1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
sdas [7]
3 years ago
12

Find cos(a) in the triangle.

Mathematics
1 answer:
kakasveta [241]3 years ago
3 0
What triangle? you need to show the triangle
You might be interested in
Part of the area under this ramp serves for storage. Items can be stored only in the unshaded part of the ramp
Mekhanik [1.2K]

Answer:

48.1 m³ available, 2.5m³ not available

Step-by-step explanation:

Tan(X) = 4.5/(5.8+1.7)

Tan X = 3/5

Hypotenuse² = 3² + 5² = 34

Sin X = 3/sqrt(34)

X = 30.96

1/sinX = 1.7/SinY

Y = 61.003°

Third angle: 180 - X - Y = 88.037

Volume

= ½ × 1.7 × 1 × sin(88.038) × 3

= .

= 2.548

= 2.5m³

Available for storage:

½×4.5×7.5×3 - 2.548

= 48.077 = 48.1 m³

7 0
3 years ago
Whats -4 1/5 times 1/3 ?
tamaranim1 [39]

Answer: -7/5

Step-by-step explanation:

-4(1/5)⋅(1/3)

Convert -4(1/5) to an improper fraction.

A mixed number is an addition of its whole and fractional parts.

− (4 +1/5) ⋅(1/3)

 Add 4 and 1/5

-21/5  ⋅ 1/3

Cancel the common factor of 3.

Move the leading negative in − 21 /5 into the numerator.

−21/5 ⋅ 1/3

Factor 3 out of −21.

3 (−7)/5 ⋅ 3 /1

Cancel the common factor.

Rewrite the expression.

−7 /5

5 0
3 years ago
Read 2 more answers
The population of Henderson City was 3,381,000 in 1994, and is growing at an annual rate 1.8%
liq [111]
<h2>In the year 2000, population will be 3,762,979 approximately. Population will double by the year 2033.</h2>

Step-by-step explanation:

   Given that the population grows every year at the same rate( 1.8% ), we can model the population similar to a compound Interest problem.

   From 1994, every subsequent year the new population is obtained by multiplying the previous years' population by \frac{100+1.8}{100} = \frac{101.8}{100}.

   So, the population in the year t can be given by P(t)=3,381,000\textrm{x}(\frac{101.8}{100})^{(t-1994)}

   Population in the year 2000 = 3,381,000\textrm{x}(\frac{101.8}{100})^{6}=3,762,979.38

Population in year 2000 = 3,762,979

   Let us assume population doubles by year y.

2\textrm{x}(3,381,000)=(3,381,000)\textrm{x}(\frac{101.8}{100})^{(y-1994)}

log_{10}2=(y-1994)log_{10}(\frac{101.8}{100})

y-1994=\frac{log_{10}2}{log_{10}1.018}=38.8537

y≈2033

∴ By 2033, the population doubles.

4 0
3 years ago
Y=3x+18 <br> -12x+4y=-24 linear system of equations
solong [7]

Answer:

Step-by-step explanation:

Use substitution, y=3x+18 so -12x+4(3x+18)=-24. From here you can try to solve for x

-12x+12x+72=-24

The xs cancel, which leaves you with 72=-24

Since 72 cannot equal -24 there is no answer to this system of equations

7 0
3 years ago
A local grocery store charges for oranges
Anon25 [30]
I think it would be 3 dollars because the 1 matches with the three.
7 0
3 years ago
Other questions:
  • If JK = 7 and JL = 12, what is JM?
    10·1 answer
  • Please help I need for graph
    5·1 answer
  • What is the recombination frequency between these genes for body color and wing size? Express your answer as a percentage.
    14·1 answer
  • Five tickets to a baseball game cost $60. How much do 3 tickets cost? A. $48 B. $36 C. $24 D. $12
    12·2 answers
  • Someone please explain this question PART "b) i" ONLY!!!
    10·1 answer
  • A pyramid with a rectangular base has a volume of 60 cubic feet and a height of 6 feet. The width of the rectangular base is 4 f
    7·1 answer
  • To prepare for virtual learning the school is purchasing extra keyboards, mice, and PC cameras. They have a budget of $1500 to s
    13·1 answer
  • URGENT CLICK TO SEE LINKS WILL GET REPORTED
    5·1 answer
  • Find the 9th term of the geometric sequence whose common ratio is
    13·1 answer
  • Evaluate g/-5, when g=-5
    15·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!