I could use some help answering this math problem!!

I Could Use Some Help Answering This Math Problem!!

Answers

Answer 1

Answer:

b

Step-by-step explanation:

2x is the same as 2 times x and if x represents the coaches uniform and the uniform is 2 times the cost of the players then the answer is b 2x


Related Questions

Use Pythagoras to find the height and hence, the area of the triangle
below. Give height to 1 decimal place and area to the nearest whole. Write
answer in format: h= A= *
20 mm
Val

Answers

Answer:

h=17.3 A=173

Step-by-step explanation:

Calculator

Answer:

height = 17.3 mm

area = 173 mm²

Step-by-step explanation:

all three sides are of the same length (20 mm).

so, the height actually splits the baseline in half

(2 × 10 mm) while hitting it at a 90 degree angle.

so, we use Pythagoras, where the full side opposite of this 90 degree angle is c (Hypotenuse), the height of the main triangle is one side, and half of the baseline is the other side.

c² = a² + b²

20² = 10² + height²

400 = 100 + height²

300 = height²

height = 17.3 mm

the area of the main triangle is baseline (20) times height divided by 2.

so,

At = 20×17.3/2 = 10×17.3 = 173 mm²

What is the answer of 2x5
pls help
#im really bad at math

Answers

Answer:

10

Step-by-step explanation:

ans is 10

because 2×5 is 10

Answer:

2x5= 10

Step-by-step explanation:

5, 10,15,20

five 2 times is 10

write an equation of the function in the form y=a(b)^x -c that has a y intercept of -6, asymptote of y= -2 and goes through (2,-18)

Answers

Step-by-step explanation:

Asymptote: y = 2 y-intercept: (0,8)

Step-by-step explanation:

The given function is

f(x) = 6(0.5)^{x} + 2f(x)=6(0.5)

x

+2

This function is of the form:

f(x) = a {b}^{x} + cf(x)=ab

x

+c

where y=c is the horizontal asymptote.

By comparing , we have c=2 hence the horizontal asymptote is

y = 2y=2

To find the y-intercept, we put x=0 into the function to get:

f(0) = 6(0.5)^{0} + 2 = 6 + 2 = 8f(0)=6(0.5)

0

+2=6+2=8

Therefore the y-intercept is (0,8).

What is the distance from the plane

Answers

Answer:

here's the answer to your question about

Chukwu, Abdul and kunle are to share $142, such that Abdul gets $11.00 less than Chukwu and $7.00 more than Kunle. How much is Chukwu's share?

Answers

Answer:

Chukwu's share is of $57.

Step-by-step explanation:

This question is solved by a system of equations.

I am going to say that:

Chukwu's share is x.

Abdul's share is y.

Kunle's share is z.

Chukwu, Abdul and kunle are to share $142

This means that [tex]x + y + z = 142[/tex]

Abdul gets $11.00 less than Chukwu

This means that:

[tex]y = x - 11[/tex]

$7.00 more than Kunle.

This means that:

[tex]y = z + 7[/tex]

So

[tex]z = y - 7[/tex]

[tex]z = x - 11 - 7[/tex]

[tex]z = x - 18[/tex]

How much is Chukwu's share?

We want to find x, so:

[tex]x + y + z = 142[/tex]

[tex]x + x - 11 + x - 18 = 142[/tex]

[tex]3x - 29 = 142[/tex]

[tex]3x = 171[/tex]

[tex]x = \frac{171}{3}[/tex]

[tex]x = 57[/tex]

Chukwu's share is of $57.

find the missing side lengths​

Answers

Answer:

b = 2√3

a = 4

Step-by-step explanation:

Here taking 60 as reference angle

so

tan60 = p/b

[tex]\sqrt{3}[/tex]    = p/2

so p = 2√3

so

b = 2√3

so

[tex]h^2 = p^2 + b^2\\h^2 = (2\sqrt{3})^2 + 2^2\\h^2 = 12 + 4\\h^2 = 16\\h = \sqrt{16} \\h = 4[/tex]

a = 4

please help! i have no idea

Answers

Answer:

θ = 51.7

SOH-CAH-TOA

CAH: COS = ADJ/HYP   cos(θ) = 19.4/29.3

θ = 48.5

93.2 + 48.5 + x = 180

x = 38.3

90 + 38.3 + θ = 180

θ = 51.7

Step-by-step explanation:

A hot air balloon is released into the air. During its straight ascent, the angle of elevation was 15° and, 3 minutes later, the angle of elevation increased 20°. How fast is the balloon traveling, in km/h, if the angle measurements were taken 300m away from the launch site?

Answers

Answer:

The speed of the balloon is 0.16 m/s.

Step-by-step explanation:

CD = 300 m

Let AD = x

AB = y

time, t = 3 min

Triangle, ADC

[tex]tan 15 = \frac{AD}{BC}\\\\0.27 \times 300 = x \\\\x = 80.4 m[/tex]

Triangle, BCD

[tex]tan 20 = \frac{BD}{BC}\\\\0.36 \times 300 = x + y \\\\x + y = 109.2 m[/tex]

So, y = 109.2 - 80.4 = 28.8 m

Speed = 28.8/180 = 0.16 m/s

A to B is 420 mile away, A travels 40 mph and B travels 80 mph. they toward to each other. how many hour will they meet ?

Answers

Answer:

Step-by-step explanation:

I hope these are people in vehicles travelling toward each other.

80 mh * t + 40 mph * t = 420 miles

The two vehicles are going to travel a total of 420 miles. B will cover a lot more ground than A, but together they will make 420 miles

80t + 40t = 420

120t = 420

t = 420/120

t = 3.5

So after driving 3.5 hours, the two cars will meet.

The diameter of the circle above is 18 cm. What is the circumference of the circle? (Use = 3.14.)

Answers

Answer:

56.57

Step-by-step explanation:

See attached photo for steps

EDIT: I forgot to add 3.14, I am so sorry

Answer:

56.52

Step-by-step explanation:

Last month, Nate spent 12 % of his paycheck on
car repairs and 25 % of the remainder on food.
He gave $ 1,320 of the remaining money to his
parents and then bought a computer on sale. If
the usual price of the computer was $ 825 and
the discount was 20 %, how much money did
Nate have in the beginning?

Answers

I think this is B…..

Nate had $3000 at the beginning.

What is the percentage?

The percentage is defined as a ratio expressed as a fraction of 100.

Let the original amount of money is P

Nate spent 12 % of his paycheck on car repairs

⇒ 12% of P = 12P/100

And he spent 25 % of the remainder on food.

⇒  0.25(88/100)P

He gave $ 1,320 of the remaining money to his parents

⇒ 1320  

If the usual price of the computer was $ 825 and the discount was 20 %

⇒ 825 - 0.20(825) = 660

P = 12P/100 + 0.25(88/100)P + 1320 + 660

P - 34P/100 = 1980

100P - 66P = 198000

P = 3000

Hence, Nate had $3000 at the beginning.

Learn more about the percentages here:

brainly.com/question/24159063

#SPJ2

Mike and Ken shared some stamps. \frac{1}{5} 5 1 ​ of Ken's stamps were \frac{1}{3} 3 1 ​ of Mike's stamps. If Mike gave Ken 24 stamps, Ken would have thrice as many stamps as Mike. Find the number of stamps each of them had in the beginning.

Answers

Answer:

Mike had 72 stamps

Ken had 120 stamps

Step-by-step explanation:

Given

[tex]M \to Mike[/tex]

[tex]K \to Ken[/tex]

[tex]\frac{1}{5} * K = \frac{1}{3} * M[/tex]

[tex]K + 24 = 3 * ( M - 24)[/tex]

Required

Find K and M

Make K the subject in: [tex]K + 24 = 3 * ( M - 24)[/tex]

[tex]K = 3 * ( M - 24) - 24[/tex]

Substitute [tex]K = 3 * ( M - 24) - 24[/tex] in [tex]\frac{1}{5} * K = \frac{1}{3} * M[/tex]

[tex]\frac{1}{5} * [3 * ( M - 24) - 24] = \frac{1}{3} * M[/tex]

Open brackets

[tex]\frac{1}{5} * [3M - 72 - 24] = \frac{1}{3} * M[/tex]

[tex]\frac{1}{5} * [3M -96] = \frac{1}{3} * M[/tex]

Multiply both sides by 15

[tex]3* [3M -96] = 5 * M[/tex]

[tex]9M -288 = 5M[/tex]

Collect like terms

[tex]9M -5M= 288[/tex]

[tex]4M= 288[/tex]

Divide both sides by 4

[tex]M= 72[/tex]

Substitute [tex]M= 72[/tex] in [tex]K = 3 * ( M - 24) - 24[/tex]

[tex]K = 3 * (72 - 24) - 24[/tex]

[tex]K = 3 * 48 - 24[/tex]

[tex]K = 120[/tex]

A student states that the translation of triangle ABC is A’B’C’. What measurements or properties of lines AA', BB', and CC' do you need to confirm that it is a translation? :D

Answers

If you can show that the following two items are true

AA' = BB' = CC'AA' || BB' || CC'

then you have shown that triangle A'B'C' is a translated or shifted copy of triangle ABC. In other words, they are the same triangle.

On Sunday Paul earns 20 cents for every newspaper he delivers if Paul receives $6.40 then how many newspapers did he deliver on Sunday

Answers

Answer:

32 newspapers

Step-by-step explanation:

6.4 / 0.2 = 32

Answer:

32 papers

Step-by-step explanation:

Take the total amount and divide by .20

6.40 / .20

Multiply the top and bottom by 10

64/2

32

a local community college has 860 students. Of these 860 students, 220 ride bicycles. Write the number of bike riders as a fraction of the number of students at the college in simplest form

Answers

Answer:

11/43

Step-by-step explanation:

A local community college has 860 students

Out of this 860 students, 220students ride bikes

Therefore the fraction of bike riders to the number of students can be calculated as follows

= 220/860

= 11/43

Help me! thank you so much

Answers

Answer:

Step-by-step explanation:

[tex]\frac{sinxcos^3x-cos xsin^3x}{cos^42x-sin^42x} \\=\frac{sin x cos x(cos^2x-sin ^2 x)}{(cis^2 2x+sin^2 2x)(cos^2 2x-sin ^22x)} \\=\frac{2sin x cos x cos 2x}{2(1)(cos 4x)} \\=\frac{sin 2x cos 2x}{2 cos 4x} \\=\frac{2 sin 2x cos 2x}{4 cos 4x} \\=\frac{sin 4x}{4 cos 4x} \\=\frac{1}{4} tan 4x[/tex]

Find the gcf for…………….

Answers

The answer would be 6uvw^2


What is 2f+4f + 2-3 evaluated at f= 3?
|

Answers

Answer:

17

Step-by-step explanation:

2f + 4f + 2 - 3 when f = 3

1) first we multiply 3 wherever f is

2 x 3 + 4 x 3 + 2 - 3 (solve)

6 + 12 + 2 - 3

18 + 2 - 3

20 - 3

17

Answer:

6+12+2-3=17

Step-by-step explanation:

Graph the line that passes through (5, 5), and is perpendicular to a line whose slope is –2.

Answers

Answer:

y = 1/2x + 5/2

Step-by-step explanation:

y = 1/2x + b

5 = 1/2(5) + b

5 = 5/2 + b

5/2 = b

Guys please help me solve this its so stressful

Answers

Answer:

Given:- [tex]y=-5x^3+10x+20[/tex]

A parabola attain maximum height at the vertex: formula to find x-coordinate of vertex is, [tex]x=-\frac{b}{2a}[/tex]

plug in a=-5 and b=10

so, [tex]x=-\frac{10}{2(-5)} =-\frac{10}{-10} =1[/tex]

Now plug in x=1 to get maximum height,

[tex]y=-5(-1)^2+10(1)+20[/tex]

[tex]=-5+10+20[/tex]

[tex]=+25[/tex]

so, maximum heigh reached by the rocket was 25 yards

Rocket will hit the ground when y=0

[tex]0=-5x^2+10x+20[/tex]

[tex]0=-5(x^2-2x-4)[/tex]

[tex]\frac{0}{-5} =\frac{-5}{-5} (x^2-2x-4)[/tex]

So, [tex]x^2-2x-4=0[/tex]

quadratic formula is,

[tex]x=\frac{-b+-\sqrt{b^2-4ac} }{2a}[/tex]

plug in a=1, b=-2,and c=-4

[tex]x=-\frac{(-2)=-\sqrt{(-2)^2-4(1)(-4)} }{2(1)}[/tex]

[tex]=\frac{2=-\sqrt{4+16} }{2}[/tex]

[tex]=\frac{2+-\sqrt{20} }{2}[/tex]

[tex]=\frac{2+-2\sqrt{5} }{2}[/tex]

[tex]=1+-\sqrt{5}[/tex]

[tex]=1-\sqrt{5} ,1+\sqrt{5}[/tex]

[tex]=1-5.236,1+2.236[/tex]

[tex]=-1.236,3.236[/tex]

So, it takes 3.2 seconds to hit the ground.

OAmalOHopeO

Hay 31 estudiantes en una clase. Ocho (8) estudiantes miden más de 6 pies de altura. Diecisiete (17) estudiantes miden 5-6 "a 6'-0" de altura. Seis (6) estudiantes miden menos de 5-2". ¿Cuál es la razón entre el número de estudiantes de menos de 5'-2" y el número de estudiantes de más de 6 'de altura?​

Answers

Answer:

4/3

Step-by-step explanation:

On the first day of travel, a driver was going at a speed of 40 mph. The next day, he increased the speed to 60 mph. If he drove 2 more hours on the first day and traveled 20 more miles, find the total distance traveled in the two days.

Answers

Answer:

380

Step-by-step explanation:

This is a bit nasty. It depends on how you read the 20 miles more and what you do with it. The best and most careful way to do it is do it a long way setting up the two equations carefully.

Second day

Let the time travelled = t

Let the speed travelled = 60 mph

d2 = 60*t

First Day

40*(t + 2) = d1

but d1 = d2 + 20 because he travelled 20 miles further on d1

40 * (t + 2) = d2 + 20

d2 however = 60*t

40*(t+2 ) = 60*t + 20          Remove the brackets

40t + 80 = 60t + 20           Subtract 20 from both sides

40t + 60 = 60t                   Subtract 40t from both sides

60 = 20*t                           Divide by 20

t = 60/20

t = 3 hours.

Day 2 = 60 + t = 180

Day 1 = 40*5  = 200

Total distance = 380

Where did that 20 miles go? It was just an observation about the difference in distance travelled between the 2 days.

The total distance the driver traveled in the two days is 260 miles

From the question, on the first day, the driver was going as a speed of 40 mph.

Let s be speed

∴ [tex]s_{1}= 40mph[/tex]

On the second day, he increased the speed to 60 mph

∴ [tex]s_{2}= 60mph[/tex]

From the statement- If he drove 2 more hours on the first day

Let time be t

Then

[tex]t_{1}= t_{2} + 2[/tex] hrs

and traveled 20 more miles

Let d be distance  

Then,

[tex]d_{1}= d_{2} + 20[/tex] miles

From the formula

Distance = Speed × Time

Then,

[tex]d = s \times t[/tex]

∴ [tex]d_{1} = s_{1} \times t_{1}[/tex]

From above,

[tex]d_{1}= d_{2}+20[/tex] miles

[tex]s_{1}= 40mph[/tex]

[tex]t_{1}= t_{2} + 2[/tex] hrs

Putting these into

[tex]d_{1} = s_{1} \times t_{1}[/tex]

[tex]d_{2} + 20 = 40\times (t_{2}+2)[/tex] ...... (1)

But,

[tex]Time = \frac{Distance}{Speed}[/tex]

∴ [tex]t_{2}= \frac{d_{2} }{s_{2} }[/tex]

From above, [tex]s_{2}= 60mph[/tex]

∴ [tex]t_{2}= \frac{d_{2} }{60}[/tex]

Put this into equation (1)

[tex]d_{2} + 20 = 40\times (t_{2}+2)[/tex]

[tex]d_{2} + 20 = 40\times (\frac{d_{2}}{60} +2)[/tex]

[tex]d_{2} + 20 = \frac{2}{3}d_{2} +80\\d_{2} = \frac{2}{3}d_{2} +80-20\\d_{2} = \frac{2}{3}d_{2} +60[/tex]

Multiply through by 3

[tex]3\times d_{2} = 3\times \frac{2}{3}d_{2} +3 \times 60\\3d_{2} = 2d_{2} + 120\\3d_{2} -2d_{2} = 120[/tex]

∴ [tex]d_{2} = 120[/tex] miles

∴The distance traveled on the second day is 120 miles

For the distance traveled on the first day,

Substitute [tex]d_{2}[/tex] into the equation

[tex]d_{1}= d_{2}+20[/tex] miles

∴ [tex]d_{1}= 120+20[/tex]

[tex]d_{1}= 140[/tex] miles

∴ The distance traveled on the first day is 140 miles

The total distance traveled in the two days = [tex]d_{1} + d_{2}[/tex]

The total distance traveled in the two days = 120 miles + 140 miles

The total distance traveled in the two days = 260 miles

Hence, the total distance the driver traveled in the two days is 260 miles

Learn more here: https://brainly.com/question/23531710

58×62 without actual multiplication​

Answers

Answer:

3596

Step-by-step explanation:

Use long multiplication to evaluate.

Answer:

3596

Step-by-step explanation:

(58*10) = 580

580*6 = 3480

58 * 2 = 116

3480 + 116 = 3596

4x+3=x+9

What is the number?

X=?

Answers

Answer:x=3

Step-by-step explanation:

4x+3=x+9

4x-x=(-3)+9

3x=6

x=6:3

x=2

Which answer shows 0.05 written in scientific notation?​

Answers

Answer:

5x10^-2 option B

Step-by-step explanation:

Answer:

5x10^-2

the second one.

One of the legs of a right triangle measures 9 cm and the other leg measures 2 cm. Find the measure of the hypotenuse. If necessary, round to the nearest tenth.

Answers

Answer:

Approximately 9.2

Step-by-step explanation:

For right triangle, if you know 2 sides, you can find the third using pythagorean theorem, a^2 + b^2 = c^2. 'a' and 'b' are the lengths of the legs, while 'c' is the hypotenuse. You can plug in what you know into this formula:

9^2 + 2^2 = c^2

81+4 = c^2

c = √85, or approximately 9.2

giving the funtions f(x)=2x-5 and g(x)=3x2=2 perform the indicated operation

Answers

Answer:3165

Step-by-step explanation:

Indicate in standard form the equation of the line through the given points, writing the answer in the equation box below.

K(6, 4), L(-6, 4)

Answers

9514 1404 393

Answer:

  y = 4

Step-by-step explanation:

The given points are on the horizontal line ...

  y = 4

This is the standard-form equation of that line.


A=E
BCE=90
BF=AC
AC=EC

Answers

Answer:  AC = EC  (Choice D)

==========================================================

Explanation:

HL stands for Hypotenuse Leg. This rule works only for right triangles.

So we must have two hypotenuses that are congruent, and we must have two legs that are congruent

The diagram shows that AB = DE are the congruent hypotenuses. So we have "H" taken care of. We just need the "L" portion.

We can see that only choice D works, which is AC = EC. If we knew that AC and EC were the congruent legs, then we have enough info to use HL.

Write an expression for the area of the square below.
4x + 2
A. 8x2 + 16x + 4
B. 16x2 + 16x + 4
C. 8x + 4
D. 16x2 + 6x + 4
Help

Answers

Area = Side^2

Area =( 4x + 2 )^2

Area = (4x)^2 + 2(4x)(2) + (2)^2

Area = 16x^2 + 16x + 4

Thus the correct answer is option B .

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