If m ZABC = 130° and mZDBC = 30°. then mZABD = [ ? ]°

The grizzly bear population in yellowstone national park in 1970 was about 270.Over the next 35 years,it increased by about 115%

yawa3891 [41]

Answer:

581

Step-by-step explanation:

Population of Grizzlys

1970 >>> 270

2005 >>> 115% INCREASE

We need to find 115% of 270 and increase 270 by that amount. Firstly, converting percentage to decimal (division by 100):

115/100 = 1.15

So, that would be now:

1.15 * 270 = 310.5

Increasing, we get:

270 + 310.5 = 580.5

We round to 581

In 2005, the population was about 581

3 0

3 years ago

Given that q(x) = 6 -5x and q(x) =21 x=

ELEN [110]

Answer:

x = -3

Step-by-step explanation:

6 - 5 ( -3 ) = 21

21 = 21 ✅

4 0

3 years ago

Please Help!! A septic tank has the shape shown in the figure. How many gallons does it hold? (1 cu ft = 7.48 gallons.) (Round t

Hoochie [10]

The overall volume is the sum of the volume of a cylinder of height 5'9" and diameter 3'6", and a sphere of diameter 3'6" (two hemispheres = full sphere).

Volume of the cylinder = (area of the base) x (height) = pi * (diameter/2)^2 * 5.75ft = 3.1415 * (3.5ft/2)^2 * 5.75 ft = 55.32 ft^3

Volume of the sphere = 4/3 * pi * (3.5ft)^3 / 8 = 22.45 ft^3

Total volume = (Volume of cylinder) + (Volume of sphere) = (55.32 + 22.45) ft^3 = 77.77 ft^3

3 0

3 years ago

If a ball is thrown into the air with a velocity of 34 ft/s, its height (in feet) after t seconds is given by y = 34t − 16t2. Fi

Finger [1]

ANSWER:

If a ball is thrown into the air with a velocity of 34 feet per second, then velocity of the ball after 1 second is 2 feet per second

SOLUTION:

Given, a ball is thrown into the air with a velocity of 34 feet per second

Initial velocity (u) = 34 feet per second

And also given a relation between displacement and time = $\mathrm{y}=34 \mathrm{t}-16 \mathrm{t}^{2}$ --- eqn 1

We need to find the velocity when t = 1 ; v = ?

We know that, v = u + at and $\mathrm{s}=\mathrm{ut}+\frac{1}{2} \mathrm{at}^{2}$

where v is instantaneous velocity and u is initial velocity

a is acceleration

t is time interval

s is displacement

using the displacement and time relation eqn (1) we get

Now, when t = 1, displacement s = 34(1) – 16(1)

$\mathrm{ut}+\frac{1}{2} \mathrm{at}^{2}=34-16$

$34 \times 1+\frac{1}{2} \times a \times 1^{2}=18$

$34+\frac{a}{2}=18$

$\begin{array}{l}{\frac{a}{2}=18-34} \\\\ {\frac{a}{2}=-16} \\ {a=-16 \times 2} \\ {a=-32}\end{array}$

here, -ve sign indicates that object is in deceleration . so acceleration is -32 ft/s

now put a value in v = u + at

v = 34 + (-32)(1)

v = 34 – 32

v = 2 ft/s

Hence, velocity of the ball after 1 second is 2 ft/s

6 0

3 years ago

X/5=1/(x+4) solve for x

solmaris [256]

ANSWER:

X is -5 and 1.

1) cross multiply

2) subtract five from both sides

3) factor

4) check for extraneous

5) ask me if you are still confused.

6 0

3 years ago