All categories

3 years ago

GEOMETRY EASY QUESTION PLEASE HELP (I'm overthinking)

Mathematics

1 answer:

Anettt [7]3 years ago

7 0

Answer: True

Step-by-step explanation: the red line is basically there to throw you off <CDA and <BDA are equal

You might be interested in

Are these ratios equivalent?

zhenek [66]

No because 5/2= 2,5 however 63/19=3,66...

7 0

3 years ago

Apr 24,8:28:53 AM<br> Find the 98th term of the arithmetic sequence -10,-8, -6, ...

Nata [24]

Answer: 184

Step-by-step explanation:

The nth term of am arithmetic sequence is calculated as:

Nth term= a+(n-1)d

where a = first term

d = common difference

a = -10

d = -8 -(-10) = -8+10 = 2

98th term= a+(n-1)d

= -10 + (98-1)(2)

= -10 + (97×2)

= -10 + 194

= 184

The 98th term of the arithmetic sequence is 184

6 0

3 years ago

???whats the answers

Aleks04 [339]

11, 3, -3, -12
Hope this helps :)

3 0

4 years ago

an inverted conical water tank with a height of 20 ft and a radius of 8 ft is drained through a hole in the vertex (bottom) at a

viktelen [127]

Answer:

the rate of change of the water depth when the water depth is 10 ft is; $\mathbf{\dfrac{dh}{dt} = \dfrac{-25}{100 \pi} \ \ ft/s}$

Step-by-step explanation:

Given that:

the inverted conical water tank with a height of 20 ft and a radius of 8 ft is drained through a hole in the vertex (bottom) at a rate of 4 ft^3/sec.

We are meant to find the rate of change of the water depth when the water depth is 10 ft.

The diagrammatic expression below clearly interprets the question.

From the image below, assuming h = the depth of the tank at a time t and r = radius of the cone shaped at a time t

Then the similar triangles ΔOCD and ΔOAB is as follows:

$\dfrac{h}{r}= \dfrac{20}{8}$ ( similar triangle property)

$\dfrac{h}{r}= \dfrac{5}{2}$

$\dfrac{h}{r}= 2.5$

h = 2.5r

$r = \dfrac{h}{2.5}$

The volume of the water in the tank is represented by the equation:

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

$V = \dfrac{1}{3} \pi (\dfrac{h^2}{6.25}) h$

$V = \dfrac{1}{18.75} \pi \ h^3$

The rate of change of the water depth is :

$\dfrac{dv}{dt}= \dfrac{\pi r^2}{6.25}\ \dfrac{dh}{dt}$

Since the water is drained through a hole in the vertex (bottom) at a rate of 4 ft^3/sec

Then,

$\dfrac{dv}{dt}= - 4 \ ft^3/sec$

Therefore,

$-4 = \dfrac{\pi r^2}{6.25}\ \dfrac{dh}{dt}$

the rate of change of the water at depth h = 10 ft is:

$-4 = \dfrac{ 100 \ \pi }{6.25}\ \dfrac{dh}{dt}$

$100 \pi \dfrac{dh}{dt} = -4 \times 6.25$

$100 \pi \dfrac{dh}{dt} = -25$

$\dfrac{dh}{dt} = \dfrac{-25}{100 \pi}$

Thus, the rate of change of the water depth when the water depth is 10 ft is; $\mathtt{\dfrac{dh}{dt} = \dfrac{-25}{100 \pi} \ \ ft/s}$

4 0

4 years ago

1/3 + a + 5/4. what is A

USPshnik [31]

Answer:

a = -1.58

Step-by-step explanation:

1/3 +a+5/4 =0

1/3+5/4= -a

4+15/12 = -a

19/12 = -a

1.58= -a

1.58/-1 = -a/-1

-1.58 = a

6 0

3 years ago