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2 years ago

What is true about the sum of the two polynomials?

Mathematics

1 answer:

const2013 [10]2 years ago

8 0

Answer:

f'(x) = -f(x) = 9xy² - 6x²y + 5x³

Step-by-step explanation:

f(x) = –9xy² + 6x²y – 5x³

additive inverse: f'(x) = -f(x) = 9xy² - 6x²y + 5x³

f(x) + f'(x) = f(x) -f(x) = (–9xy² + 6x²y – 5x³) + (9xy² - 6x²y + 5x³) = 0

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kondaur [170]

Answer:

x=3cm

Step-by-step explanation:

This is a basic proportion question.

2/18 = x/27

4 0

3 years ago

How many different pairs of faces are congruent in a rectangular prism that is not a cube

Kaylis [27]

Answer:

Step-by-step explanation:

every rectangular prism has 6 faces, 2 will be congruent together and thw other 4 will be congruent to each other

3 0

2 years ago

FIRST TO ANSWER GETS BRAINLIEST!

scZoUnD [109]

Answer:

$\huge \boxed{ \red{ \boxed{c)78 \: units ^{2} }}}$

Step-by-step explanation:

<h3>to understand this</h3><h3>you need to know about:</h3>

circle
PEMDAS

<h3>tips and formulas:</h3>

$\rm \:A_{shaded}= \dfrac{\theta}{360} \times \pi {r}^{2}$

<h3>let's solve:</h3>

$\sf sustitute \: the \: value \: of \: \theta \: and \: r : \\ A_{shaded} = \frac{140}{360} \times \pi \times {8}^{2}$
$\sf simpify \: squre : \\ A_{shaded} = \frac{140}{360} \times \pi \times 64$
$\sf simpify \: fraction: \\ A_{shaded} = \frac{7}{18} \times \pi \times 64$
$\sf reduce \: 64: \\ A_{shaded} = \frac{7}{ \cancel{ \: 18} \: ^{9} } \times \pi \times \cancel{ 64} \: ^{32} \\ A_{shaded} = \frac{7}{9} \times \pi \times 32\\ A_{shaded} = \frac{224\pi}{9}$

$\large A_{shaded}\approx 78 \: square \: units$

8 0

2 years ago

Read 2 more answers

Algebra!!!!!! Jayden Help Me!!!!!! xD

marta [7]

Under Daily flat rate, we would have 22 and 0.15x

Under Cost per mile, we would have 0.15

Under Number of Miles Driven, we would have x

4 0

2 years ago

What is the factored form of 2x2 + x-3?

Vesnalui [34]

Answer:

Step-by-step explanation:

a = 2, b = 1, c = -3

We need to factor this by finding the product of a and c, then from there find which factors of a * c will either add or subtract to give us b.

a * c = 6 and the factors of 6 and 1 and 6, 2 and 3. Well, 6 - 1 doesn't equal 1 and neither does 6 + 1. So our factors are 3 and 2. In order to combine those to get a 1 (our b), we will subtract 2 from 3 since 3 - 2 = 1. That means that 3 is positive and 2 is negative. Filling in the formula with 3 and 2 in place of 1 looks like this (always remember to put the absolute value of the largest number first):

$2x^2+3x-2x-3=0$

Group the first 2 terms together and the second 2 term together in order to factor:

$(2x^2+3x)-(2x-3)=0$ and factor out what's common in each set of parenthesis.

$x(2x+3)-1(2x+3)=0$

Notice that when we factor out a -1 from the second set of parenthesis, we can distribute it back in to get the equation we started with. We know that factoring by grouping "works" if what is inside both sets of parenthesis is exactly the same. Ours are identical: (2x + 3). That is common now, and can be factored out:

$(2x+3)(x-1)=0$

That matches your first choice

6 0

2 years ago