A = a+b+c / 3

Solve for b.
A.) 3A - a - c = b
B.) -c / 3 = b
C.) A - c / 3 = b
C.) -3 - c = b ...?

Answer: The amount she will get is $4.10

Step-by-step explanation:

My reason is because .15 times 6 since she is buying 6 pencils, it is .90. The you subtract the decimal 5.00-0.90 and you get 4.10. Hope this helps. PLease mark brainliest!!!!!

Answer:

1. Which equation can be used to find the time to complete the job when x workers are assigned to it?

A.) m(x) = 270/x

2. Suppose that H(x) varies inversely with x and H(x)=50 when x =0.25

What is H(x) when x =2?

B.) 6.25

Step-by-step explanation:

1. We know when x=3 (workers), m(x)= 90 min (time is takes to mow the lawn at a large park) and we know m(x) varies inversely with x, this means when m(x) increases so x decreases. This is obtained with a division over x.

Now, we can replace to know what is the correct answer:

[tex]m(x)=\frac{y}{x} (1) \\90=\frac{y}{3} \\90*3=y[/tex]

[tex]y=270[/tex]

The answer is A.

[tex]m(x)=\frac{270}{x}[/tex]

We can confirm if we replace x=3 in the equation before. Our answer should be 90.

[tex]m(x)=\frac{270}{3}=90[/tex]

2.  Suppose that H(x) varies inversely with x and H(x)=50 when x =0.25

We know H(x) varies inversely with x, this means when H(x) increases so x decreases. This is obtained with a division over x. Now, we need to find the constant in the numerator.

Now, we can replace in the eq (1) to know the constant in the numerator:

[tex]H(x)=\frac{y}{x}\\50=\frac{y}{0.25} \\50*0.25=y[/tex]

[tex]y=12.5[/tex]

Now, we can replace in the eq (1) to find H(x) when x=2

[tex]m(x)=\frac{y}{x}[/tex]

[tex]m(x)=\frac{12.5}{2}[/tex]

[tex]m(x)=6.25[/tex]  

The answer is B.) 6.25

cos4x = cos2x

We know that:

cos2x = 1-2cos^2 x

==> cos4x = 1-2cos^2 (2x)

Now substitute:

==> 1-2cos^2 (2x) = cos2x

==> 2cos^2 (2x) + cos2x - 1 = 0

Now factor:

==> (2cos2x -1)(cos2x + 1) = 0

==> 2cos2x -1 = 0 ==> cos2x =1/2 ==> 2x= pi/3

==> x1= pi/6 , 7pi/6

==> x1= pi/6 + 2npi

==> x2= 7pi/6 + 2npi

==> cos2x = -1 ==> 2x= pi ==> x3 = pi/2 + 2npi.

==> x= { pi/6+2npi, 7pi/6+2npi, pi/2+2npi}

The solutions to the equation cos(4x) + cos(2x) - 2 = 0 are x = nπ and x = (2n + 1)π/2, where n is an integer.

Here, we have to solve the equation cos(4x) + cos(2x) - 2 = 0, we can use trigonometric identities to simplify it.

We know the following trigonometric identities:

[tex]cos(2x) = 2cos^2(x) - 1\\cos(4x) = 2cos^2(2x) - 1[/tex]

Now, let's rewrite the equation using these identities:

[tex]2cos^2(2x) - 1 + 2cos^2(x) - 1 - 2 = 0[/tex]

Combine like terms:

[tex]2cos^2(2x) + 2cos^2(x) - 4 = 0[/tex]

Divide the entire equation by 2:

[tex]cos^2(2x) + cos^2(x) - 2 = 0[/tex]

Now, let's simplify further:

We can rewrite [tex]cos^2(2x)[/tex] as [tex](1 - sin^2(2x))[/tex]using the Pythagorean identity: [tex]sin^2(\theta) + cos^2(\theta) = 1[/tex]

So, the equation becomes:

[tex](1 - sin^2(2x)) + cos^2(x) - 2 = 0[/tex]

Rearrange the terms:

[tex]cos^2(x) - sin^2(2x) - 1 = 0[/tex]

Now, use the double-angle identity: [tex]sin^2(2x) = 2sin(x)cos(x)[/tex]

[tex]cos^2(x) - 2sin(x)cos(x) - 1 = 0[/tex]

Now, we can rewrite [tex]cos^2(x) as (1 - sin^2(x))[/tex] using the Pythagorean identity:

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

[tex](1 - sin^2(x)) - 2sin(x)cos(x) - 1 = 0[/tex]

Now, simplify further:

[tex]1 - sin^2(x) - 2sin(x)cos(x) - 1 = 0[/tex]

We can see that [tex](1 - sin^2(x))[/tex] cancels out:

-2sin(x)cos(x) = 0

Divide both sides by -2:

sin(x)cos(x) = 0

Now, there are two possible solutions for this equation:

sin(x) = 0

cos(x) = 0

Let's solve each case:

sin(x) = 0

This occurs when x is an integer multiple of π (π, 2π, 3π, ...). So the solutions are x = nπ, where n is an integer.

cos(x) = 0

This occurs when x is an odd multiple of π/2 (π/2, 3π/2, 5π/2, ...). So the solutions are x = (2n + 1)π/2, where n is an integer.

Therefore, the solutions to the equation cos(4x) + cos(2x) - 2 = 0 are x = nπ and x = (2n + 1)π/2, where n is an integer.

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The valid postulates and theorems for proving triangles congruent are ASA, AAS, and SAS. The SSA postulate is not a valid postulate.

The valid postulates and theorems for proving triangles congruent are ASA (Angle-Side-Angle), AAS (Angle-Angle-Side), and SAS (Side-Angle-Side).

The SSA (Side-Side-Angle) postulate, also known as the ambiguous case of the Law of Sines, is not a valid postulate for proving triangles congruent because it can result in two different triangles or no triangle at all.

Thus, the correct answer is A. SSA.

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Answer:

C.

Step-by-step explanation:

We are going to solve it with substitution, that is, we are going to clear one variable and then replace it in the other equation. Then,

[tex]3x + 2y = 12[/tex]

[tex]3x = 12-2y[/tex]

[tex]x = \frac{12-2y}{3}[/tex]

[tex]x = \frac{12}{3}-\frac{2y}{3}[/tex]

[tex]x = 4-\frac{2y}{3}.[/tex]

Now, we replace that x value in the equation 6x + 3y = 21:

[tex]6(4-\frac{2y}{3}) + 3y = 21[/tex]

[tex]24-\frac{12y}{3} + 3y = 21[/tex]

[tex]24-4y + 3y = 21[/tex]

[tex]24-y = 21[/tex]

[tex]-y = 21-24[/tex]

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

[tex]y = 3.[/tex]

Finally, we replace the y value found to find x.

[tex]x = 4-\frac{2*3}{3}[/tex]

[tex]x = 4-2[/tex]

[tex]x = 2.[/tex]

So, the solution is (x,y)=(2,3). Then, the answer is C.

Answer:

then f the teacher

Step-by-step explanation:

Answer:

It would take 2 seconds to the key to reach the ground.

The domain the function is: 0≤t≤2 or in interval notation [0,2].

The range of the function is: 0≤h≤64 or in interval notation [0,64].

Step-by-step explanation:

The provided function is: [tex]h=-16t^2 +64[/tex]

Where h represents the height and t represents the time.

Part(A) how long does it take the key to reach the ground?

When key will be at ground, the height will be 0.

Substitute the value of h=0 in above equation and solve for t.

[tex]-16t^2 +64=0[/tex]

[tex]-16(t^2 -4)=0[/tex]

[tex](t^2 -2^2)=0[/tex]

[tex](t+2)(t-2)=0[/tex]

[tex]t+2=0\ or\ t-2=0[/tex]

[tex]t=-2\ or\ t=2[/tex]

Ignore the negative value of t.

Hence, it would take 2 seconds to the key to reach the ground.

Part (B) what are the reasonable domain and range for the function h?

Time and Distance should be a positive number.

Thus, the value of t and h should be greater or equal to 0.

From the part (a) we know that the after 2 second the key will hit the ground so the value of t can be greater or equal to 0 but should be less or equal to 2.

Thus, the domain the function is: 0≤t≤2 or in interval notation [0,2].

The maximum height of the key will be at t=0.

Thus the maximum value of h can be 64 and minimum value can be 0.

The range of the function is: 0≤h≤64 or in interval notation [0,64].

Step-by-step explanation: First, notice that this line is in slope intercept or more commonly known as y = mx + b form because y is by itself on one side of the equation.

Our slope or m therefore is the coefficient of the x-term which in this case is 3/4. Our b or y-intercept is the constant term which in this case is -2.

To graph a line given the slope and y-intercept, we will first start with its y-intercept. The y-intercept of a line is the point where the line crosses the y-axis. Since our y-intercept is -2, we start by plotting the point that is down 2 units on the y-axis and we call that point A.

From there we take our slope of 3/4 so our rise is 3 and our run is 4 and we end up at point B. Now we can connect the two points and we have our line.

I attached the image of the line in the image provided.

Answer:

x = 3

Step-by-step explanation:

  Given:[tex]\log _{x-1}\left(16\right)=4[/tex]

We have to solve for x.

Consider the given expression [tex]\log _{x-1}\left(16\right)=4[/tex]

[tex]\mathrm{Apply\:log\:rule}:\quad \log _a\left(b\right)=\frac{\ln \left(b\right)}{\ln \left(a\right)}[/tex]

[tex]\log _{x-1}\left(16\right)=\frac{\ln \left(16\right)}{\ln \left(x-1\right)}[/tex]

Thus, the expression becomes,

[tex]\frac{\ln \left(16\right)}{\ln \left(x-1\right)}=4[/tex]

Multiply both side by [tex]\ln \left(x-1\right)[/tex]

[tex]\frac{\ln \left(16\right)}{\ln \left(x-1\right)}\ln \left(x-1\right)=4\ln \left(x-1\right)[/tex]

Simplify, we get,

[tex]\ln \left(16\right)=4\ln \left(x-1\right)[/tex]

Divide both side by 4, we have,

[tex]\frac{4\ln \left(x-1\right)}{4}=\frac{\ln \left(16\right)}{4}[/tex]

[tex]\frac{\ln \left(16\right)}{4}:\quad \ln \left(2\right)[/tex]

Thus, [tex]\ln \left(x-1\right)=\ln(2)[/tex]

[tex]\mathrm{When\:the\:logs\:have\:the\:same\:base:\:\:}\log _b\left(f\left(x\right)\right)=\log _b\left(g\left(x\right)\right)\quad \Rightarrow \quad f\left(x\right)=g\left(x\right)[/tex]

thus, x - 1 = 2

simplify for x, we have,

x = 3

Answer: 12 < x < 15

Step-by-step explanation:

Answer:

The sensory somatic nervous system controls all voluntary responses of the body.

Step-by-step explanation:

The sensory somatic nervous system controls all voluntary responses of the body.

The movements includes muscles and organs movements and reflex movements. Here the sensory neurons carry impulses to the brain and the spinal cord.

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The value of AC + BD will be 70.

A rectangle is a two dimension figure with 4 sides, 4 corners and 4 right angles. The opposite sides of the rectangle are equal and parallel to each other.

Given that;

In rectangle ABCD,

AB = 21

AD = 28

Now,

Since, ABCD is a rectangle.

Hence, We get;

⇒ AC = BD

Here, In rectangle ABCD,

AB = 21

AD = 28

So, By using definition of Pythagoras theorem, we get;

⇒ BD² = AB² + AD²

⇒ BD² = 21² + 28²

⇒ BD² = 441 + 784

⇒ BD² = 1225

⇒ BD = 35

Thus, The value of BD + AC is find as;

⇒ AC + BD = 35 + 35

                 = 70

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Answer:

Height of the kite from ground is 77 m.

Step-by-step explanation:

Given: Length of the kite string = 80 m

           Angle of elevation = 75°

To find: Height of the kite from the ground.

Figure is attached.

In ΔABC

using trigonometric ratio,

[tex]sin\,75^{\circ}=\frac{AB}{AC}[/tex]

[tex]0.97=\frac{AB}{80}[/tex]

[tex]AB=0.97\times80[/tex]

AB = 77.27 m

AB = 77 m (approx.)

Therefore, Height of the kite from ground is 77 m.