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Scorpion4ik [409]
3 years ago
11

Please please help please please ASAP

Mathematics
1 answer:
NeX [460]3 years ago
4 0

Answer:

15.09

Step-by-step explanation:

with reference angle 28°

perpendicular (p) = 8

base (b) = x

Now

tan 28° = p / b

0.53 = 8 / x

x = 8 / 0.53

x = 15.09

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Use the ruler to find 1 / 1/10 and 4 1/10
Korvikt [17]

Answer:

the first answer is 10 and the second is 40

Step-by-step explanation:

there actually asking how many meters would make 1 centimeter

On the ruler 1 whole centimeter is 10 meter. (The longer line behind the number is the 10th meter )

Dividing fractions

1 ÷ 1/10

Steps

Change the whole number which is 1 into a fraction.

1/1 ÷1/10

Flip the second fraction in the equation.

1/1 ÷ 10/1

Then change the division sign to a multiplication sign. So were multiply instead.

1/1 × 10/1

Multiply the fractions and your answer should be 10/1 .

This fraction is also changed to a whole number.

10/1 = 10

6 0
2 years ago
Carrie oakey has only nickles to spend at the fair. She buys 3 tickets for 50 cents each and has $.40 left over. How many coins
Nat2105 [25]

Answer:

she had 18 nickles

Step-by-step explanation: your so welcome

7 0
3 years ago
The accompanying data came from a study of collusion in bidding within the construction industry. No. Bidders No. Contracts 2 6
viktelen [127]

Answer:

a)

The proportion of contracts involved at most five bidders is 0.667.

The proportion of contracts involved at least five bidders is 0.51.

b)

The  proportion of contracts involved  between five and 10 inclusive bidders is 0.5.

The  proportion of contracts involved  strictly between five and 10 bidders is 0.304.

Step-by-step explanation:

No. bidders    No. contracts     Relative frequency of contracts

2                         6                       6/102=0.0588

3                         20                     20/102=0.1961

4                         24                     24/102=0.2353

5                         18                       18/102=0.1765

6                         13                        13/102=0.1275

7                          7                          7/102=0.0686

8                          5                          5/102=0.049

9                          6                          6/102=0.0588

10                         2                           2/102=0.0196

11                          1                            1/102=0.0098

Total                  102

a)

We have to find proportion of contracts involved at most five bidders.

Proportion of at most 5= Relative frequency 2+ Relative frequency 3+ Relative frequency 4+ Relative frequency 5

Proportion of at most 5=0.0588+ 0.1961+0.2353+0.1765

Proportion of at most 5=0.6667

The proportion of contracts involved at most five bidders is 0.667.

proportion of at least five bidders= proportion≥5= 1- proportion less than 5

Proportion less than 5=0.0588+ 0.1961+0.2353=0.4902

proportion of at least five bidders=1-0.4902=0.5098

The proportion of contracts involved at least five bidders is 0.51

b)

We have to find proportion of contracts involved  between five and 10 inclusive bidders.

Proportion of contracts between five and 10 inclusive= Relative frequency 5+ Relative frequency 6+ Relative frequency 7+ Relative frequency 8+ Relative frequency 9+ Relative frequency 10

Proportion of between five and 10 inclusive=0.1765 +0.1275 +0.0686 +0.049 +0.0588 +0.0196

Proportion of between five and 10 inclusive=0.5

The  proportion of contracts involved  between five and 10 inclusive bidders is 0.5

We have to find proportion of contracts involved  strictly between five and 10 bidders.

Proportion of contracts strictly between five and 10=  Relative frequency 6+ Relative frequency 7+ Relative frequency 8+ Relative frequency 9

Proportion of strictly between five and 10=0.1275 +0.0686 +0.049 +0.0588

Proportion of strictly between five and 10=0.3039

The  proportion of contracts involved  strictly between five and 10 bidders is 0.304.

8 0
3 years ago
Part A: Explain why the x-coordinates of the points where the graphs of the equations y = 4-x and y = 2x + 3 intersect are the s
ale4655 [162]
Part A: Explain why the x-coordinates of the points where the graphs of the equations y = 4-x and y = 2x + 3 intersect are the solutions of the equation 4-x = 2x + 3.

Because the point where the graphs intersect is a point that meets both rules (functions) y = 4 - x and y = 2x + 3 meaning that y from y = 4 - x equals y from 2x + 3 and also both x have the same value.

Part B: Make tables to find the solution to 4-x = 2x + 3. Take the integer values of x between -3 and 3.

x values    4 -x         2x + 3

-3              4-(-3)=7     2(-3)+3 =-3
-2              4-(-2)=6     2(-2)+3 =-1
-1              4-(-1)=5     2(-1)+3 = 1
0                4-0=4        2(0)+3 = 3
1                4-1=3         2(1)+3=5
2                4-2=2        2(2)+3 = 7
3                4-3=1        2(3)+3 = 9

The the solution is between x = 0 and x =1

Part C: How can you solve the equation 4-x = 2x + 3 graphically?

Draw in a same graph both functions  y= 4 - x and y = 2x +3.

Then read the x-coordinates of the intersection point. That is the solution.

3 0
3 years ago
The tile along the edge of a triangular community pool needs to be replaced. A right triangles, where the hypotenuse is labeled
jolli1 [7]

Perimeter of a closed figure is the sum of length of its outline. The perimeter of the consider triangular pool is: 12x^2 + 8x + 25 units.

<h3>How to calculate perimeter of a triangular figure?</h3>

Perimeter of a triangle = Sum of lengths of all its 3 sides.

For the given case, we have:

Length of first side(hypotenuse) = 8x^2 units

Height = 4x^2 + 15 units

Base is of length 8x + 10

Thus, its perimeter is calculated as:

Perimeter of a triangle = Sum of lengths of all its 3 sides.

Perimeter =  8x^2 + 4x^2 + 15 + 8x + 10 = (8+4)x^2 + 8x + 25 = 12x^2 + 8x + 25 units

Perimeter = 12x^2 + 8x + 25 units

Thus,

The perimeter of the consider triangular pool is: 12x^2 + 8x + 25 units.

Learn more about perimeter here:

brainly.com/question/10466285

3 0
2 years ago
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