If arc CE= 40°, and m∠CDE = 30°, what is arc AB ?

A) 70°
B) 100°
C) 140°
D) 20°

Answer:

The answer is C "The values of h(t) when t = 4 and 5 should be 0."

Answer:

Step-by-step explanation:

The perimeter is the sum of all sides of the figure. In this case, the perimeter of the rectangle would be: [tex]P=W+L+W+L[/tex]; where [tex]W[/tex] is width, and [tex]L[/tex] is length.

According to the problem, the width is two times the length:

[tex]W=2L[/tex]

Replacing this relation in the perimeter equation, we have:

[tex]P=2W+2L\\P=2(2L)+2L\\P=4L+2L\\P=6L[/tex]

Therefore, the perimeter is six times the length of the rectangle. The correct answer is B.

To solve the equation cos2x - sqrt 2 sinx = 1, rewrite cos2x as 2cos^2x - 1. Use the quadratic formula to solve for cosx. Substitute the values back into the equation cos2x - sqrt 2 sinx = 1 and solve for x.

To solve the equation cos2x - √2 sinx = 1, we can use trigonometric identities and equations. First, we can rewrite cos2x as 2cos^2x - 1. So, the equation becomes 2cos^2x - √2 sinx - 1 = 0. To solve this quadratic equation, let's set 2cos^2x - √2 sinx - 1 = 0 and solve for cosx.

Next, we can use the quadratic formula to solve for cosx. The quadratic formula states that x = (-b ± √(b^2 - 4ac)) / 2a. In this case, a = 2, b = -√2 sinx, and c = -1. Plugging in these values, we can solve for cosx.

After solving for cosx, we can substitute the values back into the equation cos2x - √2 sinx = 1 and solve for x. The student's question is about solving the equation cos(2x) - √2 sin(x) = 1 for all solutions. We can use the trigonometric identities cos(2x) = 1 - 2sin2(x) or cos(2x) = 2cos2(x) - 1 to rewrite the equation. Since we have a sine term in the original equation, let's use the former identity:

cos(2x) - √2 sin(x) = 1

(1 - 2sin2(x)) - √2 sin(x) = 1

2sin2(x) + √2 sin(x) - 1 = 0

This is a quadratic equation in sin(x).

We can solve this quadratic equation for sin(x), then find x using inverse trigonometric functions. By factoring the quadratic or using the quadratic formula, we get solutions for sin(x). Then, we solve for x by considering all possible angles in the unit circle that correspond to the found sine values.

The equation cos(2x) - √2 sin(x) = 1 can be solved by using trigonometric identities to simplify and factor the equation, leading to solving for sin(x) using inverse operations.

The original equation given is cos(2x) - √2 sin(x) = 1. To solve this equation, we can use trigonometric identities to simplify the cosine term. One such identity is cos(2x) = 1 - 2sin²(x), which allows us to rewrite the equation as 1 - 2sin²(x) - √2 sin(x) = 1. From there, we subtract 1 from both sides, thus isolating the sine terms on the left: - 2sin²(x) - √2 sin(x) = 0. Factoring out the common term sin(x), we get sin(x)(-2sin(x) - √2) = 0. Setting each factor equal to zero gives us two possible solutions: sin(x) = 0 and sin(x) = -√2/2. In the context of a right triangle, these solutions correspond to specific angles where the sine value is 0 and -√2/2, respectively. The solutions can be found using standard trigonometric unit circle values or by calculating the inverse sine for -√2/2.

Answer:  [tex]\overline{WX}\cong \overline{WV},\ \overline{VA}\cong\overline{AZ}[/tex]

[tex]\overline{XY}\cong\overline{YZ}[/tex]

[tex]\overline{UV}\cong \overline{UZ}\cong \overline{UX}[/tex]

Step-by-step explanation:

In the given figure we have a triangle , in which U is the circumcenter of triangle XVZ.

We know that the circumcenter is equidistant from each vertex of the triangle.

Since , the line segments which are representing the distance from the vertex and the circumcenter are [tex]\overline{UV},\ \overline{UZ},\ \overline{UX}[/tex]

Also, The circumcenter is at the intersection of the perpendicular bisectors of the triangle's sides.

Then ,  [tex]\overline{WX}\cong \overline{WV},\ \overline{VA}\cong\overline{AZ}[/tex] and [tex]\overline{XY}\cong\overline{YZ}[/tex]

Hence, the segments which are congruent are [tex]\overline{UV}\cong \overline{UZ}\cong \overline{UX}[/tex]

 [tex]\overline{WX}\cong \overline{WV},\ \overline{VA}\cong\overline{AZ}[/tex]

[tex]\overline{XY}\cong\overline{YZ}[/tex]

The value of the solution is, 3 is 1 percent of 300.

We have to give that,

To find the amount for 1 percent is 3.

Let us assume that,

3 is 1 percent of x.

Hence, It can be written as,

3 = 1% of x

Solve for x,

3 = 1/100 × x

3 × 100 = x

x = 300

Therefore,  3 is 1 percent of 300.

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Answer:  Simplified form is

[tex]1\frac{5}{6}[/tex]

Step-by-step explanation:

Since we have given that

[tex]5\frac{1}{3}-3\frac{9}{18}[/tex]

Now, we will simplify it with the following steps :

[tex]5\frac{1}{3}-3\frac{9}{18}\\\\=\frac{16}{3}-3\frac{1}{2}\\\\=\frac{16}{3}-\frac{7}{2}\\\\=\frac{32-21}{6}\\\\=\frac{11}{6}\\\\=1\frac{5}{6}[/tex]

Hence, simplified form is

[tex]1\frac{5}{6}[/tex]

Answer:

D

Step-by-step explanation:

(Michigan online school.)

Answer: [tex]x(7y)[/tex]

Step-by-step explanation:

The given expression: [tex]7(xy)[/tex]

i.e. a product of 7 and xy.

The operation used here: Multiplication.

Commutative property of multiplication  :-

[tex]a\times b=b\times a[/tex] for any numbers a and b.

Associative property of multiplication :-

[tex]a\times(b\times c)=(a\times b\times c)[/tex] for any numbers a , band c.

Now, [tex]7(xy)=(7x)y[/tex] [Associative property of multiplication]

[tex]=(x7)y[/tex]  [Commutative property of multiplication]

[tex]=x(7y)[/tex]    [Associative property of multiplication]

The solution of the system of equation [tex]5x+2y = 29[/tex] ; [tex]x+4y =13[/tex] will be (3,2) and this can be determined by using the arithmetic operations.

Given :

[tex]5x+2y = 29[/tex]  ----- (1)

[tex]x+4y =13[/tex]  ----- (2)

Now, solve for x in equation (2).

[tex]x = 13-4y[/tex]  --- (3)

Now, put the value of x obtained above in equation (1).

[tex]5(13-4y)+2y=29[/tex]

[tex]65-20y+2y=29[/tex]

[tex]36=18y[/tex]

[tex]y=2[/tex]

Now, put the value of y in equation (3).

[tex]x=13-(4\times 2)[/tex]

[tex]x=5[/tex]

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The slope of the line tangent to the curve y^2 + (xy+1)^3 = 0 at (2, -1) is -3/4.

The slope of the line tangent to the curve y^2 + (xy+1)^3 = 0 at (2, -1) can be found using the concept of implicit differentiation. To find the slope, we need to differentiate the equation with respect to x and then substitute the coordinates (2, -1) into the resulting equation. Let's solve it step by step.

First, we differentiate the equation implicitly with respect to x:

2y * dy/dx + 3(xy + 1)^2 * (y + x * dy/dx) = 0

Next, we substitute the values x = 2 and y = -1 into the equation:

2(-1) * dy/dx + 3(2(-1) + 1)^2 * (-1 + 2 * dy/dx) = 0

Simplifying the equation:

-2dy/dx - 3 * 1 * (-1 + 2dy/dx) = 0

-2dy/dx + 3 + 6dy/dx = 0

Combining like terms:

4dy/dx = -3

Finally, solving for dy/dx, we get:

dy/dx = -3/4

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The slope of the tangent line to the curve at the point \((2, -1)\) is:

[tex]\[\boxed{\frac{3}{4}}\][/tex]

To find the slope of the tangent line to the curve given by the equation [tex]\( y^2 + (xy + 1)^3 = 0 \)[/tex] at the point [tex]\( (2, -1) \)[/tex], we need to use implicit differentiation.

Given the curve:

[tex]\[y^2 + (xy + 1)^3 = 0\][/tex]

We differentiate both sides with respect to x . Using the chain rule and implicit differentiation, we get:

[tex]\[\frac{d}{dx} [y^2] + \frac{d}{dx} [(xy + 1)^3] = 0\][/tex]

First, differentiate [tex]\( y^2 \):[/tex]

[tex]\[\frac{d}{dx} [y^2] = 2y \frac{dy}{dx}\][/tex]

Next, differentiate [tex]\( (xy + 1)^3 \)[/tex] using the chain rule:

[tex]\[\frac{d}{dx} [(xy + 1)^3] = 3(xy + 1)^2 \cdot \frac{d}{dx} [xy + 1]\]\[= 3(xy + 1)^2 \cdot (y + x \frac{dy}{dx})\][/tex]

Putting it all together, we get:

[tex]\[2y \frac{dy}{dx} + 3(xy + 1)^2 (y + x \frac{dy}{dx}) = 0\][/tex]

Now, substitute [tex]\( x = 2 \) and \( y = -1 \)[/tex] into the equation:

[tex]\[2(-1) \frac{dy}{dx} + 3((2)(-1) + 1)^2 \left( -1 + 2 \frac{dy}{dx} \right) = 0\]\[-2 \frac{dy}{dx} + 3(-2 + 1)^2 \left( -1 + 2 \frac{dy}{dx} \right) = 0\][/tex]

[tex]\[-2 \frac{dy}{dx} + 3(-1)^2 \left( -1 + 2 \frac{dy}{dx} \right) = 0\]\[-2 \frac{dy}{dx} + 3(1) \left( -1 + 2 \frac{dy}{dx} \right) = 0\][/tex]

[tex]\[-2 \frac{dy}{dx} + 3(-1 + 2 \frac{dy}{dx}) = 0\]\[-2 \frac{dy}{dx} + 3(-1 + 2 \frac{dy}{dx}) = 0\][/tex]

[tex]\[-2 \frac{dy}{dx} + 3(-1) + 6 \frac{dy}{dx} = 0\][/tex]

[tex]\[-2 \frac{dy}{dx} - 3 + 6 \frac{dy}{dx} = 0\][/tex]

[tex]\[4 \frac{dy}{dx} - 3 = 0\][/tex]

[tex]\[4 \frac{dy}{dx} = 3\][/tex]

[tex]\[\frac{dy}{dx} = \frac{3}{4}\][/tex]

So, the slope of the tangent line to the curve at the point \((2, -1)\) is:

[tex]\[\boxed{\frac{3}{4}}\][/tex]

Answer:

48√3 sq. in.

Step-by-step explanation:

I know this is correct bc I just had this question and this was the correct answer

Method 1

Applying the Pythagorean Theorem

we know that

[tex]10^{2}= 5^{2} +a^{2}[/tex]

Solve for a

[tex]100= 25 +a^{2}[/tex]

[tex]a^{2}=100-25[/tex]

[tex]a^{2}=75[/tex]

[tex]a=\sqrt{75}=5 \sqrt{3}\ units[/tex]

therefore

the answer is

the length of the altitude is [tex]5 \sqrt{3}\ units[/tex]

Method 2

we know that

[tex]sin(60\°)=\frac{\sqrt{3}}{2}[/tex] -------> equation A

and

in this problem

[tex]sin(60\°)=\frac{a}{10}[/tex] --------> equation B

equate equation A and equation B

[tex]\frac{\sqrt{3}}{2}=\frac{a}{10}\\\\a=\frac{10\sqrt{3}}{2}\\\\a=5 \sqrt{3}\ units[/tex]

therefore

the answer is

the length of the altitude is [tex]5 \sqrt{3}\ units[/tex]