All categories

2 years ago

Please help !!

Mathematics

1 answer:

Bas_tet [7]2 years ago

4 0

Answer:

C, x^3+x^2−7x−15

Step-by-step explanation:

=(x+−3)(x^2+4x+5)

=(x)(x^2)+(x)(4x)+(x)(5)+(−3)(x^2)+(−3)(4x)+(−3)(5)

=x^3+4x^2+5x−3x^2−12x−15

=x^3+x^2−7x−15

You might be interested in

Find the volume of a sphere and radius is 3. please answer in terms of pi

Ghella [55]

Answer:

12 pi

Step-by-step explanation:

volume of sphere = 4/3 pi r2

4/3 pi 3 x 3

12 pi unit square

6 0

3 years ago

Read 2 more answers

In an isosceles triangle the length of the base is 10 cm

lawyer [7]

The computation shows the radius of the circle that is inscribed in the isosceles triangle will be 3.33cm.

<h3>How to calculate the radius?</h3>

From the information given, the isosceles triangle the length of a base is 10 cm and the length of a leg is 13 cm.

Let A = area of the triangle

Let S = semi perimeter of the triangle.

The radius will be: = A/S

where,

$S = \dfrac{(a + b + c)}{2} = \dfrac{(13 + 13 + 10)}{2} = 18$

The radius will be:

$=\dfrac{(\sqrt{18} - \sqrt{13})(\sqrt{18} - \sqrt{13})(\sqrt{18} - \sqrt{10})} { 18}$

= 3.33cm

In conclusion, the radius is 3.33cm.

Learn more about triangles on:

brainly.com/question/17335144

#SPJ4

5 0

2 years ago

If f(x)=2x+sinx and the function g is the inverse of f then g'(2)=

Alexxx [7]

$\bf f(x)=y=2x+sin(x)
\\\\\\
inverse\implies x=2y+sin(y)\leftarrow f^{-1}(x)\leftarrow g(x)
\\\\\\
\textit{now, the "y" in the inverse, is really just g(x)}
\\\\\\
\textit{so, we can write it as }x=2g(x)+sin[g(x)]\\\\
-----------------------------\\\\$

$\bf \textit{let's use implicit differentiation}\\\\
1=2\cfrac{dg(x)}{dx}+cos[g(x)]\cdot \cfrac{dg(x)}{dx}\impliedby \textit{common factor}
\\\\\\
1=\cfrac{dg(x)}{dx}[2+cos[g(x)]]\implies \cfrac{1}{[2+cos[g(x)]]}=\cfrac{dg(x)}{dx}=g'(x)\\\\
-----------------------------\\\\
g'(2)=\cfrac{1}{2+cos[g(2)]}$

now, if we just knew what g(2) is, we'd be golden, however, we dunno

BUT, recall, g(x) is the inverse of f(x), meaning, all domain for f(x) is really the range of g(x) and, the range for f(x), is the domain for g(x)

for inverse expressions, the domain and range is the same as the original, just switched over

so, g(2) = some range value
that means if we use that value in f(x), f( some range value) = 2

so... in short, instead of getting the range from g(2), let's get the domain of f(x) IF the range is 2

thus 2 = 2x+sin(x)

$\bf 2=2x+sin(x)\implies 0=2x+sin(x)-2
\\\\\\
-----------------------------\\\\
g'(2)=\cfrac{1}{2+cos[g(2)]}\implies g'(2)=\cfrac{1}{2+cos[2x+sin(x)-2]}$

hmmm I was looking for some constant value... but hmm, not sure there is one, so I think that'd be it

5 0

2 years ago

What is the y intercept of the line x-2y=6

gladu [14]

Y=x-3

please can i have a brainliest

7 0

2 years ago

Read 2 more answers

Parenthesis 3 plus negative square root of 16 parenthesis 6 minus negative square root of 64

inessss [21]

Answer: 2

Explanation: got off math wy in pic

6 0

2 years ago