Solve the system of equations.
8x + 5y = 24
y = -4x
x =
y =

Answer:

See explanation

Step-by-step explanation:

Function A is not clear; I will use the following in place of function A

Function A:

[tex]x \to\ 1 |\ 3 |\ 4 |\ 6[/tex]

[tex]y \to -1|\ 3|\ 5|\ 9[/tex]

Function B:

[tex]y = 2x + 4[/tex]

Required

Compare both functions

For linear functions, we often compare the slope and the y intercepts only.

Calculating the slope of function A, we have:

[tex]m = \frac{y_2 - y_1}{x_2 - x_1}[/tex]

Where:

[tex](x_1,y_1) = (1,-1)[/tex]

[tex](x_2,y_2) = (3,3)[/tex]

So, we have:

[tex]m = \frac{3 - -1}{3 - 1}[/tex]

[tex]m = \frac{4}{2}[/tex]

[tex]m = 2[/tex]

To calculate the y intercept, we set [tex]x = 0[/tex], then solve for y

i.e.[tex](x,y) = (0,y)[/tex]

Using the slope formula, we have:

[tex]m = \frac{y_2 - y_1}{x_2 - x_1}[/tex]

Where:

[tex]m = 2[/tex]

[tex](x_1,y_1) = (0,y)[/tex]

[tex](x_2,y_2) = (3,3)[/tex]

So, we have:

[tex]2 = \frac{3 - y}{3 - 0}[/tex]

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

Multiply by 3

[tex]6 = 3 - y[/tex]

Collect like terms

[tex]y = 3 - 6[/tex]

[tex]y = -3[/tex]

So, for function A:

[tex]m = 2[/tex] -- slope

[tex]y = -3[/tex] --- y intercept

For function B

[tex]y = 2x + 4[/tex]

A linear function is represented as:

[tex]y = mx + b[/tex]

By comparison

[tex]m = 2[/tex] --- slope

[tex]b = 4[/tex] --- y intercept

By comparing the results of both functions, we have the following conclusion:

Functions A and B have the same slope (i.e. 2)

Function B has a greater y intercept (i.e. 4)

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]

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:

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:

24. A)14.2

25. D)5.5

Step-by-step explanation:

I hope it help for you keep safe

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:

the answer is D

Step-by-step explanation:

Answer:

I think it's 155 cm

Step-by-step explanation:

=(5×5×3)+(5×2×4)

= 75+40

= 155 cm2

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: its 20 I think

Answer:

x = 50

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

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:

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:

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

Step-by-step explanation:

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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Answers:

3 chocolate cones

1 strawberry cones

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

Explanation:

Let's isolate the variable c in the first equation

c+s = 4

c = 4-s

we subtract s from both sides to get c all by itself. This will then be plugged into the second equation. I'm using the substitution property.

1.75c + 1.3s = 6.55

1.75(4-s) + 1.3s = 6.55 ..... replace c with 4-s

1.75*4 + 1.75*(-s) + 1.3s = 6.55

7 - 1.75s + 1.3s = 6.55

-0.45s + 7 = 6.55

-0.45s = 6.55 - 7

-0.45s = -0.45

s = -0.45/(-0.45)

s = 1

So he bought 1 strawberry cone. Use this value of s to find c

c = 4-s

c = 4-1

c = 3

This says he also bought 3 chocolate cones.

-------------------

As a check, we see that c+s = 3+1 = 4, showing that he bought 4 cones total.

Also,

1.75c + 1.3s = 1.75*3 + 1.3*1 = 6.55

indicating he spent $6.55 total for the four cones. This matches with what the instructions tell us, so the answer is confirmed.

Answer:

Step-by-step explanation:

[tex]\left \{ {{c+s=4} \atop {1.75c+1.3s=6.55}} \right.[/tex]

1.75c + 1.75s = 1.75 × 4 ........ (1)

1.75c + 1.3s = 6.55 ........ (2)

(1) - (2)

0.45s = 0.45 ⇒ s = 1

c + 1 = 4 ⇒ c = 3

[ 3 ] chocolate  [ 1 ] strawberry

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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Step-by-step explanation:

Step 2:

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

Stepc3:

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

Answer:

Radical (20)

Step-by-step explanation:

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

Answer:

[tex]m\angle A= 30^{\circ}[/tex]

Step-by-step explanation:

In the question, we're given [tex]\angle A\cong \angle D[/tex]. Therefore, the measure of these two angles must be equal.

To find the value of [tex]x[/tex], set these two angles equal to each other:

[tex]4x-26=x+16[/tex]

Add 26 and subtract [tex]x[/tex] from both sides:

[tex]3x=42[/tex]

Divide both sides by 3:

[tex]x=\frac{42}{3}=14[/tex]

Since [tex]\angle A[/tex] was labelled as [tex]4x-26[/tex], substitute [tex]x=14[/tex] to find its measure:

[tex]\angle A=4(14)-26,\\\angle A=56-26,\\\angle A=\boxed{30^{\circ}}[/tex]

You can also substitute [tex]x=14[/tex] into the label of angle D as angle A is congruent to angle D for easier calculations ([tex]14+16=\boxed{30^{\circ}}[/tex]).

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.

I think it's answer is 79 just guess

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:

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

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:

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

Given:

The inequality is:

[tex]-3x-4<2[/tex]

To find:

The values that are solutions to the given inequality.

Solution:

We have,

[tex]-3x-4<2[/tex]

Adding 4 on both sides, we get

[tex]-3x-4+4<2+4[/tex]

[tex]-3x<6[/tex]

Divide both sides by -3 and change the inequality sign because -3 is a negative value.

[tex]\dfrac{-3x}{-3}>\dfrac{6}{-3}[/tex]

[tex]x>-2[/tex]

Therefore, all the real values greater than -2 are the solutions to the given inequality.

Answer:

I think album is another name of CD.

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