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3 years ago

10

A special on a cruise to the Sahara Desert is advertised as being 1/2 of the regular price. If the regular price is $2150, what

is the sale price?

1 answer:

Anna35 [415]3 years ago

4 0

2150 * 1/2 = 1075

The sale price is 1075.

You might be interested in

En una clase de 25 alumnos hay 15 cajas de lápices de colores. Cada caja contiene 20 lápices. Queremos que todos los alumnos ten

seropon [69]

Answer: Each student will have 12 pencils.

Step-by-step explanation:

Given: Total students= 25

Total colored boxes = 15

Total pencils in 1 box =20

Then total pencils in all 15 boxes = 20 × 15 = 300

Now, the number of pencils each students will get = (Number of pencils in all 15 boxes) ÷ (Total students)

= 300÷ 25

= 12

Therefore, Each student will have 12 pencils.

7 0

3 years ago

NEED HELP BEING TIMED

Firlakuza [10]

Answer:

Yes.

Step-by-step explanation:

This problem has some different steps, so lets break it down:

Ed has 12 glasses of punch with 120 milliliters of liquid contained each.

Therefore, Ed has 12 * 120 milliliters of liquid:

$12 * 120 = 1440$

Now, we need to convert to liters, because that is what Ed's possible answer unit is in. Since there are 1000 milliliters in a liter, we will divide 1440 by 1000.

$1400 \: \div\: 1000 = 1.44.$

Thusly, Ed is correct and the answer is $\boxed{\text{Yes}}}$ .

Hope this helps! Have a nice day! ((:

6 0

3 years ago

Read 2 more answers

14, 16, and 20 using elimination method showing work. Thanks so much

Nady [450]

14) x=0, y=3, z=-2

Solution Set (0,3,-2)

16) x=1, y=1 and z=1

Solution set = (1,1,1)

20) x = -263/31, y=164/31 ,z=122/31

Solution set (-263/31, 164/31 ,122/31)

Step-by-step explanation:

14)

$x-y+2z=-7\\y+z=1\\x=2y+3z$

Rearranging and solving:

$x-y+2z=-7\,\,\,eq(1)\\y+z=1\,\,\,eq(2)\\x-2y-3z=0\,\,\,eq(3)$

Eliminate y:

Adding eq(1) and eq(2)

$x-y+2z=-7\,\,\,eq(1)\\ 0x+y+z=1\,\,\,eq(2)\\-------\\x+3z=-6\,\,\,eq(4)$

Multiply eq(2) with 2 and add with eq(3)

$0x+2y+2z=2\,\,\,eq(2)\\\\x-2y-3z=0\,\,\,eq(3)\\--------\\x-z=2\,\,\,eq(5)$

Eliminate x:

Subtract eq(4) and eq(5)

$x+3z=-6\,\,\,eq(4)\\x-z=2\,\,\,eq(5)\\-\,\,\,+\,\,\,\,\,\,-\\---------\\4z=-8\\z= -2$

So, value of z = -2

Now putting value of z in eq(2)

$y+z=1\\y+(-2)=1\\y-2=1\\y=1+2\\y=3$

So, value of y = 3

Now, putting value of z and y in eq(1)

$x-y+2z=-7\\x-(3)+2(-2)=-7\\x-3-4=-7\\x-7=-7\\x=-7+7\\x=0$

So, value of x = 0

So, x=0, y=3, z=-2

S.S(0,3,-2)

16)

$3x-y+z=3\\\x+y+2z=4\\x+2y+z=4$

Let:

$3x-y+z=3\,\,\,eq(1)\\x+y+2z=4\,\,\,eq(2)\\x+2y+z=4\,\,\,eq(3)$

Eliminating y:

Adding eq(1) and (2)

$3x-y+z=3\,\,\,eq(1)\\x+y+2z=4\,\,\,eq(2)\\---------\\4x+3z=7\,\,\,eq(4)$

Multiply eq(1) by 2 and add with eq(3)

$6x-2y+2z=6\,\,\,eq(1)\\x+2y+z=4\,\,\,eq(3)\\---------\\7x+3z=10\,\,\,eq(5)$

Now eliminating z in eq(4) and eq(5) to find value of x

Subtracting eq(4) and eq(5)

$4x+3z=7\,\,\,eq(4)\\7x+3z=10\,\,\,eq(5)\\-\,\,\,-\,\,\,\,\,\,\,\,\,\,-\\-----------\\-3x=-3\\x=-3/-3\\x=1$

So, value of x = 1

Putting value of x in eq(4) to find value of x:

$4x+3z=7\\4(1)+3z=7\\4+3z=7\\3z=7-4\\z=3/3\\z=1$

So, value of z = 1

Putting value of x and z in eq(2) to find value of y:

$x+y+2z=4\\1+y+2(1)=4\\1+y+2=4\\y+3=4\\y=4-3\\y=1$

So, x=1, y=1 and z=1

Solution set = (1,1,1)

20)

$x+4y-5z=-7\\3x+2y+2z=-7\\2x+y+5z=8$

Let:

$x+4y-5z=-7\,\,\,eq(1)\\3x+2y+2z=-7\,\,\,eq(2)\\2x+y+5z=8\,\,\,eq(3)$

Solving:

Eliminating z :

Adding eq(1) and eq(3)

$x+4y-5z=-7\,\,\,eq(1)\\2x+y+5z=8\,\,\,eq(3)\\---------\\3x+5y=1\,\,\,eq(4)$

Multiply eq(1) with 2 and eq(2) with 5 and add:

$2x+8y-10z=-14\,\,\,eq(1)\\15x+10y+10z=-35\,\,\,eq(2)\\----------\\17x+18y=-49\,\,\,eq(5)$

Eliminate y:

Multiply eq(4) with 18 and eq(5) with 5 and subtract:

$54x+90y=18\\85x+90y=-245\\-\,\,\,-\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+\\-------\\-31x=158\\x=-\frac{263}{31}$

So, value of x = -263/31

Putting value of x in eq(4)

$3x+5y=1\\3(-\frac{263}{31})+5y=1\\-\frac{789}{31}+5y=1 \\5y=1+\frac{789}{31}\\5y=\frac{820}{31}\\y=\frac{820}{31*5}\\y=\frac{164}{31}$

Now putting x = -263/31 and y=164/31 in eq(1) and finding z:

We get z=122/31

So, x = -263/31, y=164/31 ,z=122/31

Solution set (-263/31, 164/31 ,122/31)

Keywords: Solving system of Equations

Learn more about Solving system of Equations at:

#learnwithBrainly

4 0

3 years ago

The math team has 11 boys 10 girls and 1 coach. What is the ratio of coach to students?

Sedaia [141]

1:21 I believe, sorry if wrong

6 0

2 years ago

Read 2 more answers

Solve the system of equations for x and y.<br> y = x + 5<br> y = 2x + 2

DIA [1.3K]

Answer (x,y) (3, -2)

Explanation:
using the
substitution method
y
=
x
−
5
→
(
1
)
y
=
−
2
x
+
4
→
(
2
)
since both equations are expressed in terms of x we

can equate them
⇒
x
−
5
=
−
2
x
+
4
add 2x to both sides
2
x
+
x
−
5
=
−
2
x
+
2
x
+
4
⇒
3
x
−
5
=
4
add 5 from both sides
3
x
+
5
−
5
=
4
+
5
⇒
3
x
=
9
divide both sides by 3
3
x
3
=
9
3
⇒
x
=
3
substitute this value in
(
1
)
y
=
3
−
5
=
−
2
As a check
substitute these values into
(
2
)
right
=
−
6
+
4
=
−
2
=
left
⇒
point of intersection
=
(
3
,
−
2
)

3 0

3 years ago