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Answer:

the lines arent parallel, so you cant use corresponding angles theorem

The error is corresponding property does not apply.

Parallel lines are those lines that are equidistant from each other and never meet, no matter how much they may be extended in either directions. For example, the opposite sides of a rectangle represent parallel lines.

We know that when two lines are parallel they the following property:

As, there is no parallel lines line.

So, the measurement if <1 = 88 can't be true because corresponding doesn't apply.

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Answer:

=k=0.25h-7/12

Step-by-step explanation:

4(h-3k)=h+7

=4h-12k=h+7

collecting like terms and leaving characters with k on 1 side, we get;

12k=3h-7

=k=0.25h-7/12

Answer:

96 in^2

Step-by-step explanation:

area of rectangle: 24*8=192

Area of triangle: (1/2)*base*height

= .5 * 8 * 24

=192/2

=96

area of shaded=rectangle - triangle

= 192-96

which is also just 192/2

so it is 96

9/x and 20x⁹ are the monomials.

The correct option A and D.

An expression that consists of variables, constants, and exponents that is combined using mathematical operations like addition, subtraction, multiplication, and division is referred to as a polynomial.

A monomial is a polynomial with just one term.

From the given choices:

9/x and 20x⁹ are the monomials.

11x - 9 and 20x⁹ -7x are the binomial.

Therefore, A and D are the monomials.

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[tex]\overline {LM} \parallel \overline {ON} \ and \ \overline {LO} \parallel \overline {MN}[/tex] because they form congruent alternate interior angles with a common transversal, therefore, LMNO is a parallelogram because the opposite sides are parallel

The reason the above statement are correct is presented as follows:

The given parameters are;

[tex]\overline {LM}[/tex] is congruent to [tex]\overline {ON}[/tex] and [tex]\overline {LO}[/tex] is congruent to [tex]\overline {MN}[/tex]

Required:

To prove that LMNO (quadrilateral) is a parallelogram

Solution:

The two column proof is presented as follows;

Statement [tex]{}[/tex]           Reason

[tex]\overline {LM} \cong \overline {ON}[/tex]  [tex]{}[/tex]           Given

[tex]\overline {LM} = \overline {ON}[/tex]  [tex]{}[/tex]           Definition of congruency

[tex]\overline {LO}[/tex] ≅ [tex]\overline {MN}[/tex] [tex]{}[/tex]            Given

[tex]\overline {LO}[/tex] = [tex]\overline {MN}[/tex] [tex]{}[/tex]             Definition of congruency

[tex]\overline{LN}[/tex] ≅

[tex]\overline{LN}[/tex] =

ΔLNO ≅ ΔLMN [tex]{}[/tex]    Side Side Side (SSS) congruency postulate

∠NLM ≅ ∠LNO [tex]{}[/tex]    Cong. Parts of Cong. Triangles are Cong. CPCTC

∠NLM and ∠LNO are alternate interior angles

[tex]\mathbf{\overline {LM} \parallel \overline {ON}}[/tex]   [tex]{}[/tex]    Alternate interior angles between parallel lines are congruent

∠MNL ≅ ∠NLO [tex]{}[/tex]    CPCTC

∠MNL and ∠NLO are alternate interior angles

[tex]\mathbf{\overline {LO} \parallel \overline {MN}}[/tex]  [tex]{}[/tex]     Alternate interior angles formed between parallel lines are congruent

LMNO is a parallelogram because the opposite sides, [tex]\overline {LM} \parallel \overline {ON}[/tex] and [tex]\overline {LO} \parallel \overline {MN}[/tex] are parallel

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LMNO is a plane shape which has the properties of a parallelogram. The the explanations below, it has been proven to be a parallelogram.

A parallelogram is a quadrilateral which has some unique properties. These properties can be used to prove if a given shape is a parallelogram or not.

To prove that LMNO as given in the question is a parallelogram:

Thus,

/LM/ ≅ /ON/ (opposite side property)

/LO/ ≅ /MN/ (opposite side property)

        i.e LM || ON and LO || MN

        i.e <MLO + <LMN = [tex]180^{o}[/tex]

also    

<LON + <MNO = [tex]180^{o}[/tex]

LN ⊥ MO

     <LMN ≅ <LON  and <MLO ≅ <MNO

Therefore the given quadrilateral LMNO has all the properties described above, then it is a parallelogram.

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Answer:

Given equation:

To plot the line first find two points:

Plot (0, -3) and (5, -1) and connect the points with a line. This will be the required line.

Use Pythagorean to find the missing leg:

Answer:

y = 3

Step-by-step explanation:

*As seen in the photo, the fact that EF = GF and the bisector being there makes EH = HG.

5y + 10 = 28 - y

*Add y to both sides.

6y + 10 = 28

*Subtract 10 from both sides.

6y = 18

*Divide both sides by 6.

y = 3

In triangle EFG, with angle bisector FH and equal lengths for EF and GF, the value of y is 3.

Use the concept of a triangle defined as:

A triangle is a 3-sided polygon, which has three vertices and three angles which has a sum of 180 degrees.

Given that,

Angle EFG has an angle bisector FH.

EF = GF (the lengths of the corresponding sides of triangle EFG are equal).

EH = 5y + 10 (length of segment EH).

HG = 28 - y (length of segment HG).

To find the value of y,

Start by applying the angle bisector theorem in triangle EFG.

According to the theorem,

The ratio of the lengths of the segments formed by the angle bisector to the corresponding sides should be equal.

Since EF = GF,

Set up the following equation:

EH / HG = EF / FG

Substituting the given values, we have:

[tex]\dfrac{(5y + 10)}{ (28 - y)} = \dfrac{1} { 1}[/tex]

Cross-multiplying, we get:

5y + 10 = 28 - y

Combining like terms, we have:

6y = 18

Dividing both sides by 6, we find:

y = 3

Therefore, the value of y is 3.

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Answer:

[tex]c=-2[/tex]

Step-by-step explanation:

We are given the function:

[tex]f(x) = -2c + cx - x^2[/tex]

And that:

[tex]\displaystyle f^{-1} (5) = -1[/tex]

And we want to determine the value of c.

Recall that by definition of inverse functions:

[tex]\displaystyle \text{If } f(a) = b, \text{ then } f^{-1}(b) = a[/tex]

So, since f⁻¹(5) = -1, then f(-1) = 5.

Substitute:

[tex]f(-1) = 5 = -2c + c(-1) - (-1)^2[/tex]

Simplify:

[tex]5 = -2c - c - (1)[/tex]

Combine like terms:

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

And divide. Hence:

[tex]c = -2[/tex]

In conclusion, the value of c is -2.

Answer:

Step-by-step explanation:

Circle equation:

Substitute the values to get the equation:

Answer:

(x+2.4)^2 + (y+4.8)^2 =45

Step-by-step explanation:

Answer:

(x+2)^2 + (y-15)^-1 = 64

Step-by-step explanation:

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The only point that is a solution to the system of inequalities is (C) (2, 6).

To solve this problem, substitute each of the points into the inequalities.

For point (A), y < (4)² + 6 and y > (4)² − 4.

This simplifies to y < 22 and y > 12.

However, 2 is not greater than 12, so (A) is not a solution.

For point (B), y < (0)² + 6 and y > (0)² − 4.

This simplifies to y < 6 and y > −4.

However, 8 is not less than 6, so (B) is not a solution.

For point (C), y < (2)² + 6 and y > (2)² − 4.

This simplifies to y < 10 and y > 0.

6 is less than 10 and greater than 0, so (C) is a solution.

For point (D), y < (−2)² + 6 and y > (−2)² − 4.

This simplifies to y<2 and y>−8.

-4 is not greater than -8, so (D) is not a solution.

Therefore, the only point that is a solution to the system of inequalities is (C).

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to solve for y, we must multiply by t on both side to get ride of the t in the denominator. By doing this, we will get:

[tex] {t}^{3} [/tex]

on the right side.

Subtracting 3, and we have successfully isolated y.

It would be impossible to get a quantitative value for y if we don't know the value of t.

respuesta listsdafddfssdffd

Answer:

the function is a y-intercept

Step-by-step explanation:

y=mx+b is the slope intercept formula, m being the slope and b being the y-intercept, meaning y=b, so b=-9/4.

Because it says y=-9/4, we can assume that m, or the slope equals 0.

so when you plug it in, you get

y=0x-9/4

y=-9/4

Answer:

x = 8cm

Step-by-step explanation:

17² = 15² + x²

289 = 225 + x²

- x² = 225 - 289

x² = 64

x = √64

x = 8

[tex]{ \boxed{ \sf \color{blue}{answer}}}[/tex]

x = 8 cm

Step-by-step explanation:

17² = 15² + x²

289 = 225 + x²

x² = 225 - 289

x² = 64

x = √64

x = 8

Teorema pythagoras

Answer:

A. d=4p+200

Step-by-step explanation:

The reason why this is, is because d is the total amount, equaling to 4 times p due to the fact that it is $4 for each person that attends.

Answer:

4π inches

Step-by-step explanation:

Formula for finding length of arc AB = θ/360*2πr

Where,

Central angle (θ) = 120°

Radius (r) = 6 inches

Length of arc AB = 120/360*2*π*6

Length for arc AB = ⅓*12π

= 4π inches

Answer:

x = complement

x + 5.2 = first angle (5.2 more than complement)

measure of two complementary angles add up to 90.

x + x+ 5.2 = 90

2x = 84.8

x = 42.4

other angle = x + 5.2 = 42.4 + 5.2 = 47.6

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Step-by-step explanation:

Hope it will help you .

.

.

.

Hey there!

40 / 1 + 3 ‐ (3 * 7) + 7 - 5

= 40 + 3 - (3 * 7) + 7 - 5

= 43 + 7 - 5 - 3 * 7

= 50 - 5 - 3 * 7

= 45 - 3 * 7

= 45 - 21

= 24

Answer: 24

Good luck on your assignment and enjoy your day!

~Amphitrite1040:)

the speed of the boat is 5.418 km/h and the speed of the stream is 2.167 km/h

Let's define the variables:

B = speed of the boat in still water.

S = speed of the stream.

When the boat travels downstream, the total speed of the boat will be equal to the sum of the speed of the stream and the speed of the boat in still water:

speed = B + S

When the boat goes downstream, the speed will be:

speed = B - S

Now, also remember the relation:

speed*time = distance.

The given information is:

In 8 hours the boat can:

go 12 km upstream and 40km downstream.

We now need to define another variable, T, as the time that the boat travels upstream.

Then we can write this as:

12km = (B - S)*T

If the boat travels T hours upstream, and travels for a total of 8 hours, then the amount of time that travels downstream is 8h - T, then we can write:

40km = (S + B)*(8h - T)

Similarly, when we have:

"it can go 16 km upstream and 32 km downstream in the same time."

we can define a new variable T', and write:

16km = (B - S)*T'

32km = (S + B)*(8h - T')

Then we have a system of 4 equations:

16km = (B - S)*T'

32km = (S + B)*(8h - T')

40km = (S + B)*(8h - T)

12km = (B - S)*T

And we need to solve this for S and B.

To do it, we need to isolate one of the variables in one of the equations.

Let's isolate T in the last equation:

T = 12km/(B - S)

now we can replace that in the third equation to get:

40km = (S + B)*(8h - 12km/(B - S))

So now we have 3 equations:

16km = (B - S)*T'

32km = (S + B)*(8h - T')

40km = (S + B)*(8h - 12km/(S - B))

Now we need to do the same thing, this time let's isolate T' in the first equation and replace it in the second one:

T' = 16km/(B - S)

Replacing it in the second equation we get:

32km = (S + B)*(8h - T')

32km = (S + B)*(8h - 16km/(B - S))

So now we have two equations:

40km = (S + B)*(8h - 12km/(B - S))

32km = (S + B)*(8h - 16km/(B - S))

Let's simplify these:

40km = 8h*(S + B) - 12km*(S + B)/(B - S)

32km = 8h*(S + B) - 16km*(S+ B)/(B - S)

Now we can multiply both equations by (B - S) to get:

40km*(S - B) = 8h*(S + B)*(B - S) - 12km*(S + B)

32km*(S - B) = 8h*(S + B)*(B - S) - 16km*(S+ B)

Let's keep simplifying this:

40km*(B - S) + 12km*(S + B) =  8h*(S + B)*(B - S)

32km*(B - S)  + 16km*(S+ B) = 8h*(S + B)*(B - S)

Now we get:

52km*B - 28km*S = 8h*(S^2 + B^2)

48km*B - 16km*S = 8h*(S^2 + B^2)

Notice that the right side of these equations is the same thing, then we can write:

52km*B - 28km*S = 48km*B - 16km*S

(52km - 48km)*B = (28km - 18km)*S

4km*B  =  10km*S

B = (10/4)*S

B = (5/2)*S

Now we can replace this in one of our two equations, let's use the first one:

48km*B - 16km*S = 8h*(S^2 + B^2)

48km*(5/2)*S - 16km*S = 8h*( S^2 + ( (5/2)*S)^2)

Now we can solve this for S

104km*S = 8h*( S^2 + 25/4*S^2)

104km*S = 8h*(29/4*S^2) = 48h*S^2

104km*S = 48h*S^2

dividing at both sides by S we get:

104km = 48h*S

104km/48h = S = 2.167 km/h

And using B = (5/2)*S

We can find the speed of the boat:

B = (5/2)*2.167 km/h = 5.418 km/h

Then:

the speed of the boat is 5.418 km/h and the speed of the stream is 2.167 km/h

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Seashells glued on 1 photo frame = k

So, seashells glued on 5 photo frame

= Seashells glued on 1 photo frame × 5

= k × 5

= 5k

So, Erynn used 5k seashells.

Answer:

the new volume Will be 4 times the old volume

Step-by-step explanation:

I have put the photo 3

Answer:

4

Step-by-step explanation:

ln(e)=1 since ln= log_e

log(x^y)=y*log(x) so ln(e^4)=4*ln(4)=4*1=4

Answer:

D

Step-by-step explanation:

ln(e^4)

The 4 comes down in front of the natural log (ln). You get 4 ln(e)

The natural log of e (ln e )  is 1.

So the answer is 4 * 1 = 4

That may not be clear to you. We could try it another way.

e = 2.718281828

ln (2.718281828) = 1

ln(2.718281828)^4 = 4.  The calculator gags until you get it right.  You need the y^x key (or the x^y key of ^) It does come back with 4.

You have to monkey around with the calculator to get this to happen.  

Answer: -78 / 35

Step-by-step explanation:

Given

 2/5 + 3/7 + (-6/5) + (-13/7)

Combine terms that have the same denominator

= 2/5 + (-6/5) + 3/7 + (-13/7)

= -4/5 - 10/7

Convert denominator by Least Common Multiple (LCM)

= (-4 × 7) / (5 × 7) - (10 × 5) / (7 × 5)

= -28/35 - 50/35

= -78 / 35

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Answer:

[tex] \frac{2}{5} + \frac{3}{7} + ( - \frac{6}{5} ) + ( - \frac{13}{7} ) \\ \frac{2}{5} + \frac{3}{7} - \frac{6}{5} - \frac{13}{7} \\ ( \frac{2}{5} - \frac{6}{5} ) + ( \frac{3}{7} - \frac{13}{7} ) \\ \frac{2 - 6}{5} + \frac{3 - 13}{7} \\ \frac{ - 4}{5} + \frac{ - 10}{7} \\ \frac{ - 4}{5} - \frac{10}{7} \\ \frac{ - 28 - 50}{35} \\ \frac{ - 78}{35} \\ - \frac{78}{35} [/tex]

Answer:

Q1 = 87 ;

Q2 = 102 ;

Q3 = 115 ;

IQR = 28

OUTLIER = None

Step-by-step explanation:

Given the data:

111, 68, 93, 88, 74, 152, 119, 87, 88, 105, 84, 102, 151, 115, 112

Ordered data : 68, 74, 84, 87, 88, 88, 93, 102, 105, 111, 112, 115, 119, 151, 152

Sample size, n = 15

The first quartile, Q1 = 1/4(n+1)th term

Q1 = 1/4(15+1)th term

Q1 = 1/4(16) = 4th term

Q1 = 87

The Median , Q2 = 1/2(n+1)th term

Q2 = 1/2(15+1)th term

Q2 = 1/2(16) = 8th term

Q2 = 102

The third quartile, Q3 = 3/4(n+1)th term

Q3 = 1/4(15+1)th term

Q3 = 3/4(16) = 12th term

Q3 = 115

The interquartile range, IQR = Q3 - Q1

IQR = (115 - 87) = 28

Q1 = 87 ;

Q2 = 102 ;

Q3 = 115 ;

IQR = 28

OUTLIER = None

[tex]\boxed{\sf 60x^6y^4}[/tex]

Answer:

1980

Step-by-step explanation:

13 sides

Number of sides n=13

The sum of interior angles of polygon =(2n−4)×90o

=(2×13−4)×90o

=(26−4)×90o

=22×90o

We get,

=1980

Answer:

A is false

Step-by-step explanation:

notice how in both diagrams, there are 4 triangles with lengths a, b, c, but in each diagram the triangles are just rotated and positioned differently. if there are the same amount of triangles in both diagrams, and the have equal lengths then the area of the remaining figure, should both match up.

basically the area of the big square in Step 2, is equal to both of the shaded squares, combined area, in Step 1.

Answer:

[tex] {6}^{3} = 216[/tex]