A trader bought some oranges at 5c each She finds that 6 of them are rotten. She sells the rest at 8c each and makes a profit of $4,50. How many oranges did she buy? (Hint: let the number of oranges be x. Work in cents.)

can someone please help me​

Answer: (a), (b), (c), and (d)

Step-by-step explanation:

Check the options

[tex](a)\\\Rightarrow [\sin x+\cos x]^2=\sin ^2x+\cos ^2x+2\sin x\cos x\\\Rightarrow [\sin x+\cos x]^2=1+2\sin x\cos x\\\Rightarrow \Rightarrow [\sin x+\cos x]^2=1+\sin 2x[/tex]

[tex](b)\\\Rightarrow \sin (6x)=\sin 2(3x)\\\Rightarrow \sin 2(3x)=2\sin (3x)\cos (3x)[/tex]

[tex](c)\\\Rightarrow \dfrac{\sin 3x}{\sin x\cos x}=\dfrac{3\sin x-4\sin ^3x}{\sin x\cos x}\\\\\Rightarrow 3\sec x-4\sin ^2x\sec x\\\Rightarrow 3\sec x-4[1-\cos ^2x]\sec x\\\Rightarrow 3\sec x-4\sec x+4\cos x\\\Rightarrow 4\cos x-\sec x[/tex]

[tex](d)\\\Rightarrow \dfrac{\sin 3x-\sin x}{\cos 3x+\cos x}=\dfrac{2\cos [\frac{3x+x}{2}] \sin [\frac{3x-x}{2}]}{2\cos [\frac{3x+x}{2}]\cos [\frac{3x-x}{2}]}\\\\\Rightarrow \dfrac{2\cos 2x\sin x}{2\cos 2x\cos x}=\dfrac{\sin x}{\cos x}\\\\\Rightarrow \tan x[/tex]

Thus, all the identities are correct.

A. Not an identity

B. An identity

C. Not an identity

D. An identity

To check whether each expression is an identity, we need to verify if the equation holds true for all values of the variable x. If it is true for all values of x, then it is an identity. Let's check each option:

A. [tex]\((\sin x + \cos x)^2 = 1 + \sin 2x\)[/tex]

To check if this is an identity, let's expand the left-hand side (LHS):

[tex]\((\sin x + \cos x)^2 = \sin^2 x + 2\sin x \cos x + \cos^2 x\)[/tex]

Now, we can use the trigonometric identity [tex]\(\sin^2 x + \cos^2 x = 1\)[/tex] to simplify the LHS:

[tex]\(\sin^2 x + 2\sin x \cos x + \cos^2 x = 1 + 2\sin x \cos x\)[/tex]

The simplified LHS is not equal to the right-hand side (RHS) 1 + sin 2x since it is missing the sin 2x term. Therefore, option A is not an identity.

B. [tex]\(\sin 6x = 2 \sin 3x \cos 3x\)[/tex]

To check if this is an identity, we can use the double-angle identity for sine:[tex]\(\sin 2\theta = 2\sin \theta \cos \theta\)[/tex]

Let [tex]\(2\theta = 6x\)[/tex], which means [tex]\(\theta = 3x\):[/tex]

[tex]\(\sin 6x = 2 \sin 3x \cos 3x\)[/tex]

The equation holds true with the double-angle identity, so option B is an identity.

C. [tex]\(\frac{\sin 3x}{\sin x \cos x} = 4\cos x - \sec x\)[/tex]

To check if this is an identity, we can simplify the right-hand side (RHS) using trigonometric identities.

Recall that [tex]\(\sec x = \frac{1}{\cos x}\):[/tex]

[tex]\(4\cos x - \sec x = 4\cos x - \frac{1}{\cos x} = \frac{4\cos^2 x - 1}{\cos x}\)[/tex]

Now, using the double-angle identity for sine, [tex]\(\sin 2\theta = 2\sin \theta \cos \theta\),[/tex] let [tex]\(\theta = x\):[/tex]

[tex]\(\sin 2x = 2\sin x \cos x\)[/tex]

Multiply both sides by 2: [tex]\(2\sin x \cos x = \sin 2x\)[/tex]

Now, the left-hand side (LHS) becomes:

[tex]\(\frac{\sin 3x}{\sin x \cos x} = \frac{\sin 2x}{\sin x \cos x}\)[/tex]

Using the double-angle identity for sine again, let [tex]\(2\theta = 2x\):[/tex]

[tex]\(\frac{\sin 2x}{\sin x \cos x} = \frac{2\sin x \cos x}{\sin x \cos x} = 2\)[/tex]

So, the LHS is 2, which is not equal to the RHS [tex]\(\frac{4\cos^2 x - 1}{\cos x}\)[/tex]. Therefore, option C is not an identity.

D. [tex]\(\frac{\sin 3x - \sin x}{\cos 3x + \cos x} = \tan x\)[/tex]

To check if this is an identity, we can use the sum-to-product trigonometric identities:

[tex]\(\sin A - \sin B = 2\sin \frac{A-B}{2} \cos \frac{A+B}{2}\)\(\cos A + \cos B = 2\cos \frac{A+B}{2} \cos \frac{A-B}{2}\)[/tex]

Let A = 3x and B = x:

[tex]\(\sin 3x - \sin x = 2\sin x \cos 2x\)\(\cos 3x + \cos x = 2\cos 2x \cos x\)[/tex]

Now, we can rewrite the expression:

[tex]\(\frac{\sin 3x - \sin x}{\cos 3x + \cos x} = \frac{2\sin x \cos 2x}{2\cos 2x \cos x} = \frac{\sin x}{\cos x} = \tan x\)[/tex]

The equation holds true, so option D is an identity.

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Answer:

A

Step-by-step explanation:

This is because the answer you picked means congruent to, however these two triangles are not the same size

Answer:

x = 5

Step-by-step explanation:

comment if you need explanation

Answer:

Step-by-step explanation:

It just so happens that x is the hypotenuse in the right triangle with sides 3 and 4. To find x we use Pythagorean's Theorem:

[tex]x^2=3^2+4^2[/tex] and

[tex]x^2=9+16[/tex] and

[tex]x^2=25[/tex] so

x = 5

Answer:

I think it's 155 cm

Step-by-step explanation:

=(5×5×3)+(5×2×4)

= 75+40

= 155 cm2

Answer:

D

Step-by-step explanation:

The zeros from the graph, where it crosses the x- axis are

x = - 2 and x = 3 , then the corresponding factors are

(x + 2) and (x - 3) , then

y = a(x + 2)(x - 3) ( where a is a multiplier )

To find a substitute any point on the graph into the equation

Using (0, - 6 )

- 6 = a(0 + 2)(0 - 3) = a(2)(- 3) = - 6a ( divide both sides by - 6 )

1 = a

y = (x + 2)(x - 3) ← expand using FOIL

y = x² - x - 6 → D

Answer:

x = 14

Step-by-step explanation:

Since the terns form an arithmetic progression then they have a common difference d , that is

a₂ - a₁ = a₃ - a₂

x - 8 = 20 - x ( add x to both sides )

2x - 8 = 20 ( add 8 to both sides )

2x = 28 ( divide both sides by 2 )

x = 14

Answer:

90

Step-by-step explanation:

You are looking for the highest common factor for 270 and 360.  

Factor the 2 numbers

270 = 2 * 5 * 3  * 3 * 3

360 = 2 * 2 * 2 * 3 * 3 * 5

Each of the numbers has a 5

Each of the numbers has two threes

Each of the numbers has one 2

So the answer is 2 * 3*3 * 5

Answer:

x = 9

Step-by-step explanation:

The ratios of corresponding sides are equal, that is

[tex]\frac{BC}{MN}[/tex] = [tex]\frac{AB}{LM}[/tex] , substitute values

[tex]\frac{x}{15}[/tex] = [tex]\frac{6}{10}[/tex] ( cross- multiply )

10x = 90 ( divide both sides by 10 )

x = 9

Answer:

(a)

[tex]3 \sqrt{5} [/tex]

(b)

[tex] \frac{6}{ \sqrt{5} } [/tex]

Step-by-step explanation:

A(-3,0)

B(0,6)

[tex]d = \sqrt{{( - 3 - 0)}^{2} + {(0 - 6)}^{2} } = \sqrt{9 + 36} = 3 \sqrt{5} [/tex]

[tex]d = \frac{ax0 + by0 + c}{ \sqrt{ {a}^{2} + {b}^{2} } } [/tex]

2x-y+6=0

a=2, b=-1, c=6

x0=0, y0=0

[tex]d = \frac{6}{ \sqrt{4 + 1} } = \frac{6}{ \sqrt{5} } [/tex]

[tex]\huge\textsf{Hey there!}[/tex]

[tex]\large\textsf{5g + 9 + 8g + 4}\\\\\huge\textsf{COMBINE the LIKE TERMS}\\\\\large\textsf{5g + 8g + 9 + 4}\\\\\large\textsf{5g + 8g = \bf 13g}\\\\\large\textsf{9 + 4 = \bf 13}\\\\\boxed{\huge\textsf{= 13g + 13}}\large\checkmark\\\\\\\\\\\\\boxed{\boxed{\huge\textsf{Answer: \bf 13g + 13}}}\huge\checkmark\\\\\\\\\\\\\\\\\large\textsf{Good luck on your assignment and enjoy your day!}\\\\\\\\\\\frak{Amphitrite1040:)}[/tex]

I think it's answer is 79 just guess

Answer:

1×2+1=3

3×2+1=7

7×2+1=15

15x2+1=31

31×2+1=63

40÷2+4=24

24÷2+4÷16

16÷2+4=12

12÷2+4=10

10÷2+4=9

Answer:

slope = - [tex]\frac{6}{7}[/tex]

Step-by-step explanation:

Calculate the slope m using the slope formula

m = [tex]\frac{y_{2}-y_{1} }{x_{2}-x_{1} }[/tex]

with (x₁, y₁ ) = (- 2, 5) and (x₂, y₂ ) = ([tex]\frac{3}{2}[/tex], 2)

m = [tex]\frac{2-5}{\frac{3}{2}-(-2) }[/tex]

   = [tex]\frac{-3}{\frac{3}{2}+2 }[/tex]

   = [tex]\frac{-3}{\frac{7}{2} }[/tex]

  = - 3  × [tex]\frac{2}{7}[/tex]

  = - [tex]\frac{6}{7}[/tex]

Answer:

5418.54 ft

Step-by-step explanation:

So sea level is 0, okay? So Nadia (her name is more than 3 letters and I'm lazy so from now on she'll be reffered to as "N") is at -19. 26 ft. N goes up 5,437.8 ft, so we add this value on.

-19. 26+ 5437.8= 5418.54

Now just add on the units!

Hope this helps!

Answer:

Answer:

5418.54 ft

Step-by-step explanation:

Answer:

5418.54 ft

Step-by-step explanation:

So sea level is 0, okay? So Nadia (her name is more than 3 letters and I'm lazy so from now on she'll be reffered to as "N") is at -19. 26 ft. N goes up 5,437.8 ft, so we add this value on.

-19. 26+ 5437.8= 5418.54

Given:

The figure of a triangle.

The perimeter of the triangle ABC is 31.

To find:

The value of x in the given triangle.

Solution:

Three sides of the triangle ABC are AB, BC, AC are their measures are [tex]3b-4,2b+1,b+10[/tex] respectively.

The perimeter of the triangle ABC is 31.

[tex]AB+BC+AC=31[/tex]

[tex](3b-4)+(2b+1)+(b+10)=31[/tex]

[tex]6b+7=31[/tex]

Subtract 7 from both sides.

[tex]6b=31-7[/tex]

[tex]6b=24[/tex]

[tex]b=\dfrac{24}{6}[/tex]

[tex]b=4[/tex]

Now, the measures of the sides are:

[tex]AB=3b-4[/tex]

[tex]AB=3(4)-4[/tex]

[tex]AB=12-4[/tex]

[tex]AB=8[/tex]

[tex]BC=2b+1[/tex]

[tex]BC=2(4)+1[/tex]

[tex]BC=8+1[/tex]

[tex]BC=9[/tex]

And,

[tex]AC=b+10[/tex]

[tex]AC=4+10[/tex]

[tex]AC=14[/tex]

Using the law of cosines, we get

[tex]\cos A=\dfrac{b^2+c^2-a^2}{2bc}[/tex]

[tex]\cos A=\dfrac{(AC)^2+(AB)^2-(BC)^2}{2(AC)(AB)}[/tex]

[tex]\cos A=\dfrac{(14)^2+(8)^2-(9)^2}{2(14)(8)}[/tex]

[tex]\cos A=\dfrac{179}{224}[/tex]

Using calculator, we get

[tex]\cos A=0.7991[/tex]

[tex]A=\cos ^{-1}(0.7991)[/tex]

[tex]x=36.9558^\circ[/tex]

[tex]x\approx 37.0^\circ[/tex]

Therefore, the value of x is 37.0 degrees.

Step-by-step explanation:

Step 2:

[tex] ({15x}^{2} - 9x) + (5x - 3)[/tex]

Stepc3:

[tex]3x(5x - 3)[/tex]

Answer:

(11√42)/2

Step-by-step explanation:

Thanks to the given information we can say that the smaller leg of the left triangle is

(11√7)/2

while the other leg is

11/2 √7 * √3 = 11/2 √21

the triangle on the right has two congruent legs. For the Pythagorean theorem

x^2 = 2 (11/2 √21)^2

x^2 = 2  * 121/4 * 21

x^2 = 2541 / 2

x = √11^2 * 3 * 7/√2

x = 11√21/√2

x = (11√42)/2

Answer:

24. A)14.2

25. D)5.5

Step-by-step explanation:

I hope it help for you keep safe

Answer: nose

Step-by-step explanation:

Answer:

5 ..... 144/16 = 25 sqrt(25) = 5

Step-by-step explanation:

Answer:

B

Step-by-step explanation:

Option B is correct, 4 and 3/10 hours  will Mrs. Nygaard need to work over the weekend to finish grading the projects.

A fraction represents a part of a whole.

To find the total amount of time Mrs. Nygaard has available during the week to grade projects, we need to add up the hours for each day:

1 and three-fourths + 1 and one-half + 1 and one-fifth + 2 + 1 and one-fourth = 7 and five-tenths hours

This means that Mrs. Nygaard has 7.5 hours during the week to grade projects.

If she needs 12 hours to grade all the projects, then she will need to work for an additional:

12 - 7.5 = 4.5 hours over the weekend to finish grading the projects.

Therefore, 4 and 3/10 hours  required to Mrs. Nygaard need to work over the weekend to finish grading the projects.

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hello,

" a turning point is defined as the point where a graph changes from either  increasing to decreasing, or  decreasing to increasing"

a)

[tex]y=x^3+x^2-6x\\\\y'=3x^2+2x-6=0\\x=\dfrac{-2-\sqrt{76} }{6} \approx{-1.786299647...}\\or\\x=\dfrac{-2+\sqrt{76} }{6} \approx{1.1196329...}\\[/tex]

b)

Zeros are -3,0,2.

Sol={-3,0,2}

The solution of the graph function y=x³+x²-6x are -3 , 0 and 2

The link between lines and points is described by a graph, which is a mathematical description of a network. A graph is made up of certain points and the connecting lines. It doesn't matter how long the lines are or where the points are located. A node is the name for each element in a graph.

We have the function

y=x³+x²-6x

now, equating it to 0

x³+x²-6x = 0

x² + x - 6= 0

x² - 3x + 2x -6 =0

x(x -3) + 2(x -3)

x= 3 and -2

Now, ew can see from the that the equation is touching the x-axis at three points and it will represent three zeroes of the equation.

So, the solution of the graph are -3 , 0 and 2

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Answer:

Radical (20)

Step-by-step explanation:

Radical ( (4-6)² + (2-6)²)) =radical ( 4+16) = radical (20)

Answer:

52°

Step-by-step explanation:

The angle 38° and m∠1 equal 90°, so 90-38=52.

Answer:

thats is an obtuse so if 2 is 38 then 1 should be 120 but u add those together u get 158 so it shold be 120 but if not then try looking explamles up

Step-by-step explanation:

Answer:

D

Step-by-step explanation:

[tex]\int\limits^4_1 {\frac{7^{lnx}}{x} } \, dx \\put~ln~x=y\\diff.\\\frac{1}{x} dx=dy\\when~x=1,y=ln1=0\\when x=4,y=ln~4\\\int\limits^{ln4}_0 {7^y} \, dy\\=\frac{7^y}{ln7} |0 ~\rightarrow~ ln 4\\= \frac{7^{ln4}}{7}[/tex]

Answer:

Algebra

Step-by-step explanation:

(+) it's the same thing btw6x -8 = 4x

(collect like terms) meaning the numbers with x go over the " = " sign

making it = -8 = 4x -6x

the signs change when it crosses over so it becomes that

-8 = -2x

-8 ÷ -2 = 4 (cause - ÷ by - is + )

Answer:

it’s the 3rd one:)

Step-by-step explanation:

Answer:

Substitute them into the circle equation: (x - h)² + (y - k)² = r²

(x - (-2))² + (y - (-3))² = 4²

(x + 2)² + (y + 3)² = 4²

Now expand the equation:

(x + 2)² + (y + 3)² = 4²

(x + 2)(x + 2) + (y + 3)(y + 3) = 16

(x² + 2x + 2x + 4) + (y² + 3y + 3y + 9) = 16

x² + 4x + 4 + y² + 6y + 9 = 16

x² + y² + 4x + 6y + 13 - 16 = 0

x² + y² + 4x + 6y - 3 = 0

Answer: its 20 I think

Answer:

x = 50

I hope this help the side note also help me a lot as well

Answer:

[tex]\rm\displaystyle 0,\pm\pi [/tex]

Step-by-step explanation:

please note that to find but α+β+γ in other words the sum of α,β and γ not α,β and γ individually so it's not an equation

===========================

we want to find all possible values of α+β+γ when tanα+tanβ+tanγ = tanαtanβtanγ to do so we can use algebra and trigonometric skills first

cancel tanγ from both sides which yields:

[tex] \rm\displaystyle \tan( \alpha ) + \tan( \beta ) = \tan( \alpha ) \tan( \beta ) \tan( \gamma ) - \tan( \gamma ) [/tex]

factor out tanγ:

[tex]\rm\displaystyle \tan( \alpha ) + \tan( \beta ) = \tan( \gamma ) (\tan( \alpha ) \tan( \beta ) - 1)[/tex]

divide both sides by tanαtanβ-1 and that yields:

[tex]\rm\displaystyle \tan( \gamma ) = \frac{ \tan( \alpha ) + \tan( \beta ) }{ \tan( \alpha ) \tan( \beta ) - 1}[/tex]

multiply both numerator and denominator by-1 which yields:

[tex]\rm\displaystyle \tan( \gamma ) = - \bigg(\frac{ \tan( \alpha ) + \tan( \beta ) }{ 1 - \tan( \alpha ) \tan( \beta ) } \bigg)[/tex]

recall angle sum indentity of tan:

[tex]\rm\displaystyle \tan( \gamma ) = - \tan( \alpha + \beta ) [/tex]

let α+β be t and transform:

[tex]\rm\displaystyle \tan( \gamma ) = - \tan( t) [/tex]

remember that tan(t)=tan(t±kπ) so

[tex]\rm\displaystyle \tan( \gamma ) = -\tan( \alpha +\beta\pm k\pi ) [/tex]

therefore when k is 1 we obtain:

[tex]\rm\displaystyle \tan( \gamma ) = -\tan( \alpha +\beta\pm \pi ) [/tex]

remember Opposite Angle identity of tan function i.e -tan(x)=tan(-x) thus

[tex]\rm\displaystyle \tan( \gamma ) = \tan( -\alpha -\beta\pm \pi ) [/tex]

recall that if we have common trigonometric function in both sides then the angle must equal which yields:

[tex]\rm\displaystyle \gamma = - \alpha - \beta \pm \pi [/tex]

isolate -α-β to left hand side and change its sign:

[tex]\rm\displaystyle \alpha + \beta + \gamma = \boxed{ \pm \pi }[/tex]

when is 0:

[tex]\rm\displaystyle \tan( \gamma ) = -\tan( \alpha +\beta \pm 0 ) [/tex]

likewise by Opposite Angle Identity we obtain:

[tex]\rm\displaystyle \tan( \gamma ) = \tan( -\alpha -\beta\pm 0 ) [/tex]

recall that if we have common trigonometric function in both sides then the angle must equal therefore:

[tex]\rm\displaystyle \gamma = - \alpha - \beta \pm 0 [/tex]

isolate -α-β to left hand side and change its sign:

[tex]\rm\displaystyle \alpha + \beta + \gamma = \boxed{ 0 }[/tex]

and we're done!

Answer:

-π, 0, and π

Step-by-step explanation:

You can solve for tan y :

tan y (tan a + tan B - 1) = tan a + tan y

Assuming tan a + tan B ≠ 1, we obtain

[tex]tan/y/=-\frac{tan/a/+tan/B/}{1-tan/a/tan/B/} =-tan(a+B)[/tex]

which implies that

y = -a - B + kπ

for some integer k. Thus

a + B + y = kπ

With the stated limitations, we can only have k = 0, k = 1 or k = -1. All cases are possible: we get k = 0 for a = B = y = 0; we get k = 1 when a, B, y are the angles of an acute triangle; and k = - 1 by taking the negatives of the previous cases.

It remains to analyze the case when "tan "a" tan B = 1, which is the same as saying that tan B = cot a = tan(π/2 - a), so

[tex]B=\frac{\pi }{2} - a + k\pi[/tex]

but with the given limitation we must have k = 0, because 0 < π/2 - a < π.

On the other hand we also need "tan "a" + tan B = 0, so B = - a + kπ, but again

k = 0, so we obtain

[tex]\frac{\pi }{2} - a=-a[/tex]

a contradiction.