Answer:

**Answer:**

18√2

**Step-by-step explanation:**

The area of the smaller triangle is 1/2 that of the larger one. Since the triangles are similar, the dimensions of the smaller triangle are √(1/2) those of the larger one.

36 · √(1/2) = 36 · (√2)/2 = **18√2 . . . . length of line dividing the triangle**

Look back at the plans these students used to solve the word problem below.Who found a correct solution?- STEP 4: LOOK BACKAccording to soap box derby rules, a racer must weigh 250pounds or less. The Math Club's car weighed in at 266pounds on the day of the derby. How many pounds did theMath Club need to remove from their soap box racer?Dana added the weight limit to the Hector subtracted the weight limitracer's weight. Sincefrom the racer's weight. Since250 +266 = 516, the Math Club 266 - 250 = 16, the Math Clubneeded to remove 516 pounds from needed to remove 16 poundsthe racerfrom the racerO A. DanaB. Hector

The fraction of defective integrated circuits produced in a photolithography process is being studied. A random sample of 300 circuits is tested, revealing 12 defectives.(a) Calculate a 95% two-sided confidence interval on the fraction of defective circuits produced by this particular tool. Round the answers to 4 decimal places.< p\l>(b) Calculate a 95% upper confidence bound on the fraction of defective circuits. Round the answer to 4 decimal places

In ΔBCD, the measure of ∠D=90°, CB = 25, BD = 7, and DC = 24. What is the value of the sine of ∠C to the nearest hundredth?

consider the quadratic form q(x,y,z)=11x^2-16xy-y^2+8xz-4yz-4z^2. Find an orthogonal change of variable that eliminates the cross product in q(x,y,z) and express q in the new variables.

A ladder leans against a building that angle of elevation of the latter is 70° the top of the ladder is 25 feet from the ground. to the nearest 10th of a foot how far from the building is the base of the ladder a. 20.5 feet b. 30.5 feet C.32.3’ or D.39.5 feet

The fraction of defective integrated circuits produced in a photolithography process is being studied. A random sample of 300 circuits is tested, revealing 12 defectives.(a) Calculate a 95% two-sided confidence interval on the fraction of defective circuits produced by this particular tool. Round the answers to 4 decimal places.< p\l>(b) Calculate a 95% upper confidence bound on the fraction of defective circuits. Round the answer to 4 decimal places

In ΔBCD, the measure of ∠D=90°, CB = 25, BD = 7, and DC = 24. What is the value of the sine of ∠C to the nearest hundredth?

consider the quadratic form q(x,y,z)=11x^2-16xy-y^2+8xz-4yz-4z^2. Find an orthogonal change of variable that eliminates the cross product in q(x,y,z) and express q in the new variables.

A ladder leans against a building that angle of elevation of the latter is 70° the top of the ladder is 25 feet from the ground. to the nearest 10th of a foot how far from the building is the base of the ladder a. 20.5 feet b. 30.5 feet C.32.3’ or D.39.5 feet

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9514 1404 393

**Explanation:**

a) The velocity curve is linearly increasing from 0 to 6 m/s over a period of 2 seconds, then linearly decreasing from 6 m/s to 0 over the same period. The acceleration is the rate of change of velocity, so for the first half of the motion the acceleration is a constant (6 m/s)/(2 s) = 3 m/s². Similarly, over the second half of the motion, the acceleration is a constant (-6 m/s)/(2 s) = -3 m/s².

The distance traveled is the integral of the velocity, so the linearly increasing velocity will cause the distance vs. time curve to have a parabolic shape. The shape will likewise be parabolic, but with decreasing slope, as the velocity ramps down to zero. Overall, the distance versus time curve will have an "S" shape.

The motion (position and velocity) will be continuous, but the acceleration will not be. There will be a significant "j.erk" at the 2-second mark where acceleration abruptly changes from increasing the velocity to braking (decreasing the velocity).

__

b) The attachment shows the (given) velocity curve in meters per second and its integral, the position curve, in meters.

The integral in the attached works nicely for machine evaluation. For hand evaluation, it is perhaps best written piecewise:

number am I?

**Answer:**

- 865

**Step-by-step explanation:**

Let the 3-digit number is __abc__ = 100a + 10b + c.

__We have:__

- 100a + 10b + c - 100c - 10b - a = 297
- b = a - 6
- a = 2c - 2

__Simplify the first equation:__

- 99a - 99c = 297
- a - c = 3
- a = c + 3

__Solve for c by substitution:__

- 2c - 2 = c + 3
- 2c - c = 3 + 2
- c = 5

__Find a:__

- a = 3 + 5 = 8

__Find b:__

- b = 8 - 2 = 6

__The number is:__

- 865

**Answer:**

- (a) minor arc: XY; major arc: XVY
- (b) XVY = 248°
- (c) tangent: UV; secant: UX
- (d) UV = 6√5

**Step-by-step explanation:**

**(a)** Any pair of points on the circle that are separated by less than the diameter will define a minor arc. (The minor arc is the shortest arc of the circle between the points.) Possible minor arcs in this diagram are ...

VX, VY, XY

The corresponding major arc is usually named by adding the name of a point between the two endpoints that is not on the minor arc. For the minor arcs above, the corresponding major arcs are ...

VYX, VXY, XVY

Given that part (b) tells us the **minor arc** of interest is 112°, we assume that arc is the one subtended by the chord: **XY**.

Then, per the discussion above, the corresponding **major arc is XVY**.

___

**(b)** The sum of major and minor arcs is the whole circle, 360°. So, the **measure of the major arc **is ...

360° -112° = **248°**

___

**(c)** A tangent line intersects a circle at exactly one point. It is perpendicular to a radius to that point of intersection. The **tangent line** in this diagram is **UV**.

A secant intersects a circle in two places. The portion of the secant between the points of intersection is called a *chord*. The **secant line UX **contains the chord XY.

___

**(d)** A rule of secants (and chords) is that the product of distances from where the secants (or chords) meet to the two intersection points with the circle is the same. For a tangent line, effectively, the two points of intersection are at the same distance. This means ...

UV·UV = UX·UY

UV² = 9·(9+11) = 180

UV = √180 = √(6²·5)

** UV = 6√5 ≈ 13.42**

_____

The attached figure is drawn to scale with arc XY being 112°.

**Answer:**

(a) minor arc: arc VX

major arc: arc VYX

(b) 248 degrees

(c) Tangent: UV

Secant: UY

**Step-by-step explanation:**

(b) 360 - 112 = 248

(c) UV crosses on the circumference of the circle at exactly one point

UY crosses through the circle at exactly two points

**Answer:**

y = -2

**Step-by-step explanation:**

To find the equation of the tangent we apply implicit differentiation, and then we take apart dy/dx

The equation is

implicit differentiation give us

But we know that

Hence, for the point (0,-2) and by replacing for dy/dx

**Hence m=0**, that is, the tangent line to the point is a horizontal line that cross the y axis for y=-2. The equation is:

**y=(0)x+b = -2**

**HOPE THIS HELPS!! **

In order to find the equation of the tangent line to the curve y²(y² - 4) = x²(x² - 5) at the point (0, -2), we will use the method of implicit differentiation. Here are the steps:

Step 1: Differentiate Each Side of the Given Equation with Respect to x

Applying the chain rule to differentiate y²(y² - 4) with respect to x gives:

2y*y'(y² - 4) + y²*2y*y' = d/dx [y²(y² - 4)]

The chain rule is also applied to differentiate x²(x² - 5) with respect to x, yielding:

2x(x² - 5) + x²*2x = d/dx [x²(x² - 5)]

Step 2: Equate the Two Expressions Found from Step 1 and Solve for y'

2y*y'(y² - 4) + y²*2y*y' = 2x(x² - 5) + x²*2x

This equation can be solved by isolating y' (the derivative of y with respect to x), which represents the slope of the tangent line.

Step 3: Use the Given Point (0, -2) to Find the Slope of the Tangent Line

Substitute x = 0 and y = -2 into the equation found in Step 2 to get the specific value for the slope at the given point.

Step 4: Use the Point-Slope Form of the Line to Write the Equation of the Tangent Line

The point-slope form of the line** y - y₁ = m(x - x₁)** can be used to write the equation of the tangent line. We substitute for x₁ and y₁ with the coordinates of the **given point (0, -2)**, and m with the slope found from Step 3.

The resulting equation represents the tangent line to the curve at the given point (0, -2). Please note that the full calculation may result in a complex slope due to the nature of the given curve equation. Nonetheless, this process illustrates the application of implicit differentiation and the point-slope form of a line in finding the equation of a tangent line to a curve.

#SPJ3

The answer would be $198.93