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
netineya [11]
4 years ago
6

Mike drank more than half the juice in his glass.what fraction of the juice could Mike have drunk?

Mathematics
1 answer:
jekas [21]4 years ago
7 0
9/16 or anything else greater than 1/2 and less than 1

You might be interested in
Help help help pls :)
kykrilka [37]

Answer:

opposite\approx 70.02

Step-by-step explanation:

The triangle in the given problem is a right triangle, as the tower forms a right angle with the ground. This means that one can use the right angle trigonometric ratios to solve this problem. The right angle trigonometric ratios are as follows;

sin(\theta)=\frac{opposite}{hypotenuse}\\\\cos(\theta)=\frac{adjacent}{hypotenuse}\\\\tan(\theta)=\frac{opposite}{adjacent}

Please note that the names (opposite) and (adjacent) are subjective and change depending on the angle one uses in the ratio. However the name (hypotenuse) refers to the side opposite the right angle, and thus it doesn't change depending on the reference angle.

In this problem, one is given an angle with the measure of (35) degrees, and the length of the side adjacent to this angle. One is asked to find the length of the side opposite the (35) degree angle. To achieve this, one can use the tangent (tan) ratio.

tan(\theta)=\frac{opposite}{adjacent}

Substitute,

tan(35)=\frac{opposite}{100}

Inverse operations,

tan(35)=\frac{opposite}{100}

100(tan(35))=opposite

Simplify,

100(tan(35))=opposite

70.02\approx opposite

4 0
2 years ago
What Does Y intercept =​
anastassius [24]

The y-intercept of a line is where the line goes across the y line.

Does that answer your question?

4 0
3 years ago
Un móvil azul de 573 g se encuentra en movimiento sobre un carril metálico. Si sobre el móvil, el resorte ejerce una fuerza haci
ad-work [718]

Answer:

The total mechanical energy is 0.712 J.

Step-by-step explanation:

A blue 573 g mobile is moving on a metal rail. If on the mobile, the spring exerts a force to the right of modulus 0.85 N and in the indicated position the kinetic energy of the mobile-spring system is 0.47 J and its elastic potential energy is 0.242 J. Determine the mechanical energy of the system mobile-spring in the position shown as indicated in the figure.

The total mechanical energy is given by the sum of the kinetic energy and the potential energy.

Kinetic energy = 0.47 J

Potential energy = 0.242 J

The total mechanical energy is

T = 0.47 + 0.242 = 0.712 J

4 0
3 years ago
Write an expression in expression in simplest form that represents the total amount in the situation. You rent x pairs of shoes
lesya692 [45]
2x+(1.50+9)= $10.50×2= 21 so x=21
8 0
3 years ago
Can someone help me please
Lady bird [3.3K]
Answer: 12.50$


Explanation: 5.25$ (per pound) and he wants 3 pounds so we multiply 5.25x3 which equals 15.75 and then we subtract 3.25 (because of the coupon) which makes your final answer 12.50$
8 0
3 years ago
Other questions:
  • Q5 Q3.) How​ far, to the nearest​ foot, is the ship from the​ statue's base?
    7·1 answer
  • Rosen 15 how many solutions are there to the equation x1 x2 x3 x4 x5
    9·1 answer
  • John took four tests.
    14·2 answers
  • A line with slope of -3 passes through the points ( x , -6 ) and ( -2 , 6 ). What is the missing x-coordinate?
    8·1 answer
  • The skateboard is 800cm. How many meters would this be?
    12·1 answer
  • Anyone please help!!!​
    13·1 answer
  • Please help!!!
    6·1 answer
  • Anyone want to help me with my math assignments ill post another question with the assignments on it
    10·1 answer
  • What is the solution to 3.25k + 1 - 4.25k = -2
    11·2 answers
  • Simplify<br> 21x^6 y^5<br> 14x^2y^9
    8·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!