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3 years ago

6

The sum of two numbers is 20 the difference between three times the first number and twice the second is 40 find the smallest nu

mber

1 answer:

Dmitriy789 [7]3 years ago

7 0

Answer:

The smallest number is 0

Step-by-step explanation:

x + y = 20

3x - 3y = 40

solving these two equations

x = 0

y = 20

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Help this is hard and I will thank you

Basile [38]

F= G(m1m2/d^2)

First divide all terms by G

G(m1m2/d^2) / G = F/G

Cancel common Factor:

m1m2 / d^2 = F/G

Multipl;y both sides by d^2:

m1m2 = F/G * d^2

Divide both sides by m2:

m1 = F /G * d^2 /m2

Simplify:

m1 = Fd^2/Gm2

7 0

4 years ago

Finding the size of angle x on a straight line. 2 of the angles are 29 degree and 82 degrees

Arte-miy333 [17]

Answer:

x = 111°

Step-by-step explanation:

We know that angles in a triangle add up to 180°

Step 1: Find angle supplementary to <em>x</em>

82 + 29 + x = 180

111 + x = 180

x = 69°

Step 2: Find angle <em>x</em>

Both the angles are supplementary, meaning they add up to 180°.

x + 69 = 180

x = 111°

4 0

4 years ago

PLEASE HELP!!!!!!<br> Counts for most of my grade

MrMuchimi

Answer:

the answer is 9/ 15

Step-by-step explanation:

Sin y = p / h

= 9/ 15 (from y reference )

cos x =b/h = 9 / 15

as from x reference b= 9

I hope this will be helpful to you

if yes give 5 star

5 0

3 years ago

Read 2 more answers

Sketch the following to help answer the question. Kite QRST has a short diagonal of QS and a long diagonal of RT. The diagonals

BigorU [14]

Kite has two pairs of adjacant sides equal. So side QR is either equal to QT or to RS. If side QR equals QT. So then QT = 17

Diagonals in kite cross each other at right angle. So in right trinagle QPT, QT is hypotenuse. Now if QT = 17, and we are given PT = 25. We cannot have hypotenuse smaller than other side. So QR cannot be equal to side QT. So QR has to be equal to other adjacent side RS. So we will have kite QRST as shown in figure.

So RS = 17. If side QR and RS are equal so diagonal RT will bisect(divide in half) diagonal QS then.

AS QS = 30 (given) so PS = PQ = $\frac{30}{2} = 15$

Now in right triangle RPS we have

RS = 17

PS = 15

We have to find other side RP.

For that we will use pythagorean formula

$c^{2} = a^{2} + b^{2}$ where c is the hypotenuse side and a and b are the other two sides of triangle

So in right triangle RPS we will have hypotenuse as RS (opposite ot right angle) so we will have

$RS^{2} = PS^{2} + RP^{2}$

Now plug values in equation. Plug 17 in RS place, 15 in PS place and RP is known so let that be x

$17^{2} = x^{2} + 15^{2}$

$289 = x^{2} + 225$

now solve for x as shown

$289 - 225 = x^{2} + 225 - 225$

$64 = x^{2}$

$\sqrt{64} = \sqrt{x^2}$

8 = x

So RP = 8

Also we are given PT = 25

So diagonal RT= RP + PT = 8 + 25 = 33m

Area of kite is given by formula

$\frac{pq}{2}$ where p and q are the diagonals in kite

Here we have digonal RT= 33m and diagonal QS = 30m

So area = $\frac{33 \times 30}{2} = \frac{990}{2} =495 m^{2}$

so final answer for area of kite is 495 $m^{2}$

4 0

3 years ago

Can someone plz help again? sorry

umka2103 [35]

Answer:

Subtraction property of equality

Step-by-step explanation:

$x+1=-2\\$

Move 1 to the $x = -2-1\\simplify\\\\x = -3$ right and change its sign

8 0

4 years ago

Read 2 more answers