PLEASE HELP ILL GIVE BRAINLIESTTTT

Answer:

$0.10

Step-by-step explanation:

Answer:

10.

Step-by-step explanation:

X is often associated with the value of "independent variable"

y=20-2(5)

y=20-10

y=10

Answer:

6x+20 cm

Step-by-step explanation:

-formula : perimeter of rectangle=2(L+W)

L=2x+10  W=x

2(2x+10+x)

=2(3x+10) *distribute 2 on 3x ,10*

=6x+20 cm

Answer:

c

Step-by-step explanation:

Keep in mind, whatever is inside of the square root, stays inside of the square root, and whatever is outside, stays outside, they don't multiply together. With parenthesis, exponents multiply, if there's no parenthesis, the exponents don't multiply. In this case, the exponents add.

Multiply each one together:

sqrt(6x^2) x sqrt(9x) = sqrt(54x^3)

sqrt(6x^2) x -(x sqrt(5x^5)) = -x sqrt(30x^7)

4 sqrt(8x^3) x sqrt(9x) = 4 sqrt(72x^4)

4 sqrt(8x^3) x -(x sqrt(5x^5)) = -4x sqrt(40x^8)

Then simplify:

sqrt(54x^2) = 3x sqrt(6x)

x sqrt(30x^7) = -x^4 sqrt(30x)

4 sqrt(72x^4) = 24x^2 sqrt(2)

-4x sqrt(40x^8) = -8x^5 sqrt(10)

Answer:the first option

Step-by-step explanation:

Answer:

25.

Step-by-step explanation:

5 ft 7 in = 5.583 ft.

We have 2 similar triangles here.

If h is the height of the tree, then:

h / 5.583 = 18/4

h = 18*5.583/4

h = 25.12 ft

The height of the tree is approximately 25.12 feet.

The ratio is defined as a relationship between two quantities, it is expressed as one divided by the other.

The ratio of the shadow lengths is equal to the ratio of the object heights.

In other words, if we let H be the height of the tree in feet and h be the height of the person in feet, we have:

H/h = 18/4

We can find h by converting the person's height to feet. Since 1 inch is equal to 1/12 foot, the person's height in feet is 5 + 7/12 = 5.58 feet.

Substituting this value into the equation above and solving for H, we get:

H = (18/4) x 5.58 feet

= 25.12 feet

Therefore, the tree is around 25.12 feet tall.

Learn more about the Ratio here:

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[tex]sin^2(\theta)+cos^2(\theta)=1\qquad \qquad sin(2\theta)=2sin(\theta)cos(\theta) \\\\[-0.35em] \rule{34em}{0.25pt}[/tex]

[tex]tan(x)+cot(x)=\cfrac{2}{sin(2x)} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{doing the left-hand-side}}{\cfrac{sin(x)}{cos(x)}+\cfrac{cos(x)}{sin(x)}\implies \cfrac{sin^2(x)+cos^2(x)}{\underset{\textit{using this LCD}}{sin(x)cos(x)}}} \implies \cfrac{1}{sin(x)cos(x)}[/tex]

now, let's recall that anything times 1 is just itself, namely 5*1 =5, 1,000,000 * 1 = 1,000,000, "meow" * 1 = "meow" and so on, so we can write anything as time 1.

let's recall something else, that same/same = 1, so

[tex]\cfrac{cheese}{cheese}\implies \cfrac{spaghetti}{spaghetti}\implies \cfrac{horse}{horse}\implies \cfrac{butter}{butter}\implies \cfrac{25^7}{25^7}=1[/tex]

therefore

[tex]\cfrac{1}{sin(x)cos(x)}\cdot \cfrac{2}{2}\implies \cfrac{2}{2sin(x)cos(x)}\implies \cfrac{2}{sin(2x)}[/tex]

Answer:

-5

Step-by-step explanation:

We use the slope formula to find the slope

m = ( y2-y1)/(x2-x1)

    = ( 7-2)/(5-6)

    = 5/-1

    = -5

Answer: 3y = 9x -12

Step-by-step explanation:

Answer:

A system of two equations can be classified as follows: If the slopes are the same but the y-intercepts are different, the system has no solution. If the slopes are different, the system has one solution. If the slopes are the same and the y-intercepts are the same, the system has infinitely many solutions.

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The best way to do this is to convert the mixed numbers into improper fractions. Improper fractions are when the numerator is greater than the denominator. An example of an improper fraction would be 5/4. Since 4/4 is one whole that means 5/4 as a mixed number would be 1 1/4. I'll use the expression 3 1/4 + 5 2/9 as an example.

3 1/4 is 3 wholes and 1/4 left over. We know that 4/4 is 1 whole so to find 3 holes we multiply the 1 by 3. 4 x 3 =12 so there are 12/4 in 3 holes. We then add the leftover 1/4 to get 13/4 as our first number.

For the second number, we have 5 2/9. This one is a bit harder. We know that 9/9 is in one whole, but how do we find how many are in 5 wholes? We can use multiplication. 1x 5=5. 5 x 9 =45. That means we have 45/9 in 5 holes. Add the leftover 2/9 to have 47/9 as our 2nd number.

Now that we have these two improper fractions, what do we do with them? The two numbers we are working with are 13/4 and 47/9. We need to find the common denominator. The common denominator is when both fractions have the same bottom number. If two fractions have different denominators we cannot add them.

To make the denominators the same, let's find a common multiple. A common multiple is a number that is a product of all chosen numbers. This is kind of confusing but another way to think of a common multiple is a number that is divisible by all chosen numbers. Take the two denominators and multiply them. 9x4=36. 36 is a common multiple of both 9 and 4 because both of them can go into 36 without being a decimal or a fraction.

To convert the fractions we multiply again. Whatever you do to one side you must always do to the other. In 13/4 we multiplied 4x9 to get 36. So now we multiply the numerator 13x9 to get 117. That means we have 117/36. We multiplied 9x4 to also get 36. So our numerator for our second number is 47x4=188 or 188/36.

Now we have 117/36 and 188/36. We add them together like normal and get 305/36. In some cases you'd need to simplify it, but for the example equation I used this is already the fraction's lowest form. I hope this helped! Comment if you need another example or have any questions! :)

Answer:

6 times or 6.4

Answer:

The answer is 20,880

Step-by-step explanation:

This won't be a very long explanation. What I did was add 18,000 plus 16% and I got 20,880.

Hope this helps!  

Have a nice day/night/afternoon.  

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[tex]|m| -54\\\\\text{When}~ m=45,\\\\|45| -54 = 45 -54 = -9[/tex]

[tex]\huge\textbf{Hey there!}\\\\\\\\\\\\\\\huge\boxed{\mathbf{|m| - 54 = }}\\\huge\boxed{\rightarrow \mathbf{|45| - 54 = }}\\\huge\boxed{\mathbf{\rightarrow 45 - 54}}\\\\\huge\text{Start at 45 and go back 54 to the left.}\\\huge\text{You should be left on \boxed{\bf -9}}\\\\\\\huge\textbf{Therefore, your answer is: \boxed{\mathsf{-9}}}\checkmark\\\\\\\\\\\\\\\huge\textbf{Good luck on your assignment \& enjoy}\\\huge\textbf{your day!}\\\\\\\\\\\\\\\frak{Amphitrite1040:)}[/tex]

There's something very off about this question.

In spherical coordinates,

x² + y² + z² = ρ²

so that

f(x, y, z) = 6 - (x² + y² + z²)

transforms to

g(ρ, θ, φ) = 6 - ρ²

When transforming to spherical coordinates, we also introduce the Jacobian determinant, so that

dV = dx dy dz = ρ² sin(φ) dρ dθ dφ

Since we integrate over a sphere with radius 4, the domain of integration is the set

E = {(ρ, θ, φ) : 0 ≤ ρ ≤ 4 and 0 ≤ θ ≤ 2π and 0 ≤ φ ≤ π}

so that the integral is

[tex]\displaystyle \int_{\phi=0}^{\phi=\pi} \int_{\theta=0}^{\theta=2\pi} \int_{\rho=0}^{\rho=4} (6 - \rho^2) \rho^2 \sin(\phi) \, d\rho \, d\theta \, d\phi[/tex]

Computing the integral is simple enough.

[tex]\displaystyle = \int_{\phi=0}^{\phi=\pi} \int_{\theta=0}^{\theta=2\pi} \int_{\rho=0}^{\rho=4} (6 \rho^2 - \rho^4) \sin(\phi) \, d\rho \, d\theta \, d\phi[/tex]

[tex]\displaystyle = 2\pi \int_{\phi=0}^{\phi=\pi} \int_{\rho=0}^{\rho=4} (6 \rho^2 - \rho^4) \sin(\phi) \, d\rho \, d\phi[/tex]

[tex]\displaystyle = 2\pi \left(\int_{\phi=0}^{\phi=\pi} \sin(\phi) \, d\phi\right) \left(\int_{\rho=0}^{\rho=4} (6 \rho^2 - \rho^4) \, d\rho\right)[/tex]

[tex]\displaystyle = 2\pi \cdot 2 \cdot \left(-\frac{384}5\right) = \boxed{-\frac{1536\pi}5}[/tex]

but the mass can't be negative...

Chances are good that this question was recycled without carefully changing all the parameters. Going through the same steps as above, the mass of a spherical body with radius R and mass density given by

[tex]\delta(x, y, z) = k - (x^2 + y^2 + z^2)[/tex]

for some positive number k is

[tex]\dfrac{4\pi r^3}{15} \left(5k - 3r^2\right)[/tex]

so in order for the mass to be positive, we must have

5k - 3r² ≥ 0   ⇒   k ≥ 3r²/5

In this case, k = 6 and r = 4, but 3•4²/5 = 9.6.

Answer:

1003

Step-by-step explanation:

The problem is a classic example of a telescoping series of products, a series in which each term is represented in a certain form such that the multiplication of most of the terms results in a massive cancelation of subsequent terms within the numerators and denominators of the series.

The simplest form of a telescoping product[tex]a_{k} \ = \ \displaystyle\frac{t_{k}}{t_{k+1}}[/tex], in which the products of n terms is

[tex]a_{1} \ \times \ a_{2} \ \times \ a_{3} \ \times \ \cdots \times \ a_{n-1} \ \times \ a_{n} \ = \ \displaystyle\frac{t_{1}}{t_{2}} \ \times \ \displaystyle\frac{t_{2}}{t_{3}} \ \times \ \displaystyle\frac{t_{3}}{t_{4}} \ \times \ \cdots \ \times \ \displaystyle\frac{t_{n-1}}{t_{n}} \ \times \ \displaystyle\frac{t_{n}}{t_{n+1}} \\ \\ \-\hspace{5.55cm} = \ \displaystyle\frac{t_{1}}{t_{n+1}}.[/tex].

In this particular case, [tex]t_{1} \ = \ 2[/tex] , [tex]t_{2} \ = \ 3[/tex], [tex]t_{3} \ = \ 4[/tex], ..... , in which each term follows a recursive formula of [tex]t_{n+1} \ = \ t_{n} \ + \ 1[/tex]. Therefore,

[tex]\displaystyle\frac{t_{2}}{t_{1}} \times \displaystyle\frac{t_{3}}{t_{2}} \times \displaystyle\frac{t_{4}}{t_{3}} \times \cdots \times \displaystyle\frac{t_{n}}{t_{n-1}} \times \displaystyle\frac{t_{n+1}}{t_{n}} \ = \ \displaystyle\frac{3}{2} \times \displaystyle\frac{4}{3} \times \displaystyle\frac{5}{4} \times \cdots \times \displaystyle\frac{2005}{2004} \times \displaystyle\frac{2006}{2005} \\ \\ \-\hspace{5.95cm} = \ \displaystyle\frac{2006}{2} \\ \\ \-\hspace{5.95cm} = 1003[/tex]

Answer:

25

Step-by-step explanation:

x+7 = 32 (similar triangles)

x = 25

Answer:

This a vector addition problem and you need to know the law of cosines and law of sines to solve this problem.

Two vector forces of 80lbs and 70 lbs act on an object at 40 degree angle between them.

The magnitude of the resultant force can be found by applying the law of cosines:

c2 = a2 + b2 - 2abcos140°

where a = 80, b = 70 and c is the resultant force vector you are asked to find

Once you find the magnitude of the resultant force, then you can find the angle it makes wrt the 70lb force using the law of sines:

sinα/a = sin140°/c

where α is the angle opposite side a, or the angle between the resultant and the 70lb vector that you are asked to find. so it would be 140lbs

Step-by-step explanation

Answer:

Answer:

the slope of the line is -2.

Step-by-step explanation:

Let X = total cost of game

          (4/5) X = $25

Multiply both sides of equation by 5/3

   (5/3)(3/5) X = (5/3) $25

                   X = $35

Answer:

solution given :

Since opposite side and side of a congruent triangle are equal.

Now

the statement which is true is:

<HDG=~ <HFE

Answer:

559

Step-by-step explanation:

Answer:

(1/2)x + 4 - 4 = 10 - 4

2(1/2)x = 6 * 2

x = 12

Answer:

Goofy unlimited lighting 12 pack of light bulbs

Step-by-step explanation:

12 = 8.40/12 = .70 cents

19 = 15.35 = .81 cents

Answer:

540 last week

52X540.=$28,080 a year

Step-by-step explanation:

Answer:

The sum is a trinomial. It has three terms of different powers in variable "x" and three non zero constants as coefficients.

Step-by-step explanation:

(5x⁴ + x³) + (2x³ + x²) = 5x⁴ + 3x³ + x²

The sum of 5x⁴ + x³ and 2x³ + x² is a trinomial expression that is  5x⁴ + 3x³ + x²

Expression in maths is defined as the collection of numbers variables and functions by using signs like addition, subtraction, multiplication, and division.

The sum is a trinomial. It has three terms of different powers in variable "x" and three non zero constants as coefficients.

We are given the expression as;

5x⁴ + x³ and 2x³ + x²

We need to find of the sum of the expression as;

(5x⁴ + x³) + (2x³ + x²)

= 5x⁴ + 3x³ + x²

The sum of 5x⁴ + x³ and 2x³ + x²  is  5x⁴ + 3x³ + x²

To know more about an expression follow;

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Step-by-step explanation:

log 10/log7=a

log10=alog7

log10=log7^a

10=7^a

log 5/log7=b

log5=blog7

log5=log7^b

5=7^b

[tex]\dfrac{\sin(x+y)}{\cos x \cos y}\\\\\\=\dfrac{\sin x \cos y + \sin y \cos x}{\cos x \cos y}\\\\\\=\dfrac{\tfrac{\sin x \cos y}{\cos x \cos y} + \tfrac{\sin y \cos x}{\cos x \cos y}}{\tfrac{\cos x \cos y}{\cos x \cos y}}\\\\\\=\dfrac{\tan x + \tan y}1\\\\\\= \tan x + \tan y[/tex]