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Wewaii [24]
3 years ago
5

A pitcher could hold two-twelfths of a gallon of water.If Roger filled up nine pitches,how much water would he have?

Mathematics
2 answers:
11111nata11111 [884]3 years ago
8 0
1 1/2 gallons of water
OLga [1]3 years ago
7 0
Roger would then have 1 1/2 gallons of water.
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Solve using substitution. <br> 7x-4y=-12<br> 9x-4y=-20
MrRissso [65]

Answer:y= -4

Step-by-step explanation:

1 Solve for xx in 7x-4y=-127x−4y=−12.

x=\frac{4(y-3)}{7}

x=

7

4(y−3)

​

2 Substitute x=\frac{4(y-3)}{7}x=

7

4(y−3)

​

 into 9x-4y=-209x−4y=−20.

\frac{36(y-3)}{7}-4y=-20

7

36(y−3)

​

−4y=−20

3 Solve for yy in \frac{36(y-3)}{7}-4y=-20

7

36(y−3)

​

−4y=−20.

y=-4

y=−4

4 Substitute y=-4y=−4 into x=\frac{4(y-3)}{7}x=

7

4(y−3)

​

.

x=-4

x=−4

5 Therefore,

\begin{aligned}&x=-4\\&y=-4\end{aligned}

​

 

x=−4

y=−4

​

3 0
3 years ago
The lengths of the diagonals of a rhombus are 2x and 10x. What expression gives the perimeter of the rhombus?
True [87]

Answer:

P=4x\sqrt{26}\ units

Step-by-step explanation:

we know that

The sides of a rhombus are all congruent and the diagonals are perpendicular bisectors of each other

so

Applying the Pythagorean Theorem

c^2=a^2+b^2

where

c is the length side of the rhombus      

a and b are the semi-diagonals

we have

a=2x/2=x\ units\\b=10x/2=5x\ units

substitute the values

c^2=x^2+(5x)^2

c^2=26x^2

c=x\sqrt{26}\ units

To find out the perimeter of the rhombus multiply the length side by 4

P=(4)(x\sqrt{26})

P=4x\sqrt{26}\ units

6 0
3 years ago
the mean mark got by 6 pupil's in a cat is 75.5 of them scored 83 73 71 67 and 69. what was the score of the sixth pupil?.what w
vichka [17]

Answer: score if the sixth pupil is 90

Step-by-step explanation:

Mean mark-75.5

6pupil

83+73+71+67+69+x= 75.5

363+x=75.5×6

×=453-363

×=90

8 0
3 years ago
Help Pleasee<br><br> Reflect shape A in the line x = -2
Cerrena [4.2K]

Answer: Check out the diagram below

The reflected image is shown in red.

=================================================

Explanation:

Draw a vertical line through -2 on the x axis. This is the mirror line.

Now focus on the upper right corner of the figure, which is at (-3, -1). Notice how the horizontal distance from this corner point to the mirror line is exactly 1 unit. If we move another 1 unit to the right, then we'll arrive at (-1,-1) which is where the reflected point lands or ends up.

In short, the upper right corner point (-3,-1) reflects over x = -2 to land on (-1,-1)

----------------------

As another example, the upper left corner point (-5, -1) will move exactly 4 spaces to the right to get to the mirror line. Then we move another 4 spaces to the right to get to (2,-1).

So the upper left corner (-5,-1) will ultimately move to (2,-1) after the reflection over x = -2.

Apply these steps to the other corner points and you'll end up with what is shown below.

Take note that a point like A(-5,-1) moves to A'(1,-1), and similar to the other points as well. Also, notice that when going from A to B to C, etc we are moving clockwise. We move counterclockwise when going from A' to B' to C' etc. Reflections always swap the orientation.

4 0
2 years ago
45 shirts, 6 shirts in each box. What is the fewest boxes you will need to pack all the shirts
ycow [4]
45/ 6=  7.5

you can not have half a box

so 8
4 0
3 years ago
Read 2 more answers
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