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2 years ago

Question 4 of 17

Mathematics

1 answer:

Vlada [557]2 years ago

4 0

Answer:

I think it is A OR D

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What is the inverse to 2^x-1

nydimaria [60]

Replace x with y and solve for y~

y=2*

x=2^y

log x = y log 2

y=log x / log 2

Hope this helps and leave a brainliest to help me reach expert ;)

4 0

3 years ago

HELP ASAP

denis-greek [22]

A) I included a graph, look below.

B)

Input the y in x = y + 3.

x = (-4x - 3) + 3
x = -4x + 0
Add 4x to both sides.
5x = 0
Divide both sides by 5.
x = 0

Input that x value in y = -4x - 3

y = -4(0) - 3
y = 0 - 3
y = -3

(0, -3)

C)

Convert both equations to Standard Form.

x = y + 3
Subtract y from both sides.
x - y = 3

y = -4x - 3
Add 4x to both sides.
4x + y = -3

Add the equations together.

4x + y = -3
x - y = 3
equals
5x = 0
Divide both sides by 5.
x = 0

Input that into one of the original equations.

0 = y + 3
Subtract 3 from both sides.
-3 = y

(0, -3)

7 0

3 years ago

How do I do this. Please help

MakcuM [25]

Answer:

you gotta go to school

Step-by-step explanation:

8 0

3 years ago

Three security cameras were mounted at the corners of a triangles parking lot. Camera 1 was 110 ft from camera 2, which was 137

Nata [24]

Answer:

<em>Camera 2nd has to cover the maximum angle, i.e. </em> $78.70^\circ$ .

Step-by-step explanation:

Please have a look at the triangular park represented as a triangle $\triangle ABC$ with sides

a = 110 ft

b = 158 ft

c = 137 ft

1st camera is located at point C, 2nd camera at point B and 3rd camera at point A respectively.

We can use law of cosines here, to find out the angles $\angle A, \angle B, \angle C$

As per Law of cosine:

$cos C = \dfrac{a^{2}+b^2-c^2 }{2ab}\\cos B = \dfrac{a^{2}+c^2-b^2 }{2ac}\\cos A = \dfrac{b^{2}+c^2-a^2 }{2bc}$

Putting the values of a,b and c to find out angles $\angle A, \angle B, \angle C$ .

$cos C = \dfrac{110^{2}+158^2-137^2 }{2\times 110 \times 158}\\\Rightarrow cos C = \dfrac{12100+24964-18769 }{24760}\\\Rightarrow cos C =0.526\\\Rightarrow C = 58.24^\circ$

$cos B = \dfrac{110^{2}+137^2-158^2 }{2\times 110 \times 137}\\\Rightarrow cos B = \dfrac{12100+18769 -24964}{30140}\\\Rightarrow cos B = \dfrac{5905}{30140}\\\Rightarrow cos B =0.196\\\Rightarrow B = 78.70^\circ$

$cos A = \dfrac{158^{2}+137^2-110^2 }{2\times 158 \times 137}\\\Rightarrow cos A = \dfrac{24964+18769-12100}{43292}\\\Rightarrow cos A = \dfrac{31633}{43292}\\\Rightarrow cos A = 0.731\\\Rightarrow A = 43.05^\circ$

<em>Camera 2nd has to cover the maximum angle</em>, i.e. $78.70^\circ$ .

6 0

3 years ago

Write a function rule for the table- please help

Ipatiy [6.2K]

Answer: The 3rd option

Step-by-step explanation: f(x) aka 5 is = to x aka 2+3 (2+3=5)

Plz make brainliest

7 0

3 years ago

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