All categories

2 years ago

What’s the answer I need it ASAP

Mathematics

1 answer:

mestny [16]2 years ago

3 0

Answer:

(3) HL is an angle bisector.

You might be interested in

U0ukj787y8fr7hjhy7 which one

Sedbober [7]

If ABC and DEF are complementary, then the value of x will be,

90 - ABC = x

90 - 53 = x

x = 37°

5 0

1 year ago

Read 2 more answers

A local health clinic surveys its patients about their waterdrinking habits. It found the data is normally distributed,the mean

satela [25.4K]

In this case, we'll have to carry out several steps to find the solution.

Step 01:

Data:

water drinking habits:

mean = 62 ounces

standard deviation = 5.2 ounces

Step 02:

normally distribution:

95% ===> 2 SD

(62 + 5.2 + 5.2) ounces = 72.4 ounces ==> + 2 SD

(62 - 5.2 - 5.2) ounces = 51.6 ounces ==> - 2 SD

The answer is:

51.6 ounces - 72.4 ounces

3 0

10 months ago

the sum of the measurement of two angles is 183.61 angle one has a measure of 54 what is the measure in degrees of the second an

Dennis_Churaev [7]

Hfhfjhfhchfjfnnfnfncncbc

7 0

2 years ago

PLS TRY TO ANSWER ASAP.

Harrizon [31]

Answer:

0.075 m/min

Explanation:

You need to use derivatives which is an advanced concept used in calculus.

1. Write the equation for the volume of the cone:

$V=\dfrac{1}{3}\pi r^2h$

2. Find the relation between the radius and the height:

r = diameter/2 = 5m/2 = 2.5m

h = 5.2m

h/r =5.2 / 2.5 = 2.08

3. Filling the tank:

Call y the height of water and x the horizontal distance from the axis of symmetry of the cone to the wall for the surface of water, when the cone is being filled.

The ratio x/y is the same r/h

x/y=r/h

y = x . h / r

The volume of water inside the cone is:

$V=\dfrac{1}{3}\pi x^2y$

$V=\dfrac{1}{3}\pi x^2(2.08)\cdot x\\\\\\V=\dfrac{2.08}{3}\pi x^3$

4. Find the derivative of the volume of water with respect to time:

$\dfrac{dV}{dt}=2.08\pi x^2\dfrac{dx}{dt}$

5. Find x² when the volume of water is 8π m³:

$V=\dfrac{2.08}{3}\pi x^3\\\\\\8\pi=\dfrac{2.08}{3}\pi x^3\\\\\\ 11.53846=x^3\\ \\ \\ x=2.25969\\ \\ \\ x^2=5.1062$ m²

6. Solve for dx/dt:

$1.2m^3/min=2.08\pi(5.1062m^2)\dfrac{dx}{dt}$

$\dfrac{dx}{dt}=0.03596m/min$

7. Find dh/dt:

From y/x = h/r = 2.08:

$y=2.08x\\\\\\\dfrac{dy}{dx}=2.08\dfrac{dx}{dt}\\\\\\\dfrac{dy}{dt}=2.08(0.035964m/min)=0.0748m/min\approx0.075m/min$

That is the rate at which the water level is rising when there is 8π m³ of water.

4 0

2 years ago

Solve for u 6≤ -9/4u

masya89 [10]

if you mean.

$6 \leqslant \frac{4u}{ - 9}$

then the answer would be,

$u \leqslant - \frac{27}{2}$

7 0

2 years ago

Read 2 more answers