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3 years ago

What is the product of any integer and –1?

Mathematics

2 answers:

motikmotik3 years ago

6 0

It will be the negative of that integer. For example, 5*-1 = -5

scZoUnD [109]3 years ago

5 0

The integer's opposite form meaning if you multiply a negative integer with -1, you get the same integer in its positive form.
Ex: -2 x -1 = 2
If you multiply a positive integer with -1, you get the same integer in its positive form.
Ex: 2 x -1 = -2

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How many liters are there in a tank of volume 7.32m³?

Ivenika [448]

Answer: 3 liters in a tank of volume 7.32m^3

Step-by-step explanation:

7 0

2 years ago

HELP ASAP ~ WILL GIVE BRAINLIEST ASAP

dimaraw [331]

Answer:

Domain [-5,∞)

Range [0,∞)

Step-by-step explanation:

Part 1) Find the domain

we have

$f(x)=\frac{1}{2}\sqrt{x+5}$

we know that

The radicand must be greater than or equal to zero

$x+5\geq 0$

solve for x

subtract 5 both sides

$x\geq -5$

The solution for x is the interval [-5,∞)

All real numbers greater than or equal to -5

Remember that

The domain of a function is the set of all possible values of x

therefore

The domain of the function f(x) is the interval [-5,∞)

Part 2) Find the range

we have

$f(x)=\frac{1}{2}\sqrt{x+5}$

Find the value of f(x) for the minimum value of x

For x=-5

$f(x)=\frac{1}{2}\sqrt{-5+5}$

$f(x)=0$

The minimum value of f(x) is equal to zero

The solution for f(x) is the interval [0,∞)

All real numbers greater than or equal to 0

Remember that

The range of a function is the complete set of all possible resulting values of y, after we have substituted the domain.

therefore

The range of the function is the interval [0,∞)

8 0

3 years ago

<img src="https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7By%7D%20%287%5Csqrt%5B3%5D%7B8y%5E2%7D-%5Csqrt%5B3%5D%7By%5E5%7D%20-4y%5Csqrt%

noname [10]

Answer:

$\huge\boxed{-y^2+2y}$

Step-by-step explanation:

$\sqrt[3]y\cdot\left(7\sqrt[3]{8y^2}-\sqrt[3]{y^5}-4y\sqrt[3]{27y^2}\right)\\\\=(\sqrt[3]y)(7\sqrt[3]{8y^2})-(\sqrt[3]y)(\sqrt[3]{y^5})-(\sqrt[3]y)(4y\sqrt[3]{27y^2})\\\\=7\sqrt[3]{(y)(8y^2)}}-\sqrt[3]{(y)(y^5)}-4y\sqrt[3]{(y)(27y^2)}\\\\=7\sqrt{8y^3}-\sqrt{y^6}-4\sqrt{27y^3}\\\\=7\sqrt[3]{2^3y^3}-\sqrt{y^{2\cdot3}}-4\sqrt{3^3y^3}\\\\=7\sqrt[3]{(2y)^3}-\sqrt{(y^2)^3}-4\sqrt{(3y)^3}\\\\=7\cdot2y-y^2-4\cdot3y\\\\=14y-y^2-12y\\\\=-y^2+2y$