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3 years ago

2x(5+3x)-x find the coefficient of x

Mathematics

1 answer:

s344n2d4d5 [400]3 years ago

4 0

To start, we need to distribute

multiply 2x by <em>everything</em> in the parenthesis

2x (5 + 3x) - x

2x × 5 = 10x

2x × 3x = 6x²

now we have...

10x + 6x² - x

<em>simplify </em>

the coefficient of x is 9!

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Whats 7, −6, 4, −10, 3 least to greatest

exis [7]

Answer:

-10, -6, 3, 4, 7

Step-by-step explanation:

Hope it helps ^^

4 0

2 years ago

Simplify the logarithm log2(32)

Rom4ik [11]

let log2 32 = x

then 2^x = 32

but we know 2^5 = 32

therefore, x = 5

6 0

2 years ago

Read 2 more answers

A student claims that the rectangular point below can be represented by the

Sladkaya [172]

Since we don't have a figure we'll assume one of them is right and we're just being asked to check if they're the same number. I like writing polar coordinates with a P in front to remind me.

It's surely false if that's really a 3π/7; I'll guess that's a typo that's really 3π/4.

P(6√2, 7π/4) = ( 6√2 cos 7π/4, 6√2 sin 7π/4 )

P(-6√2, 3π/4) = ( -6√2 cos 3π/4, -6√2 sin 3π/4 )

That's true since when we add pi to an angle it negates both the sine and the cosine,

cos(7π/4) = cos(π + 3π/4) = -cos(3π/4)

sin(7π/4) = sin(π + 3π/4) = -sin(3π/4)

Answer: TRUE

3 0

3 years ago

Let ρ = x3 + xe−x for x ∈ (0, 1), compute the center of mass.

hram777 [196]

The center of mass is mathematically given as

$\bar{x}=\left(\frac{44 e-100}{25 e-40}\right)\end{aligned}$

<h3>What is the center of mass.?</h3>

Determine the center of mass in one dimension:

Represent the masses at the respective distances.

$\begin{|c|c|} Masses \ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ Located at \\\rho=x^{3}+x \cdot e^{-x} & \ \ \ \ x \in(0,1)$ \\\end$

We calculate the total mass of the system.

$\begin{aligned}m &=\int_{0}^{1} \rho \cdot d x \\& m =\int_{0}^{1}\left(x^{3}+x \cdot e^{-x}\right) \cdot d x \\&m =\left|\frac{x^{4}}{4}-(x+1) e^{-x}\right|_{0}^{1} \\&m =\left(\frac{5}{4}-\frac{2}{e}\right)\end{aligned}$

Step 03: Calculate the moment of the system.

$\begin{aligned}M &=\int_{0}^{1}(\rho \cdot x) \cdot d x \\& M=\int_{0}^{1}\left(x^{4}+x^{2} \cdot e^{-x}\right) \cdot d x \\&M =\left|\frac{x^{5}}{5}-\left(x^{2}-2 x+2\right) \cdot e^{-x}\right|_{0}^{1} \\&M=\left(\frac{11}{5}-\frac{5}{e}\right)\end{aligned}$

we calculate the center of mass.

$\begin{aligned}\bar{x} &=\left(\frac{M}{m}\right) \\& \bar{x}=\left\{\left(\frac{\left.11-\frac{5}{5}\right)}{\left(\frac{5}{4}-\frac{2}{e}\right)}\right\}\right.\\& \bar{x}=\left(\frac{11 e-25}{5 e}\right) \cdot\left(\frac{4 e}{5 e-8}\right) \\&\bar{x}=\left(\frac{44 e-100}{25 e-40}\right)\end{aligned}$

Read more about the center of mass.

brainly.com/question/27549055

#SPJ1

8 0

1 year ago

How do I find each measure? Please I need answers ASAP it's due at 1:50

Norma-Jean [14]

Answer:

Angle CAD is 44 degrees

Angle ACD is 44 degrees

Angle ACB is 136 degrees

Angle ABC is 22 degrees

Explanation:

29. Triangle ADC is an isosceles triangle because it has two equal sides.

If segments AD and DC are congruent, then segment AC is the base and the base angles of an isosceles triangle are equal.

Let x be angle CAD.

Let's go ahead x;

$\begin{gathered} 92+x+x=180\text{ (sum of angles in a triangle)} \\ 92+2x=180 \\ 2x=180-92 \\ 2x=88 \\ x=\frac{88}{2} \\ \therefore x=44^{\circ} \end{gathered}$

Therefore, measure of angle CAD is 44 degrees.

30. Measure of angle ACD is 44 degrees (Base angles of an isosceles triangle are equal)

31. Let angle ACB be y,

Let's go ahead and find measure of angle ACB;

$\begin{gathered} 44+y=180\text{ (angles on a straight line)} \\ y=180-44 \\ \therefore y=136^{\circ} \end{gathered}$

So measure of angle ACB is 136 degrees.

32. Let angle ABC be z.

Triangle ACB is also an isosceles triangle so the base angles are the same.

Let's go ahead and find z;

$\begin{gathered} 136+z+z=180_{}\text{ (sum of angles in a triangle)} \\ 138+2z=180 \\ 2z=180-136 \\ 2z=44 \\ z=\frac{44}{2} \\ \therefore z=22^{\circ} \end{gathered}$

So measure of angle ABC is 22 degrees.

3 0

8 months ago