A cyclist traveled kilometers per hour faster than an in-line skater. In the time it took the cyclist to travel kilometers, the skater had gone kilometers. Find the speed of the skater.

Answer:

4 units

Step-by-step explanation:

Hello!

d₁ = distance covered in Part 1

d₂ = distance covered in Part 2

d₃ = distance covered in Part 3

Distance in Part 2 :

⇒ d₃ = 15 x 300 = 4500 m

⇒ d₂ = 2/5d₃

⇒ d₂ = 2/5 x 4500

⇒ d₂ = 1800 meters

Speed in Part 2 :

⇒ v₂ = 1800 m / 12 min

⇒ v₂ = 150 meters per minute

Answer:

Step-by-step explanation:

Argument

I think the answer is A = N or D. It that is true, then M = C because all three angles are equal which gives similarity. But there is a side involved. OM = CB. Therefore the two triangles are congruent by ASA

Answer

ASA

There are 50-50 chances that you will get a head or a tail

Answer:

29u

Step-by-step explanation:

a = (V-U)/t = (1.2u-u)/10 = 0.02u

D1: V₀t+0.5at² = (u)(10)+(0.5)(0.02u)+(10)(10) = 11u

D2: (V)(t) = (1.2u)(15) = 18u

D = D1 + D2 = 29u

For first half

We need acceleration first

So distance traveled (Apply second equation of kinematics

For second half

No acceleration

Distance=Speed×Time

Total distance

Answer:

b. The graph of g(x) is a translation 10 units right and 8 units up from the graph of f(x).

Step-by-step explanation:

See attached image.

The quantity that is present after 2000 years will be A. 7.851 grams.

From the information given, initially at t = 0, the quantity is 10 grams.

After 2000 years, the appropriate quantity will be:

Q(2000) = 10 (1/2) × (2000/5730)

Q(2000) = 7.851 grams.

In conclusion, the correct option is A.

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[tex] \qquad \qquad \bf \huge\star \: \: \large{ \underline{Answer} } \huge \: \: \star[/tex]

[ b ] is the Correct one

[tex]\textsf{ \underline{\underline{Steps to solve the problem} }:}[/tex]

[tex] \qquad❖ \: \sf \:4 \: log_{3}(y) - 6 \: log_{3}(x) + 7 \: log_{3}(z) [/tex]

[tex] \qquad❖ \: \sf \: \: log_{3}(y {}^{4} ) - \: log_{3}(x {}^{6} ) +\: log_{3}(z {}^{7} ) [/tex]

[tex] \qquad❖ \: \sf \: \: log_{3}(y {}^{4} ) + \: log_{3}(z{}^{7} ) - \: log_{3}(x {}^{6} ) [/tex]

([tex] \sf a\: log (b) = log {b}^{a}[/tex])

[tex] \qquad❖ \: \sf \: log_{3}( {y}^{4} {z}^{7} ) - log_{3}( {x}^{6} {}^{} ) [/tex]

([tex] \sf{log (a) + log (b) = log (ab)} [/tex])

[tex] \qquad❖ \: \sf \: log_{3} \bigg( \dfrac{y {}^{4} {z}^{7} }{ {x}^{6} } \bigg ) [/tex]

([tex] \sf{log (a) - log (b) = log (a/b)} [/tex])

[tex] \qquad \large \sf {Conclusion} : [/tex]

[ b ] is the Correct choice

The correct match for the graph of the total cost function is the option B graph.

The function is a type of relation, or rule, that maps one input to specific single output.

Given;

y = 2x is function

so, our graph should be of function y = 2x

Here, x is the number of pounds of rice scooped and purchased from a bulk bin at the grocery store and y is the total cost of the rice.

it means x is a number like 0, 1, 2, 3

so, the graph of the total cost function is a discrete function in nature.

Hence, graph-B is correct answer.

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The system of linear equations that represents this scenario is: q + d = 16

0.25q + 0.10d = 3.10

To represent the scenario described, we can set up a system of linear equations using the given variables:

Let q represent the number of quarters.

Let d represent the number of dimes.

We are given two conditions:

1) The total number of coins is 16:

q + d = 16

2) The total value of the coins is $3.10:

0.25q + 0.10d = 3.10

Therefore, the system of linear equations that represents this scenario is:

q + d = 16

0.25q + 0.10d = 3.10

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Answer:

-70x, 26y^3, -3z

Step-by-step explanation:

The way to solve this is by combining like terms. The terms we have given to us include x, y^3, and z. As a result, we will group terms based on these variables.

1. Find all the terms with each variable.

* -35x, -35x

* 23y^3, 3y^3

* -3z

2. Combine like terms.

* -35x + (-35x) = -35x - 35x = -70x

* 23y^3 + 3y^3 = 26y^3

* -3z + 0 = -3z

Answer:

Hello! The answer is: [tex]-70x - 3z + 26y^3[/tex]

Step-by-step explanation:

First, rearrange to combine like terms:

[tex]-35x - 35x - 3z + 23y^3 + 3y^3[/tex]

Now, combine like terms:

[tex]-70x - 3z + 26y^3[/tex]

Answer:

angle angle theorem.

Step-by-step explanation:

If you want to determine whether a few angles are similar you can use formulas such as SAS AAA ASA etc.

this means

side angle side

angle angle anlge

angle side angle

The multiplicative rate of change of the function will be 0.5. Then the correct option is C.

Let b is the base and x is the power of the exponent function and a is the leading coefficient. The exponent is given as

y = a(b)ˣ

The missing table is attached below.

For x = 2, y will be 0.125

0.125 = ab²  ...1

For x = 3, y will be 0.0625

0.0625 = ab³   ...2

Divide equation 2 by equation 1, we have

b = 0.0625 / 0.125

b = 0.5

Then the correct option is C.

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Answer:165

Step-by-step explanation:

Answer:

<C = 45 degrees.

Step-by-step explanation:

This is a right triangle (indicated by the square representing <A). All angles in a triangle add up to 180 degrees

180 - 90 is 90. Knowing that there are only two angles left, you can divide 90 by 2, resulting in 45.

<C = 45.

The measure of each interior angle of a regular polygon is [tex]\frac{180(n-2)}{n}[/tex] degrees, where n is the number of sides.

Question 1

[tex]144=\frac{180(n-2)}{n}\\\\144n=180(n-2)\\\\144n=180n-360\\\\-36n=-360\\\\n=10 \text{ sides}[/tex]

Question 2

[tex]\frac{180(n-2)}{n}=135\\\\180(n-2)=135n\\\\180n-360=135n\\\\-360=-45n\\\\n=8 \text{ sides}[/tex]

Given that DE||AC, by similar triangles, we have;

A part of the question that appears missing is; side DE||AC

The completed two column proof is presented as follows;

Statement. Reason

1. DE||AC 1. Given

2. <EDB congruent to <CAB

<DEB congruent to <ACB

Which can be written as; <CAB congruent to <EDB, <ACB congruent to <DEB

2. If two parallel lines are cut by a common transversal, the corresponding angles are congruent

3. ∆ABC ~ ∆DBE 3. by AA criterion for similarity

4.

[tex] \mathbf{\frac{ AB }{ CB} } = \frac{ DB }{ EB} [/tex]

4. Corresponding sides of similar triangles are proportional

5. AB = AD + DB

CB = CE + EB 5. Segment addition postulate

6. [tex] \mathbf{\frac{ AD + DB }{ CE + EB} } = \frac{ DB }{ EB} [/tex]

6. Substitution property of equality

7. [tex] \frac{ AD + DB }{ DB} = \frac{ CE + EB }{ EB} [/tex]

7. Division property

8. [tex] \frac{ AD }{ DB} + 1 = \frac{ CE }{ EB} + 1 [/tex]

[tex] \frac{ AD }{ DB} = \frac{ CE }{ EB} [/tex]

8. Subtraction property

Therefore, the missing step in the proof is the option;

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Answer:

The answer is C = -1

2x = 10

☆...hope this helps...☆

_♡_mashi_♡_

Answer:

A. 2x = -10

E. 2x = 10

Explanation:

[tex]\sf |2x| = 10[/tex]

If |a| = b then a = b or a = -b

[tex]\sf 2x = 10 \quad or \quad 2x = -10[/tex]

divide both sides by 2

[tex]\sf x = 5 \quad or \quad x = -5[/tex]

Answer:

78.5

Step-by-step explanation:

pi(r^2)

3.14(5^2)

3.14(25)

78.5

pls kindly see the attached picture for the answer and explanation thank you

The chocolate will last for 1 day, if she gets 4 more dragons

The expression can be represented using the following ratio:

Ratio = Dragons : Days : Chocolate

For the 2 dragons she has, we have:

Ratio = 2 : 12 : 300

When she gets 4 more dragons, we have:

Ratio = 6 : days : 300

Equate both ratios

6 : days = 12 : 2

Express as fraction

days/6 = 2/12

Multiply both sides by 6

days = 6 * 2/12

Evaluate

days = 1

Hence, the chocolate will last for 1 day

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Answer:

a=223.57

c=181.43

Step-by-step explanation:

a=405-c

substitute for a in the equation :

12a+5c=3590

12(405-c)+5c=3590

4860-12c+5c=3590

-7c=3590-4860

-7c=-1270

c=1270/7=181.43

a=405-1270/7

a=1565/7=223.57

y = 2x² - 16x - 32 is the quadratic function of function passes through the points (-2,0), (4,0), and (0,-32).

y=ax²+bx+c is called as Quadratic function

y=ax²+bx+c

The three points are (-2,0), (4,0), and (0,-32)

Let,

(-2,0) in equation

0=a(-2)²+b(-2)+c

0=-2b+c...(1)

(4,0)

0=a(4)²+b(4)+c

0=16a+4b+c..(2)

(0,-32)

-32=a(0)²+b(0)+c

c=-32...(3)

Hence put c value in (1)

0=-2b-32

2b=-32

b=-16

Now put b and c value in 2

0=16(a)+4(-16)+32

0=16a-64+32

0=16a-32

a=2

Hence y = 2x² - 16x - 32 is the quadratic function of function passes through the points (-2,0), (4,0), and (0,-32).

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Current year's debt ratio is 2.50

Last year's debt ratio is 2.44

Compute Dan's Dependable Delivery debt ratio for two consecutive years?

The debt ratio is the total liabilities divided by the total assets

The debt ratio for the current year =  2.50

The debt ratio for last year =2.44(find attached excel file and formula sheet)

Debt ratio means the portion of the firm's total assets funded using borrowing(external funding), the lower the ratio, the better since having a higher debt means a higher interest would be paid, the higher the debt, the higher the chance of the firm defaulting on paying interest and repaying the principal borrowed.

The fact that the debt ratio has increased in the current year, means that the ratio has worsened because more debt has been taken.

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Answer:

3/25

Step-by-step explanation:

0.12=12/100

12/100 can be simplified 6/50 = 3/25

the quadratic equation is:

y = 0.2*x*(x - 28)

If we know that the roots are J and K, and the leading coefficient is a, then the quadratic equation is:

y = a*(x - K)*(x - J)

Here we know that the roots are at x = 0 and x =28, then:

y = a*(x - 0)*(x - 28).

We also know that the vertex is a minimum, so a is positive, and we know that the vertex is (14, -40), then we have:

-40 = a*(14)*(14 - 24) = -a*196

a = 40/196 = 0.20

Then the quadratic equation is:

y = 0.2*x*(x - 28)

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The y-intercept of the line is 0.75 or 3/4

The given parameters are:

Slope, m = 1/4

Point (x, y) = (0.2, 4/5)

A linear equation is represented as:

y = mx + b

This gives

4/5 = 1/4 * 0.2 + b

Evaluate the product

4/5 = 0.05 + b

Subtract 0.05 from both sides

b = 0.75

Hence, the y-intercept of the line is 0.75 or 3/4

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The equation of the hyperbola with foci (-2, 0), (2, 0), and vertices, (-1, 0), (1, 0) is [tex] \underline{\frac{ {x}^{2} }{ {1}^{2} } - \frac{ {y}^{2} }{ 3} } = 1[/tex]

The general form of the equation of an hyperbola is presented as follows;

[tex] \frac{ {x}^{2} }{ {a}^{2} } - \frac{ {y}^{2} }{ {b}^{2} } = 1[/tex]

Where a is obtained from the vertex written as (-a, 0), (a, 0)

The given vertex are (-1, 0), (1, 0), which gives;

a = 1

The general form of the foci of a hyperbola are; (-c, 0), (c, 0)

The given foci of the hyperbola are; (-2, 0), (2, 0)

Where;

Therefore;

By comparison, we have;

c² = 2²

Which gives;

The equation of the hyperbola is therefore;

[tex] \underline{\frac{ {x}^{2} }{ {1}^{2} } - \frac{ {y}^{2} }{ 3} } = 1[/tex]

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The diameter of the milky way is given as 4.3 x 10¹¹m while the diameter of the radius of the earth is 0.00276

The transformed system is given as 23.18 x 10⁸ in real length

Hence the diameter is

10 x 10²⁰ / 23.18 x 10⁸

= 4.3 x 10¹¹ m

Distance from the sun to the earth = 1.496 x 10¹¹m

Transformed system = 1.496 x 10¹¹m/23.18 x 10⁸

= 64.5 m

Actual radius = 6.4 x10⁶ / 23.18 x 10⁸

= 0.00276

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