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3 years ago

6

Chase makes 2 gallons of soup for a dinner party. He serves 10 cups of soups to his guests. How many cups of soup will he have l

eft over?

2 answers:

MArishka [77]3 years ago

5 0

Answer:

Step-by-step explanation:

1 gallon = 16 cups

2 gallons = 16 x 2 = 32

32-10=22

there will be 22 cups left over

iVinArrow [24]3 years ago

4 0

1 gallon = 16 cups

2 gallons = 16 x 2 = 32

32-10=22

there will be 22 cups left over

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3 years ago

The sum of a whole number and twice the square of the number is 10. Find the number.

sveticcg [70]

Answer:

x = $\frac{10}{1 + 2x}$

Step-by-step explanation:

Let the number be x

x + 2x² = 10

Factorize x + 2x²

x(1 + 2x) = 10

Divide both sides by (1 + 2x)

x = $\frac{10}{1 + 2x}$

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3 years ago

Find the y-intercept and x-intercept of the line.<br> -x+4y = 8

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Answer:

x-intercept(s): ( − 8 , 0 )

y-intercept(s): ( 0 , 2 )

Step-by-step explanation:

To find the x-intercept, substitute in 0 for y and solve for x . To find the y-intercept, substitute in 0 for x and solve for y.

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2 years ago

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Step-by-step explanation:

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2 years ago

Factor f(x) = 15x^3 - 15x^2 - 90x completely and determine the exact value(s) of the zero(s) and enter them as a comma separated

Illusion [34]

Answer:

$x=-2,0,3$

Step-by-step explanation:

We have been given a function $f(x)=15x^3-15x^2-90x$ . We are asked to find the zeros of our given function.

To find the zeros of our given function, we will equate our given function by 0 as shown below:

$15x^3-15x^2-90x=0$

Now, we will factor our equation. We can see that all terms of our equation a common factor that is $15x$ .

Upon factoring out $15x$ , we will get:

$15x(x^2-x-6)=0$

Now, we will split the middle term of our equation into parts, whose sum is $-1$ and whose product is $-6$ . We know such two numbers are $-3\text{ and }2$ .

$15x(x^2-3x+2x-6)=0$

$15x((x^2-3x)+(2x-6))=0$

$15x(x(x-3)+2(x-3))=0$

$15x(x-3)(x+2)=0$

Now, we will use zero product property to find the zeros of our given function.

$15x=0\text{ (or) }(x-3)=0\text{ (or) }(x+2)=0$

$15x=0\text{ (or) }x-3=0\text{ (or) }x+2=0$

$\frac{15x}{15}=\frac{0}{15}\text{ (or) }x-3=0\text{ (or) }x+2=0$

$x=0\text{ (or) }x=3\text{ (or) }x=-2$

Therefore, the zeros of our given function are $x=-2,0,3$ .

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3 years ago