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3 years ago

Harry has a cube with a side length of 7cm. What is the volume of Harry’s cube?

Mathematics

2 answers:

tresset_1 [31]3 years ago

3 0

343cm^3

as 7x7x7 = 343

BartSMP [9]3 years ago

3 0

Answer:

$\huge\boxed{343 \ cm^3}$

Step-by-step explanation:

We know that the side length of a cube is 7 cm.

The formula to find the volume of a cube is $l^3$ , where l is the length of the cube.

So we can substitute our length in (7) into the formula.

$7^3$

Now, what does it mean to cube something? To cube something is to multiply it by itself 3 times.

If we had the number n cubed, it would be

$n \cdot n \cdot n$

So we can do the same if we cube 7.

$7 \cdot 7 \cdot 7 = 49 \cdot 7 = 343$

Hope this helped!

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2 years ago

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Given:

$1. -3 x^{2}\left(4 x^{3}-7\right)\\$2. $(6 x-5)(2 x+3)$\\3. $(3 x-1)\left(x^{2}+5 x-2\right)$$

To find:

The product of the polynomials.

Solution:

1. $-3 x^{2}\left(4 x^{3}-7\right)$ $=-3 x^{2}(4 x^{3}) -3 x^{2}(-7)$

Multiply the numerical coefficient and add the powers of x.

$=-12 x^{5}+21 x^{2}$

2. $(6 x-5)(2 x+3)=6 x(2 x+3)-5(2x+3)$

Multiply each term of first polynomial with each term of 2nd polynomial.

Multiply the numerical coefficient and add the powers of x.

$=12 x^2+18x-10 x-15$

$=12 x^2+8x-15$

3. $(3 x-1)\left(x^{2}+5 x-2\right)=3 x(x^{2}+5 x-2)-1(x^{2}+5 x-2)$

Multiply each term of first polynomial with each term of 2nd polynomial.

Multiply the numerical coefficient and add the powers of x.

$=3x^{3}+15 x^2-6x-x^{2}-5 x+2$

Add or subtract like terms together.

$=3x^{3}+14 x^2-11x+2$

The answer for multiplying polynomials:

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3 years ago

A quadratic equation is shown below: 2x2 − 10x − 8 = 0 Part A: Describe the solution(s) to the equation by just determining the

dangina [55]

PART A:

The given quadratic equation is 2x²-10x-8=0

The radicand is given by b²-4ac where a, b, and c are the constants in a quadratic form ax²+bx+c

From the given equation, we have
a = 2
b = -10
c = -8

Radicand b²-4ac = (-10)² - 4(2)(-8) = 100 + 64 = 164

The radicand is >0 hence the quadratic equation has two distinct roots

PART B:

4x²-12x+5 = 0

We can use the factorization method to solve the equation
Firstly, we multiply 4 by 5 to get 20
Then we find the pair of numbers that multiply gives 20 and sum gives -12

The pair of number is -2 and -10
Rewriting the equation
4x²-2x-10x+5 = 0
2x(2x-1)-5(2x-1) = 0
(2x-1)(2x-5) = 0

2x-1 = 0 and 2x-5 = 0
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4 years ago