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3 years ago

6

There are 134 Girl Scouts going camping each cabin can hold 8 girls. How many cabins will they need for all the Girl Scouts? Sho

w all the work.

1 answer:

anyanavicka [17]3 years ago

6 0

134 divided by 8 = 16.75, round it to nearest number
they would need 17 cabins

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Amy wants to sell at most 25 cookies and at least 10 sprites

Inga [223]

Answer:

$x\leq 25\\y\geq 10\\.50x + y\geq 30$

Step-by-step explanation:

It is given that,

Amy wants to sell at most 25 cookies ,

Let number of cookies be ' $x$ ' , then $x$ should be less than or equal to 25

Which is represented by $x\leq 25$ ,

Further,

Let number of sprites be ' $y$ ', then ' $y$ ' should be greater than or equal to 10

Which is represented by $y\geq 10$

Also,

Given,

she profits $.50 on every cookie

and $1.00 on every sprite.

So total profit on selling cookies would be

= Profit on each cookie times the number of cookies

= $.50\times x$

= $.5x$

Similarly, total profit on selling sprites would be

=profit on each sprite times the number of sprites

= $1.00\times y$

= $y$

Now, combining these two should be greater than or equal to $30

representing this through an equation,

$.50x+y\geq 30$

Thus the sysytem of equations is given as:

$x\leq 25\\y\geq 10\\.50x + y\geq 30$

6 0

3 years ago

Is this right. If it is thanks

Natali [406]

Answer:

yes

Step-by-step explanation:

yes

4 0

3 years ago

Please help me ASAP .......

mariarad [96]

Answer:

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line I || line m - Transitive Property

3 0

3 years ago

A destroyer on patrol spots an aircraft at an altitude of 10,000 ft and an angle of elevation of 20⁰. How far is the aircraft fr

Reika [66]

$sin= \frac{opp}{hyp} \\\\sin (20) = \frac{10000}{hyp}\\\\hyp= \frac{10000}{sin(20)} \\\\hyp= 29238 ft$

The aircraft is 29238 ft from the ship

//

Note: this is based on my assumption that we are looking for the hypotenuse.

If we are looking for the adjacent side, then the answer would be 27475 ft.

4 0

3 years ago

1/2x+1/4y=6. 1/5x+1/3y=1

Ierofanga [76]

$\frac{1}{2}x+\frac{1}{4}y=6 \ \ \ |\times 2 \\
\frac{1}{5}x+\frac{1}{3}y=1 \ \ \ |\times (-5) \\ \\
x+\frac{1}{2}y=12 \\
\underline{-x-\frac{5}{3}y=-5} \\
\frac{1}{2}y-\frac{5}{3}y=12-5 \\
\frac{3}{6}y-\frac{10}{6}y=7 \\
-\frac{7}{6}y=7 \\
y=7 \times (-\frac{6}{7}) \\
y=-6 \\ \\
x+\frac{1}{2}y=12 \\
x+\frac{1}{2} \times (-6)=12 \\
x-3=12 \\
x=12+3 \\
x=15 \\ \\
\boxed{(x,y)=(15,-6)}$

3 0

3 years ago

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