meet at point P.

___

A graphing calculator can make use of the mod function to show when the bugs meet at point P (total is displacement of the two bugs from P is zero). It shows the meeting occurs after 40 minutes.

5 0

3 years ago