I need some help with this. I’m not sure if I’m right

Answer:

m=105

Step-by-step explanation:

The required, measure of angle m∠8 is also 110 degrees as lines JL and MP are parallel.

Parallel lines are defined as identical lines with identical orientations, parallel lines never intersect each other but meet at infinity.

Here,
Since lines JL and MP are parallel, we know that the alternate interior angles are congruent. Therefore:

m∠5 = m∠8

We are given that m∠5 = 110°, so:

m∠8 = 110°

Therefore, m∠8 is also 110 degrees.

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Given that the diagonals of a rhombus are always perpendicular bisectors of each other, what is the area of a rhombus with side length  \(\sqrt{89}\) units and diagonals that differ by 5 units.

Diagonals of Rhombus:

A diamond's Rhombus is a line segment that connects two non-adjacent vertices of the diamond. A rhombus has two diagonals that bisect it at right angles. That is, they form four congruent right triangles. The rhombic diagonal formula is based on the area of ​​the diagonal, expressed as p = (2 × area)/q. where 'p' and 'q' are the two diagonals of the rhombus.

Now,

Label each half of the short diagonal as X.  

Label each half of the long diagonal as X+ 3.

The sides of the rhombus as \(\sqrt{89}\).

Each of the internal triangles will be figured the same.

We have a right triangle with one side X, one side X+3, and the hypotenuse \(\sqrt{89}\) .

According to the Pythagorean theorem, in a right triangle (90 degrees), the square of the hypotenuse equals the sum of the squares of the other two sides. The triangle ABC where BC2 = AB2 + AC2. where AB is the base, AC is the height (height), and BC is the hypotenuse. Note that the hypotenuse is the longest side of a right triangle.

Therefore,

Pythagoras' Theorem  :

(X)² + (X+3)² =   \((\sqrt{89} )^2\)

X² + X² + 6X + 9 = 89

2X² + 6X – 80 = 0  

This will factor to          

(2X + 16)(X – 5) = 0

Disregard the negative solution.  

Half of the short diagonal is 5 units.  

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Solution

For this case we can do the following proportion:

Solving for x

Then the answer would be:

0.019 kg * (1000gr/1 kg)= 19 grams

Answer:

Step-by-step explanation:

                  Statements                  Reasons

1). EF ≅ RQ                            1). Given

2). FD ≅ QP                          2). Given

3). ∠EFD ≅ ∠PQR                 3). Given

4). ΔDFE ≅ ΔPQR                 4). By SAS postulate of congruence

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4/14 is equivalent. This was found by multiplying both the numerator and denominator by 2.

Answer:

15 °C

Step-by-step explanation:

°C = (°F - 32) * (5/9)

Given that the initial temperature was 77 °F and it dropped by 10 °C, we can calculate the final temperature.

Initial temperature: 77 °F

Converting to Celsius:

°C = (77 - 32) * (5/9)

°C ≈ 25

The temperature dropped by 10 °C, so the final temperature is:

Final temperature = Initial temperature - Temperature drop

Final temperature ≈ 25 - 10 = 15 °C

Therefore, the temperature after the storm was approximately 15 °C.

The shaded region of the rectangle is 242.9 cm² and the shaded region of the sector is 7.1 square units.

Question 17) is a figure of a rectangle and two inscribed circles.

The area of a rectangle is expressed as: A = length × width

The area of a circle is expressed as: A = πr²

Where r is the radius.

To determine the area of the shaded region, we simply subtract the areas of the two circles from the area of the rectangle.

Area = ( Length × width ) - 2( πr² )

Area = ( 40 × 10 ) - 2( π × 5² )

Area = ( 400 ) - 2( 25π )

Area = 400- 50π

Area = 242.9 cm²

Area of the shaded region is 242.9 squared centimeters.

Question 18) is the a figure a sector of a circle and a right triangle.

The area of a sector is expressed as: A = (θ/360º) × πr²

The area of a triangle  is expressed as: A = 1/2 × base × height

To determine the area of the shaded region, we simply subtract the areas of the triangle from the area of the sector.

Hence:

Area = ( (θ/360º) × πr² ) - ( 1/2 × base × height )

Plug in the values:

Area = ( (90/360º) × π × 5² ) - ( 1/2 × 5 × 5 )

Area = 25π/4 - 12.5

Area = 7.1

Therefore, the area of the shaded region is 7.1 square units.

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Answer:

30 miles

Step-by-step explanation:

Given that :

Scale = 2 inches represents 5 miles

This means that 2 inches in the map equals to 5 miles on ground ;

Hence, if the trail measures 12 inches on the map, the length on ground will be ; x

2 inches = 5 miles

12 inches = x miles

Cross multiply :

2x = (12 * 5)

2x = 60

x = 60 / 2

x = 30 miles

The correct option is (2)

For any quadratic equation: ax2 + bx + c = 0.

The complex roots are expressed in the form a ± bi.

where a is the real part

& b is the imaginary part.

Since, one root is √7 + 5i, therefore the other root will be of the form √7 - 5i.

Hence, option (2) is correct.

Answer:

you can earn up to $5 a day

Step-by-step explanation:

50 X .10 is 5

:))

Using translation concepts, we have that:

A translation is represented by a change in the function graph, according to operations such as multiplication or sum/subtraction either in it’s definition or in it’s domain. Examples are shift left/right or bottom/up, vertical or horizontal stretching or compression, and reflections over the x-axis or the y-axis.

On the translation of △ABC to △A′B′C′, the rule applied to the vertices is:

(x,y) -> (x + 4, y).

Hence it was translated 4 units right.

On the translation of △A′B′C′​ to △A′′B′′C′', the rule is given by:

(x,y) -> (-y,x).

Hence it was rotated 90 degrees counterclockwise about the origin.

The triangle keeps the same size after the translations, hence:

Because the transformations are both rigid, the pre-image and the image are congruent.

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The quadrilateral formed would be a square and the diagonal of the square would have a length of 12 inches

A rhombus is a quadrilateral (has four sides and four angles) in which opposite sides are parallel to each other. All the sides of rhombus are equal. Opposite angles are also equal.

ABCD is a rhombus. If the midpoints of its sides are joined to form a quadrilateral, the quadrilateral formed would be a square and the diagonal of the square would have a length of 12 inches (same as the rhombus length).

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Answer:

Area of table with leaf = 28.56 feet²

Area of table without leaf = 16 feet²

Step-by-step explanation:

Given:

Side of square table = 4 feet

Diameter of side leaf = 4 feet

Number of semicircle leaf = 2

Find:

Area of table without leaf

Area of table with leaf

Computation:

Area of square = Side x Side

Area of table without leaf = 4 x 4

Area of table without leaf = 16 feet²

Radius of side leaf = Diameter of side leaf / 2

Radius of side leaf = 4 / 2

Radius of side leaf = 2 feet

Area of table with leaf = Area of table without leaf + 2[πr²/2]

Area of table with leaf = 16+ 2[πr²/2]

Area of table with leaf = 16+ (3.14)(2)²

Area of table with leaf = 16+ (3.14)(4)

Area of table with leaf = 16+ 12.56

Area of table with leaf = 28.56 feet²

Answer:

krkk4k4k

Step-by-st man this really is an annoying question

Answer:

Tyler would get $110 and Kyle would get $10

Step-by-step explanation:

Have a nice day/night :)

And if you can may you please mark me brainliest

Answer:

sqr of 161

Step-by-step explanation:

8^2 + x^2 = 225

64+x^2 = 225

161=x^2

sqr of 161 = x

hope i helped!!!

Step-by-step explanation:

\( {15}^{2} - {8}^{2} = (15 + 8)(15 - 8) \\ = 23 \times 7 = 161\)^-2

Answer:

Try contacting the Help center located at the bottom part of the home page of this site. It is recoverable as long as no violation committed

Based on the mean of the data, the percent of the data that is within the interval of being less than 52 is 50%.

What percent of the data is less than 52?

The standard normal distribution (z distribution) is a normal distribution with a mean of 0 and a standard deviation of 1. Any point (x) from a normal distribution can be converted to the standard normal distribution (z) with the formula z = (x-mean) / standard deviation.

The data is said to be a normal distribution which means that 50% of the data lies on either side of the mean of 52.

This means that the percentage of the data that would lie below 52, would be 50%.

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The Length of the rectangle is: B. 6 2/3.

The area of a rectangle is calculated as: Area = (length)(width). Therefore, if we known the area of a rectangle and its width, we can find the length of the rectangle using:

Length of rectangle = area of rectangle / width of rectangle.

Given the following:

Area of a rectangle = 21 and 2 over 3 square inches

Width = 3 and 1 over 4 inches

Length of the rectangle = 21 2/3 ÷ 3 1/4

Length of the rectangle = 65/3 ÷ 13/4

Length of the rectangle = 65/3 × 4/13

Length of the rectangle = 260/39 = 6 2/3.

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The missing angle is <B= 40 degree and missing side length is AB = 12.25 and AC = 19.068.

To find the missing angle and side measures of ΔABC, we can use the properties of a triangle.

Given:

∠A = 50°

∠C = 90°

CB = 16

We can start by finding the measure of ∠B:

∠A + ∠B + ∠C = 180° (Sum of angles in a triangle)

50° + ∠B + 90° = 180°

∠B + 140° = 180°

∠B = 180° - 140°

∠B = 40°

Now, using Sine law

CB/ sin A = AB / sin C

16 / sin 50 = AB / sin 90

16 / 0.766044 = AB

AB = 12.25

Again 12.25 = AC/ sin B

12.25 = AC / sin 40

AC = 19.068

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Here are the correct matches to the expressions to their solutions.

The GCF of 28 and 60 is 4.

(-3/8)+(-5/8) = -4/4 = -1.

-1/6 DIVIDED BY 1/2 = -1/6 X 2 = -1/3.

The solution of 0.5 x = -1 is x = -2.

The solution of 1/2 m = 0 is m = 0.

-4 + 5/3 = -11/3.

-2 1/3 - 4 2/3 = -10/3.

4 is not a solution of -4 < x.

1. The GCF of 28 and 60 is 4.

The greatest common factor (GCF) of two numbers is the largest number that is a factor of both numbers. To find the GCF of 28 and 60, we can factor each number completely:

28 = 2 x 2 x 7

60 = 2 x 2 x 3 x 5

The factors that are common to both numbers are 2 and 2. The GCF of 28 and 60 is 2 x 2 = 4.

2. (-3/8)+(-5/8) = -1.

To add two fractions, we need to have a common denominator. The common denominator of 8/8 and 5/8 is 8. So, (-3/8)+(-5/8) = (-3 + (-5))/8 = -8/8 = -1.

3. -1/6 DIVIDED BY 1/2 = -1/3.

To divide by a fraction, we can multiply by the reciprocal of the fraction. The reciprocal of 1/2 is 2/1. So, -1/6 DIVIDED BY 1/2 = -1/6 x 2/1 = -2/6 = -1/3.

4. The solution of 0.5 x = -1 is x = -2.

To solve an equation, we can isolate the variable on one side of the equation and then solve for the variable. In this case, we can isolate x by dividing both sides of the equation by 0.5. This gives us x = -1 / 0.5 = -2.

5. The solution of 1 m = 0 is m = 0.

To solve an equation, we can isolate the variable on one side of the equation and then solve for the variable. In this case, we can isolate m by dividing both sides of the equation by 1. This gives us m = 0 / 1 = 0.

6. -4 + 5/3 = -11/3.

To add a fraction and a whole number, we can convert the whole number to a fraction with the same denominator as the fraction. In this case, we can convert -4 to -4/3. So, -4 + 5/3 = -4/3 + 5/3 = -11/3.

7. -2 1/3 - 4 2/3 = -10/3.

To subtract two fractions, we need to have a common denominator. The common denominator of 1/3 and 2/3 is 3. So, -2 1/3 - 4 2/3 = (-2 + (-4))/3 = -6/3 = -10/3.

8. 4 is not a solution of -4 < x.

The inequality -4 < x means that x must be greater than -4. The number 4 is not greater than -4, so it is not a solution of the inequality.

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To find the angle B we can use the tangent property:

Rounding it will be 62.7 degrees.

Answer:

A) (x, y, z) = ( 7/4 , \(\frac{7\sqrt{3} }{4} , \frac{7\sqrt{3} }{2}\)  )

B)  (x, y, z) = ( 6 , \(\frac{\pi }{2}, \frac{3\pi }{4}\) )

Step-by-step explanation:

A) ( 7, π/3, π/6)

(x, y, z) = ( 7/4 , \(\frac{7\sqrt{3} }{4} , \frac{7\sqrt{3} }{2}\)  )

B) (6, π/2, 3π/4)

(x, y, z) = ( 6 , \(\frac{\pi }{2}, \frac{3\pi }{4}\) )

attached below is the plot of the points and the detailed solution

Answer:

A) Most points are between these limits, so the process appears to be in control with respect to variability.

D) The value of S on the 5th day lies above the UCL, so an out-of-control signal is generated.

Step-by-step explanation:

We have the data of 30 days of daily samples, of size n=200. The data is expressed in "number of non-conforming chips". To graph the data in a p-chart, we have to calculate the proportion, so we have to divide the number of non-conforming chips in each sample by the sample size in order to get the proportion.

The data of non-conforming chips es:

[9 16 24 18 36 19 6 26 8 25 30 17 16 19 15 20 13 22]

The proportion of non-conforming chips is calculated as:

\(p_i=\dfrac{x_i}{n}=\dfrac{x_i}{200}\)

Then, the proportions of non-conforming chips are:

[0.045 0.08 0.12 0.09 0.18 0.095 0.03 0.13 0.04 0.125 0.15 0.085 0.08 0.095 0.075 0.1 0.065 0.11 0.045 0.115 0.09 0.075 0.075 0.125 0.16 0.095 0.06 0.115 0.075 0.13

]

For the p-chart we calculate the average proportion:

\(\bar{p}=\dfrac{1}{30}\sum p_i=\dfrac{2.855}{30}\approx0.095\)

Then, we calculate the control limits as:

\(\bar{p}\pm\sqrt{\dfrac{\bar{p}(1-\bar{p})}{m}}\\\\\\0.095\pm\sqrt{\dfrac{0.095\cdot0.905}{30}}=0.095\pm0.054\\\\\\LCL=0.095-0.054=0.041\\\\UCL=0.095+0.054=0.149\)

With these values, we can graph the data in the p-chart.

We can see that most points are within these limits, as it is expected (we are calculating the control limits with the same data we are charting). There are 5 out of 30 (one sixth of the samples) that are outside the control limits but there are not more than one consecutive points outside the contorl limits, so we can conclude that the process is in control.

The value of the 5th day (p=0.180) is over the UCL (UCL=0.149), so an out-of-control signal is generated. As the 6th day sample is in control (p=0.095), we could conclude that a special condition makes the 5th day sample lies outside the control limits.

Trigonometric identity is tanθ ≅ 0.4364

\($\sin A=\frac{2}{5}$\)

\($\cos ^2 A=1-\sin ^2 A=\frac{21}{25}$\)

\($\cos A=\frac{\sqrt{21}}{5}$\)

\($\tan A=\frac{\sin A}{\cos A}=\frac{\left(\frac{2}{5}\right)}{\left(\frac{\sqrt{21}}{5}\right)}=\frac{2}{\sqrt{21}} \cong 0.4364$\)

Sine, cosine, tangent, cosecant, secant, and cotangent are the functions. All of these trigonometric ratios are defined using the sides of a right triangle, specifically the adjacent, opposite, and hypotenuse sides.

The even-odd identities relate the value of a trigonometric function at a given angle to the value of the function at the opposite angle. tan ( − θ ) = − tan θ tan ( − θ ) = − tan θ cot ( − θ ) = − cot θ cot ( − θ ) = − cot θ sin ( − θ ) = − sin θ sin ( − θ ) = − sin θ csc ( − θ ) = − csc θ csc ( − θ ) = − csc θ

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Answer:

basta Mao nana siya

And ayw pug sa Kung wlay answer

The upper and lower bound range of the capacity of a tank is 0.55<x<0.65.

Given that, the capacity of a tank is 0.6 cm³.

Upper and lower bounds are the maximum and minimum values that a number could have been before it was rounded. They can also be called limits of accuracy. The upper and lower bounds can be written using error intervals.

To find the upper and lower bound add and subtract 0.05 to the given number.

Lower bound =0.6-0.05

= 0.55

Upper bound =0.6+0.05

= 0.65

So, the upper and lower bound range is 0.55<x<0.65

Therefore, the upper and lower bound range of the capacity of a tank is 0.55<x<0.65.

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