The amount Cami raised during last year’s charity walk, $45.50, is StartFraction 7 over 10 EndFraction of the amount she raised this year. Which equation represents n, the number of dollars she raised this year?
45.50 (StartFraction 7 over 10 EndFraction) = n
45.50 = 0.7 n
45.50 + n = StartFraction 7 over 10 EndFraction
45.50 = 0.7 + n

Answer:

20

11

Step-by-step explanation:

10^1+10 steps:

10^1+10

Calculate 10 to the power of 1 and get 10.

10+10

Add 10 and 10 to get 20.

20

10^0+10 steps:

10^0 +10

Calculate 10 to the power of 0 and get 1.

1+10

Add 1 and 10 to get 11.

11

Answer:

with 128 legsin in all 56 160 one of them

Answer: 113 inches

Step-by-step explanation:

To solve the question, we have to find the perimeter of a circle since the skirt is circular. This will be:

Perimeter = 2πr

where

π = 3.142

r = diameter/2 = 36/2 = 18

The closest measurement needed for the sequin trim will then be:

= 2πr

= 2 × 3.142 × 18

= 113.112

= 113 inches

Answer:

b and c

Step-by-step explanation:

Answer:

(-4,-2)

Step-by-step explanation:

Point f is on x -4, which is in the first slot and y -2 which is your second slot

Answer: D. 15 square root with 3 inside

Step-by-step explanation:

Just try it

Answer:

c

Step-by-step explanation:

We know, 1 year = 12 month.

So, 3 month = 3/12 = 0.25 year.

Now, principle amount, P = $400 .

Rate is, r = 8% = 0.08

Time period, t = 0.25 year.

Now, Interest is given by :

I = P × r × t

I = $( 400 × 0.08 × 0.25 )

I = $8

Therefore, she will pay $8 interest in total.

Answer:

35: x=10

36: 9:100

37: 2/3

38: 2 1/42

Step-by-step explanation:

Answer:

The flagpole is 54.5 inches tall.

Step-by-step explanation:

We solve this question using proportions, by rule of three.

Each feet has 12 inches. So

A 6 foot man measured his shadow at13 feet 9 inches.

So when the real height is of 6 feet, the shadow is of 13 feet + (9/12) feet = 13.75 feet.

When the shadow is of 125 feet, the height is x. So

6 feet - 13.75 feet

x feet - 125 feet

Applying cross multiplication

[tex]13.75x = 6*125[/tex]

[tex]x = \frac{6*125}{13.75}[/tex]

[tex]x = 54.5[/tex]

The flagpole is 54.5 inches tall.

Answer:

c

Step-by-step explanation:

3X-x+2=14

3*6-6+2=14

14=14

plz make me brainliest

Answer:

yes

Step-by-step explanation:

i used the pythagorean theorem and used 50 and 35 as the legs and the diagonal as the hypotenuse

35² + 50² = c²

c² = 3725

c = [tex]\sqrt{3725}[/tex], which is 61.03

since 61.03 is less than 62" then it should fit in

No, Given Mirror will be not fit in the wall.

Here, we use Pythagoras theorem

Base is 50 inch , Height is 35 inch

Now we have to find hypotenuse

[tex]h^{2} =(50)^{2}+(35)^{2} \\\\h^{2}=2500+1225=3725\\\\h=\sqrt{3725}=61.03[/tex]

Since, Diagonal length of wall is 61.03 inch which is less than mirror diagonal length.

Therefore, mirror is not fit in the wall.

Learn more:

https://brainly.com/question/13391411

Answer:

-15

Step-by-step explanation:

-2 to the power 0 = -1

-2 to the power 1 = -2

-2² = -4

-2³ = -8

-1 + -2 + -4 + -8 = -15

Answer:

x <= 10

<= means less or equal to

Step-by-step explanation:

At least means less or equal to

Answer:

after it is necessary to rearrange and equations for that one variable is expressed in terms de in term of tea to express it in terms of d we need to solve the equation c is equal to fir for our in terms of c solve the equation p is equal to 2 w plus two hours for w in terms of p and h

Answer:

after it is necessary to rearrange and equations for that one variable is expressed in terms de in term of tea to express it in terms of d we need to solve the equation c is equal to fir for our in terms of c solve the equation p is equal to 2 w plus two hours for w in terms of p and h

Answer:

Follows are the solution to this question:

Step-by-step explanation:

[tex]T: 1 \ R^n \to 1 \ R^n[/tex] is invertible lines transformation

[tex]S[T(x)]=x \ and \ V[T(x)]=x \\\\t'x \ \varepsilon\ 1 R^n\\\\[/tex]

T is invertiable linear transformation means that is  

[tex]T(x) =A x \\\\ where \\\\ A= n \times n \ \ matrix[/tex]

and [tex]\ det(A) \neq 0 \ \ that \ is \ \ A^{-1} \ \ exists[/tex]

Let

[tex]V \varepsilon\ a\ R^{n} \ consider \ \ u= A^{-1} v \varepsilon 1 R^n\\\\T(u)= A(A^{-1} v)=(A \ A^{-1}) \\\\ v= I_{n \times n} \cdot v = v[/tex]

so,  

[tex]s[T(u)]=v[T(u)]\\\\s(v)=v(v) \ \ \forall \ \ v \ \ \varepsilon \ \ 1 R^n[/tex]

s(3)=4(3)-7+(3+1)^2

s(3)=12-7+16

s(3)=21

Answer:

21

Step-by-step explanation:

We find that  s(x) = 4x - 7 + (x + 1)^2. Expanding, we get s(x) = x^2 + 6x - 6. Plugging in x = 3, we have s(3) = 3^2 + 6 x 3 - 6 = 21.

Alternatively, we can compute that f(3) = 5 and g(3) = 16, so  s(3) = f(3) + g(3) = 21.

Answer:

#8. 2

#9. 7

#10. 2

#11. 2

#12. 0

#13. -4

#14. 17

#15. 1

Answer:#8=2

#9=-11

#10=8

#11=2

#12=0

#14=17

#15=1

Step-by-step explanation: #8-13 (Subtract the numbers using a number line, than add the numbers using a integer chips, after that calculate the sum of the differences) #14-15( normal solving stepping for integers(recommend using a T1-84 Calculator)

Answer:

[tex]F_x=43.30\ N[/tex]

Step-by-step explanation:

Given that,

Force, F = 50 pounds

It is directed at 30° from the horizontal.

The force will be given by :

[tex]F_x=F\cos\theta\\\\=50\times \cos(30)\\\\=43.30\ N[/tex]

So, the required force is 43.30 N.

Answer:

The area would reduce to 1/4 the original area

Step-by-step explanation:

The area of a triangle = (1/2)(base)(height) or bh/2

If the triangle base and height were to reduce by one-half, the formula would change to:

Area = (1/2)(1/2 base)(1/2 height) or bh/8. Which is a reduction by 1/4

Answer:

I believe you can set the equations equal to each other and solve for x.

Step-by-step explanation:

reason being is that by definition, a line can be infinite, so in theory, those 2 lines can be equal.

9514 1404 393

Answer:

  1a: x+3 = 5

  1c: 6 = 2z

  2b: x = 2

  2d: 3 = z

  3: the solutions make the hangars balance

Step-by-step explanation:

1. We can write the equations by listing the contents of the hangar and using an equal sign to show the balance between left side and right side. It can work well to put left side contents of the hangar on the left side of the equal sign.

  A: x + 3 = 5

  C: 1 + 1 + 1 + 1 + 1 + 1 = z + z  simplifies to  6 = 2z

__

2. B: We can subtract 3 from both sides of the hangar (and equation) to find the value of x.

  (x +3) -3 = 5 -3

  x = 2 . . . . . hangar balances with 2 on the right

D: We can divide both sides of the hangar by 2, splitting the content into two equal parts. Then one of those parts can be removed from each side.

  2(3) = 2(z)

  3 = z . . . . . . hangar balances with 3 on the left

__

3. The found values will keep the hangar in balance when they are substituted for the corresponding variables.

  A: 2 + 3 = 5

  C: 1 + 1 + 1 + 1 + 1 + 1 = 3 + 3

Problem 15

Answers:

Each interval is interval notation.

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Explanation:

To get the graph, you can plug in various x values to find their paired y values, then draw a curve through those points. You can only plug in x values that are 3 or larger, as I'll mention later in the next paragraph. A quicker way to get the graph is to use technology. I used GeoGebra to generate the graphs below.

To get the domain, we need to ensure that the stuff under the square root is never negative. So we need to make the x-3 to be 0 or larger. Solving [tex]x-3 \ge 0[/tex] leads to [tex]x \ge 3[/tex] showing that 3 is the smallest value we can plug in. The domain is the interval from 3 to positive infinity. We can write that as [tex]3 \le x < \infty[/tex] which condenses to the interval notation [3, infinity). Note how the square bracket is used to include the endpoint.

The range can be determined from the graph. The lowest point is when y = 2, so the range consists of y outputs that are 2 or larger. We write the interval notation [2, infinity) to mean [tex]2 \le y < \infty[/tex]

The graph also helps us see where the curve is increasing or decreasing. In this case, the curve goes uphill as we move from left to right. Therefore, the graph is increasing over its entire domain. We write the domain as the answer here. Because the function increases over the entire domain, there's no room for the function to decrease.

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

Problem 16

Answers:

----------------------------------

Explanation:

We follow the same idea as the previous problem.

This time we want the x+1 under the square root to be 0 or larger, so [tex]x+1 \ge 0[/tex] solves to [tex]x \ge -1[/tex] telling us the smallest input allowed. The value of x can be this or larger.

Since this is an increasing function throughout the domain (similar to the previous problem), this means that the smallest domain value corresponds exactly to the smallest range value. Plugging in x = -1 leads to y = -3 which is the smallest possible output. As x gets bigger, so does y. The graph shows that the lowest point occurs when y = -3 to visually confirm this.

The increasing interval is over the entire domain, so we just write the domain again for the increasing interval. This means we write "none" for the decreasing interval.

Side note: The graphs are shown together on the same xy coordinate axis, but for your hw problem, you'll have the graphs on their own separate grid.

Answer:

the square root is 2.2

Answer:

2.2

Step-by-step explanation:

make me brainestly

the guy who did the second answer wants beef

Answer:

Total cost of cementing the path = Rs. 17,000.

Step-by-step explanation:

Let the square garden be PQRS

Let the region inside the garden (PQRS) be KLMN.

Given the following data;

Length of sides of PQRS = 90m

Width of path = 5m

Cost = Rs. 10 per m²

Area of PQRS = 90 * 90

Area of PQRS = 8100m²

To find the area of KLMN;

KL = KN = 90 - (5 + 5)

KL = KN = 90 - 10

KL = KN = 80m

Area of KLMN = KL * KN

Substituting into the equation, we have;

Area of KLMN = 80 * 80

Area of KLMN = 6400 m²

Area of path = Area of PQRS - Area of KLMN

Area of path = 8100 - 6400

Area of path = 1700 m²

Total cost of cementing the path = Area of path * Cost

Total cost of cementing the path = 1700 * 10

Total cost of cementing the path = Rs. 17,000

9514 1404 393

Answer:

  C.  161.459 square feet

Step-by-step explanation:

In feet, the dimensions of the pool are ...

  (5 m)(3.28084 ft/m) = 16.4042 ft

  (3 m)(3.28084 ft/m) = 9.84252 ft

Then the area of the pool is ...

  (16.4042 ft)(9.84252 ft) = 161.458666584 square feet

  about 161.459 square feet

_____

Additional comment

If you consider that a square meter is slightly less than 11 square feet, the pool area can be estimated to be (5 m)(3 m)(11 ft²/m²) = 165 ft². This is close enough to point you to the correct answer choice.

Answer:

0.028 ; 0.649 ; 4

Step-by-step explanation:

P(correctly activation), p = 0.7

Number of sprinklers, n = 10

q = 1 - p = 1 - 0.7 = 0.3

Using the binomial probability relation :

P(x =x) = nCx * p^x * (1 - p)^(n - x)

Probability that all sprinklers activate correctly;

P(x = 10) = 10C10 * 0.7^10 * 0.3^0

P(x = 10) = 1 * 0.0282475249 * 1

P(x = 10) = 0.028

Probability that atleast 7 will operate correctly :

P(x ≥ 7) = p(x = 7) + p(x = 8) + p(x = 9) + p(x = 10)

P(x ≥ 7) = 0.267 + 0.233 + 0.121 + 0.028

P(x ≥ 7) = 0.649

3.)

Probability of atleast 1 operates :

P(x ≥ 1) = 0.98

1 - probability of 0 operates

1 - p(x =0)

P(x = 0) = nC0 * 0.7^0 * 0.3^n = 0.02

Recall :

nC0 = 1 ;

1 - p(x = 0)

P(x = 0) = 1 * 1 * 0.3^n = 0.02

0.3^n = 0.02 - - - (1)

0.3^3.24 = 0.02 - - - (2)

Comparing (1) and (2)

n = 3.24

Since, n cannot be a fraction ;

Then n is rounded to the next whole number = 4

Answer:

Your parents will retire in 18 years. ... You expect to earn 12% annually on the account. ... An investment will pay $100 at the end of each of the next 3 years, $200 at the end of Year 4, ...