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velikii [3]
3 years ago
11

Angle rqs and tqs are linear pair where angle rqs equal 2x+4 and angle tqs equal 6x+20 solve for x

Mathematics
1 answer:
Ulleksa [173]3 years ago
8 0
The answer is x = -2 you move the all x to one side and all numbers to the other side. then divide to find x

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The function f is defined by f(x) = 2x²-1.<br> Find f (2x).
umka21 [38]

Answer: f(2x)=8x^2-1

Step-by-step explanation:

All that is required here is to substitute x = 2x into the right side in the same way that is done for a numeric value.

f(2x) = 2(2x)^2-1\\f(2x) = 2(4)x^2-1\\f(2x) = 8x^2-1\\

5 0
3 years ago
Given the function f(x) = x^4 + 3x^3 - 2x^2 - 6x - 1, use intermediate theorem to decide which of the following intervals contai
marta [7]

f(x) = x^4 + 3x^3 - 2x^2 - 6x - 1

Lets check with every option

(a) [-4,-3]

We plug in -4  for x  and -3 for x

f(-4) = (-4)^4 + 3(-4)^3 - 2(-4)^2 - 6(-4) - 1= 55

f(-3) = (-3)^4 + 3(-3)^3 - 2(-3)^2 - 6(-3) - 1= -1

f(-4) is positive and f(-3) is negative. there is some value at x=c on the interval [-4,-3] where f(c)=0. so there exists atleast one zero on this interval.

(b) [-3,-2]

We plug in -3  for x  and -2 for x

f(-3) = (-3)^4 + 3(-3)^3 - 2(-3)^2 - 6(-3) - 1= -1

f(-2) = (-2)^4 + 3(-2)^3 - 2(-2)^2 - 6(-2) - 1= -5

f(-2) is negative and f(-3) is negative. there is no value at x=c on the interval [-3,-2] where f(c)=0.  

(c) [-2,-1]

We plug in -2  for x  and -1 for x

f(-2) = (-2)^4 + 3(-2)^3 - 2(-2)^2 - 6(-2) - 1= -5

f(-1) = (-1)^4 + 3(-1)^3 - 2(-1)^2 - 6(-1) - 1= 1

f(-2) is negative and f(-1) is positive. there is some value at x=c on the interval [-2,-1] where f(c)=0. so there exists atleast one zero on this interval.

(d) [-1,0]

We plug in -1  for x  and 0 for x

f(-1) = (-1)^4 + 3(-1)^3 - 2(-1)^2 - 6(-1) - 1= 1

f(0) = (0)^4 + 3(0)^3 - 2(0)^2 - 6(0) - 1= -1

f(-1) is positive and f(0) is negative. there is some value at x=c on the interval [-1,0] where f(c)=0. so there exists atleast one zero on this interval.

(e) [0,1]

We plug in 0  for x  and 1 for x

f(0) = (0)^4 + 3(0)^3 - 2(0)^2 - 6(0) - 1= -1

f(1) = (1)^4 + 3(1)^3 - 2(1)^2 - 6(1) - 1= -5

f(0) is negative and f(1) is negative. there is no value at x=c on the interval [0,1] where f(c)=0.  

(f) [1,2]

We plug in 1  for x  and 2 for x

f(1) = (1)^4 + 3(1)^3 - 2(1)^2 - 6(1) - 1= -5

f(2) = (2)^4 + 3(2)^3 - 2(2)^2 - 6(2) - 1= 19

f(-4) is positive and f(-3) is negative. there is some value at x=c on the interval [-4,-3] where f(c)=0. so there exists atleast one zero on this interval.

so answers are (a) [-4,-3], (c) [-2,-1],  (d) [-1,0], (f) [1,2]

8 0
3 years ago
Read 2 more answers
What does k equal if <br>485= 1/((k)(5.99x10^-3))
Verizon [17]
k\ne 0

485=\frac{1}{(k)(5.99*10^{-3})}\\\\ \frac{485}{1}= \frac{1}{0.00599k}\\\\ 485*0.00599k=1\\\\ k=\frac{1}{485*0.00599}\\\\k\approx 0.3442
7 0
2 years ago
A 228<br> B 328<br> C 488<br> D 572
Aleksandr [31]

Answer:

G 328 ft

Step-by-step explanation:

\cos(angleBAC)  =  \frac{AB}{AC}

\cos(55°)  =  \frac{AB}{400}

AB = 229.431 ft

AB² + BC² = AC²

229.431² + h² = 400²

h = 327.659 ft ≈ 328 ft

<em>H</em><em>O</em><em>P</em><em>E</em><em> </em><em>T</em><em>H</em><em>I</em><em>S</em><em> </em><em>H</em><em>E</em><em>L</em><em>P</em><em>S</em><em> </em><em>A</em><em>N</em><em>D</em><em> </em><em>H</em><em>A</em><em>V</em><em>E</em><em> </em><em>A</em><em> </em><em>N</em><em>I</em><em>C</em><em>E</em><em> </em><em>D</em><em>A</em><em>Y</em><em> </em><em><</em><em>3</em>

7 0
1 year ago
Simplify the quotient. Write the answer in sn.
MA_775_DIABLO [31]

Answer:

2.4*10^4

Step-by-step explanation:

write in scientific notation

first i would simplify the 3.6/1.5

=2.4

then you get (2.4*10^7)/10^3

formula: (m^x)/(m^y)= m^(x-y)

(2.4*10^7)/10^3

=2.4*10^(7-3)/1

=2.4*10^4

3 0
2 years ago
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