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soldier1979 [14.2K]

3 years ago

15

You are making identical snack bags. You have 18 fruit-chew snacks and 24 granola snacks. What is the greatest amount of snack b

ags that you can make with no snacks left over?

2 answers:

Basile [38]3 years ago

6 0

Answer:

16 love

Step-by-step explanation:

Marysya12 [62]3 years ago

4 0

Answer: 18 24 Prime factorization:

6*3 12*2 24:2,2,2,3

2*3 6*2 18:3,2,3

2*3 GCF:2*3=6

Answer=6. Or subtraction: 24-18=16 but most probably GCF.

Step-by-step explanation:Find the gcf of 18 and 24

You might be interested in

What the solution 2(4+2x)≥5x+5

olga nikolaevna [1]

Answer:

x ≤ 3

Step-by-step explanation:

2(4+2x) ≥ 5x+5 (by PEDMAS, expand parentheses first)

4(2) + 2x(2) ≥ 5x + 5

8 + 4x ≥ 5x + 5 (subtract 8 from each side)

4x ≥ 5x + 5 - 8

4x ≥ 5x - 3 (subtract 5x from both sides)

4x -5x ≥ - 3

-x ≥ - 3 (multiply both sides by -1, remember to flip the inequality when multiplying both sides by a negative number)

x ≤ 3

4 0

2 years ago

Read 2 more answers

Lakelyn is training for a triathlon she is practising her swiming running and biking skills so far she can swim at a rate of 250

Nezavi [6.7K]

Answer:

The answer to the question is

It would take Lalelyn bout 3 hours to run the 18m miles, and it will take her

$\frac{x}{1.7} +\frac{y}{6} +\frac{z}{12} = t$ and x +y+z =18 where x, y, and z are the distannce covered swimming, running and bikinr respectifully

Step-by-step explanation:

To solve the question, we lisrt out the variables as follows

Swimming speed of Lakelyn = 250 yards every 5 minutes or 250×12 = 3000 yards/hour = 1.7 miles/hr

Running speed of Lakelyn = 6 miles/hour

Biking speed = 12 miles/hour

The duration it will take for Lakelyn to run 18 miles is given by

Timee taken to run 18 miles t = (Distance covered)/ (Speed of Lakelyn))

or t = D/S

Substituting the known values of such as

Running speed of Lakelyn = 6 miles/hour

From which we have the time taken to run 18 miles = 18 miles /(6 miles/hour) = 3 hours

should the 18 miles form part of the triathlon then, we have

$\frac{x}{1.7} +\frac{y}{6} +\frac{z}{12} = t$ and x +y+z =18

Where x, y and z are the distance covered Swimming, Running and Biking respectively

4 0

3 years ago

Read 2 more answers

find the number of running and passing plays using substitution method 110 plays 3 yards per run 7 yards per pass total yards 37

Zinaida [17]

Let x be the number of running plays

Y be the number of passing plays

Since the total plays is 110

X + y = 110

And total yard is 378

3x + 7y = 378

Using the first equation

X = 110 – y and substitute to the 2nd equation

3(110 – y) + 7y = 378

And solve for y

Y = 12

Substitute to eqution 1

X = 110 – 12 = 98

7 0

2 years ago

The variables x and y have a proportional relationship: when x is 5, y is 8.

Alenkasestr [34]

Answer:

1.6

Step-by-step explanation:

Since y = 8

And x = 5,

You divide both values by the X Value

8 divided by 5 = 1.6

5 divided by 5 = 1

the ratio is 1:1.6

So the constant is 1.6

6 0

2 years ago

What is the derivative of x times squaareo rot of x+ 6?

Dafna1 [17]

Hey there, hope I can help!

$\mathrm{Apply\:the\:Product\:Rule}: \left(f\cdot g\right)^'=f^'\cdot g+f\cdot g^'$
$f=x,\:g=\sqrt{x+6} \ \textgreater \ \frac{d}{dx}\left(x\right)\sqrt{x+6}+\frac{d}{dx}\left(\sqrt{x+6}\right)x \ \textgreater \ \frac{d}{dx}\left(x\right) \ \textgreater \ 1$

$\frac{d}{dx}\left(\sqrt{x+6}\right) \ \textgreater \ \mathrm{Apply\:the\:chain\:rule}: \frac{df\left(u\right)}{dx}=\frac{df}{du}\cdot \frac{du}{dx} \ \textgreater \ =\sqrt{u},\:\:u=x+6$
$\frac{d}{du}\left(\sqrt{u}\right)\frac{d}{dx}\left(x+6\right)$

$\frac{d}{du}\left(\sqrt{u}\right) \ \textgreater \ \mathrm{Apply\:radical\:rule}: \sqrt{a}=a^{\frac{1}{2}} \ \textgreater \ \frac{d}{du}\left(u^{\frac{1}{2}}\right)$
$\mathrm{Apply\:the\:Power\:Rule}: \frac{d}{dx}\left(x^a\right)=a\cdot x^{a-1} \ \textgreater \ \frac{1}{2}u^{\frac{1}{2}-1} \ \textgreater \ Simplify \ \textgreater \ \frac{1}{2\sqrt{u}}$

$\frac{d}{dx}\left(x+6\right) \ \textgreater \ \mathrm{Apply\:the\:Sum/Difference\:Rule}: \left(f\pm g\right)^'=f^'\pm g^'$
$\frac{d}{dx}\left(x\right)+\frac{d}{dx}\left(6\right)$

$\frac{d}{dx}\left(x\right) \ \textgreater \ 1$
$\frac{d}{dx}\left(6\right) \ \textgreater \ 0$

$\frac{1}{2\sqrt{u}}\cdot \:1 \ \textgreater \ \mathrm{Substitute\:back}\:u=x+6 \ \textgreater \ \frac{1}{2\sqrt{x+6}}\cdot \:1 \ \textgreater \ Simplify \ \textgreater \ \frac{1}{2\sqrt{x+6}}$

$1\cdot \sqrt{x+6}+\frac{1}{2\sqrt{x+6}}x \ \textgreater \ Simplify$

$1\cdot \sqrt{x+6} \ \textgreater \ \sqrt{x+6}$
$\frac{1}{2\sqrt{x+6}}x \ \textgreater \ \frac{x}{2\sqrt{x+6}}$
$\sqrt{x+6}+\frac{x}{2\sqrt{x+6}}$

$\mathrm{Convert\:element\:to\:fraction}: \sqrt{x+6}=\frac{\sqrt{x+6}}{1} \ \textgreater \ \frac{x}{2\sqrt{x+6}}+\frac{\sqrt{x+6}}{1}$

Find the LCD
$2\sqrt{x+6} \ \textgreater \ \mathrm{Adjust\:Fractions\:based\:on\:the\:LCD} \ \textgreater \ \frac{x}{2\sqrt{x+6}}+\frac{\sqrt{x+6}\cdot \:2\sqrt{x+6}}{2\sqrt{x+6}}$

$Since\:the\:denominators\:are\:equal,\:combine\:the\:fractions$
$\frac{a}{c}\pm \frac{b}{c}=\frac{a\pm \:b}{c} \ \textgreater \ \frac{x+2\sqrt{x+6}\sqrt{x+6}}{2\sqrt{x+6}}$

$x+2\sqrt{x+6}\sqrt{x+6} \ \textgreater \ \mathrm{Apply\:exponent\:rule}: \:a^b\cdot \:a^c=a^{b+c}$
$\sqrt{x+6}\sqrt{x+6}=\:\left(x+6\right)^{\frac{1}{2}+\frac{1}{2}}=\:\left(x+6\right)^1=\:x+6 \ \textgreater \ x+2\left(x+6\right)$
$\frac{x+2\left(x+6\right)}{2\sqrt{x+6}}$

$x+2\left(x+6\right) \ \textgreater \ 2\left(x+6\right) \ \textgreater \ 2\cdot \:x+2\cdot \:6 \ \textgreater \ 2x+12 \ \textgreater \ x+2x+12$
$3x+12$

Therefore the derivative of the given equation is
$\frac{3x+12}{2\sqrt{x+6}}$

Hope this helps!

8 0

3 years ago