For which equations below is x = –3 a possible solution? Check all that apply.

Shelly a 4.2 oz of Trail Mix Marsh Marshall ate 4.25 oz of Trail Mix how much more trail mix did Marshall eat

dimulka [17.4K]

With just some easy subtraction we can see that he ate 0.05 more trail mix than shelly. please mark brainliesy if correct, hope this helps!

8 0

3 years ago

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Sam drinks water at an incredible rate. She drinks 2 1/5 liters of water every 3/4 of an hour. Sam drinks water at a constant ra

lora16 [44]

Answer:

2.9 litres

or

2 14/15 litres

Step-by-step explanation:

To determine the amount of water Sam drinks in an hour, divide the total amount drunk by the number of hours

Amount drank in an hour = total litres drunk / time

$2\frac{1}{5}$ ÷ $\frac{3}{4}$

$\frac{11}{5}$ × $\frac{4}{3}$ = $\frac{44}{15}$ = 2 $\frac{14}{15}$ litres

7 0

3 years ago

An office suply store sold 410,309 pencils last year. what is the expanded form of 410,309?

lions [1.4K]

Answer:

400000 + 10000 + 300 + 9

Step-by-step explanation:

Just simply take each place value and turn the rest after to zeros, but DO NOT include the zeros that are within the number. Zeros are unnecessary.

I am joyous to assist you anytime.

8 0

3 years ago

A box with a square base and an open top is being constructed out of A cm2 of material. If the volume of the box is to be maximi

viktelen [127]

Answer:

Side length = $\sqrt{\frac{A}{3} }$ cm , Height = $\frac{1}{2} \sqrt{\frac{A}{3} }$ cm , Volume = $\frac{A\sqrt{A}}{6\sqrt{3} }$ cm³

Step-by-step explanation:

Assume

Side length of base = x

Height of box = y

total material required to construct box = A ( given in question)

So it can be written as

A = x² + 4xy

4xy = A - x²

$y = \frac{A - x^{2} }{4x}$

Volume of box = Area x height

V = x² ₓ y

V = x² ₓ ( $\frac{A - x^{2} }{4x}$ )

V = $\frac{Ax - x^{3} }{4}$

To find max volume put V' = 0

So taking derivative equation becomes

$\frac{A - 3 x^{2} }{4} = 0$

A = 3 $x^{2}$

$x^{2}$ = $\frac{A}{3}$

x = $\sqrt{\frac{A}{3\\} }$

put value of x in equation 1

y = $\frac{A - \frac{A}{3} }{4\sqrt{\frac{A}{3} } }$

y = $\frac{2 \sqrt{\frac{A}{3} } }{4 \sqrt{\frac{A}{3} } }$

y = $\frac{1}{2} \sqrt{\frac{A}{3} }$

So the volume will be

V = $x^{2}$ × y

Put values of x and y from equation 2 & 3

V = $\frac{A}{3} (\frac{1}{2} \sqrt{\frac{A}{3} } )$

V = $\frac{A\sqrt{A}}{6\sqrt{3} }$

8 0

4 years ago

A recipe requires 3/4 cup of nuts for one cake.

lana66690 [7]

10. How I would solve it, is I would convert them into decimals so it is 7.5/.75.

6 0

3 years ago

Read 2 more answers