Please help need asap

In order to estimate the height of a building, two students stand at a certain distance from the building at street level. from

enyata [817]

Let h represent the height of the building. Then the difference between the two angle measurement locations is
.. h/tan(39°) -h/tan(50°) = 300 ft
.. h = (300 ft)/(1/tan(39°) -1/tan(50°))
.. h ≈ 758 ft

The height of the building is about 758 feet.

4 0

3 years ago

List the integers that satisfy both of these inequalities 2x+9<0 x>-12 put the answers on one line

lbvjy [14]

Answer:

Step-by-step explanation:

Given the inequality expressions

2x+9<0 and x>-12

2x+9<0

2x< 0-9

2x < -9

x < -9/2

x < -4.5

Since x >-12

-12 < x

Combining both inequalities

-12 <x < 4.5

<em>The integers between the intervals are -11, -10, -9, -8, -7, -6 and -5 </em>

6 0

3 years ago

Please help hurry!!!!!!

nordsb [41]

You do C = πd which it would be 69.08

3 0

4 years ago

A food packet is dropped from a helicopter and is modeled by the function f(x) = −15x2 + 6000. The graph below shows the height

melisa1 [442]

General Idea:

Domain of a function means the values of x which will give a DEFINED output for the function.

Applying the concept:

Given that the x represent the time in seconds, f(x) represent the height of food packet.

Time cannot be a negative value, so

$x\geq 0$

The height of the food packet cannot be a negative value, so

$f(x)\geq 0$

We need to replace $-15x^2+6000$ for f(x) in the above inequality to find the domain.

$-15x^2+6000\geq 0 \; \; [Divide \; by\; -15\; on\; both\; sides]\\ \\ \frac{-15x^2}{-15} +\frac{6000}{-15} \leq \frac{0}{-15} \\ \\ x^2-400\leq 0\;[Factoring\;on\;left\;side]\\ \\ (x+200)(x-200)\leq 0$

The possible solutions of the above inequality are given by the intervals $(-\infty , -2], [-2,2], [2,\infty )$ . We need to pick test point from each possible solution interval and check whether that test point make the inequality $(x+200)(x-200)\leq 0$ true. Only the test point from the solution interval [-200, 200] make the inequality true.

The values of x which will make the above inequality TRUE is $-200\leq x\leq 200$

But we already know x should be positive, because time cannot be negative.

Conclusion:

Domain of the given function is $0\leq x\leq 200$

4 0

4 years ago

Read 2 more answers

Simplify: (8t4 + 6 )+(914 – 2ť + 6t? – 6 )

vekshin1

Answer:t

2⋅(17t 2 −2t+6)

Step-by-step explanation:

v(8•(t4))+6)+((((9•(t4))-(2•(t3)))+(2•3t2))-6)

STEP

2

:

Equation at the end of step

2

:

((8•(t4))+6)+((((9•(t4))-2t3)+(2•3t2))-6)

STEP

3

:

Equation at the end of step

3

:

((8•(t4))+6)+(((32t4-2t3)+(2•3t2))-6)

STEP

4

:

Equation at the end of step

4

:

(23t4 + 6) + (9t4 - 2t3 + 6t2 - 6)

STEP

5

:

STEP

6

:

Pulling out like terms

6.1 Pull out like factors :

17t4 - 2t3 + 6t2 = t2 • (17t2 - 2t + 6)

(8•(t4))+6)+((((9•(t4))-(2•(t3)))+(2•3t2))-6)

STEP

2

:

Equation at the end of step

2

:

((8•(t4))+6)+((((9•(t4))-2t3)+(2•3t2))-6)

STEP

3

:

Equation at the end of step

3

:

((8•(t4))+6)+(((32t4-2t3)+(2•3t2))-6)

STEP

4

:

Equation at the end of step

4

:

(23t4 + 6) + (9t4 - 2t3 + 6t2 - 6)

STEP

5

:

STEP

6

:

Pulling out like terms

6.1 Pull out like factors :

17t4 - 2t3 + 6t2 = t2 • (17t2 - 2t + 6)

(8•(t4))+6)+((((9•(t4))-(2•(t3)))+(2•3t2))-6)

STEP

2

:

Equation at the end of step

2

:

((8•(t4))+6)+((((9•(t4))-2t3)+(2•3t2))-6)

STEP

3

:

Equation at the end of step

3

:

((8•(t4))+6)+(((32t4-2t3)+(2•3t2))-6)

STEP

4

:

Equation at the end of step

4

:

(23t4 + 6) + (9t4 - 2t3 + 6t2 - 6)

STEP

5

:

STEP

6

:

Pulling out like terms

6.1 Pull out like factors :

17t4 - 2t3 + 6t2 = t2 • (17t2 - 2t + 6)

5 0

3 years ago