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3 years ago

What is the product? (-2d^2+s)(5d^2-6s)

Mathematics

1 answer:

wlad13 [49]3 years ago

5 0

The answer is −10d4+17d2s−6s2

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What is the probability of getting eiather getting blue or green spinner that is 3/10 green and 1/3 blue

inysia [295]

The probability is 50% or half or 1/2

8 0

3 years ago

When simplifying 4x^3-10x^2+6x/2x^3+x^2-3x, what are the term(s) that can be cancelled

brilliants [131]

Answer:

x can be cancelled

Step-by-step explanation:

we are given

$\frac{4x^3-10x^2+6x}{2x^3+x^2-3x}$

Firstly, we will factor numerator and denominator

and then we can factor it

$4x^3-10x^2+6x=2x(2x^2-5x+3)$

$4x^3-10x^2+6x=2x(2x+1)(x-3)$

now, we can factor denominator

$2x^3+x^2-3x=x(2x^2+x-3)$

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

now, we can replace it

$\frac{4x^3-10x^2+6x}{2x^3+x^2-3x}=\frac{2x(2x+1)(x-3)}{x(x-1)(2x+3)}$

we can see that

both terms are having x common

so, x can be cancelled

So,

x can be cancelled

3 0

4 years ago

What is the median of this data set?<br> 22, 37, 49, 15, 12

blsea [12.9K]

Answer:

the answer is 22

Step-by-step explanation:

12,15,22,37,49

since 22 is in the middle its the median (make sure you put the #'s from least to greatest)

6 0

2 years ago

Read 2 more answers

12.21 is 3.7 times the number p.

ASHA 777 [7]

P3.7 = 12.21 (p × 3.7 = 12.21)

p÷3.7 = 12.21÷3.7 (p÷3.7 cancels out p×3.7 and leaves you left with the equation p = 12.21÷3.7)

p=3.3

4 0

3 years ago

Write a equation of a hyperbola given the foci and the asymptotes

professor190 [17]

Solution:

The standard equation of a hyperbola is expressed as

$\begin{gathered} \frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1\text{ \lparen parallel to the x-axis\rparen} \\ \frac{(y-k)^2}{a^2}-\frac{(x-h)^2}{b^2}=1\text{ \lparen parallel to the y-axis\rparen} \end{gathered}$

Given that the hyperbola has its foci at (0,-15) and (0, 15), this implies that the hyperbola is parallel to the y-axis.

Thus, the equation will be expressed in the form:

$\frac{(y-k)^2}{a^2}-\frac{(x-h)^2}{b^2}=1\text{ ----equation 1}$

The asymptote of n hyperbola is expressed as

$y=\pm\frac{a}{b}(x-h)+k$

Given that the asymptotes are

$y=\frac{3}{4}x\text{ and y=-}\frac{3}{4}x$

This implies that

$a=3,\text{ and b=4}$

To evaluate the value of h and k,

$undefined$

3 0

1 year ago