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Anna35 [415]
3 years ago
11

How many solutions are there to the equation below |x|=77?

Mathematics
2 answers:
Pavlova-9 [17]3 years ago
5 0

C.

Solve for x by simplifying both sides of the equation, then isolating the variable.

x = 77, -77

erik [133]3 years ago
3 0

Solution, \left|x\right|=77\quad :\quad x=-77\quad \mathrm{or}\quad \:x=77

Steps:

\left|x\right|=77

|f\left(x\right)|=a\quad \Rightarrow \:f\left(x\right)=-a\quad \mathrm{or}\quad \:f\left(x\right)=a, |f\left(x\right)|=a\quad \Rightarrow \:f\left(x\right)=-a\quad \mathrm{or}\quad \:f\left(x\right)=a

The correct answer is <u><em>C. 2</em></u>

Hope this helps!!!

<3 -austint1414

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A good-quality measuring tape can be off by 0.49 cm over a distance of 23 m. What is the percent uncertainty?33% Part (a) If 44
V125BC [204]

Step-by-step explanation:

First:

(Error/Total )x 100%

(0.49/23) x100%= 2.13%

Second:

(Absolute uncertainty/mean) x100%

(3/44)x100%= 6.81%

8 0
3 years ago
Y= 2x + 6 for x = 0,1,2,3,4
algol [13]

x=0    y=6

x=1     y=8

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8 0
3 years ago
F(3)= 5x^2 - 4x + 9<br> evaluate this quadratic function
Neporo4naja [7]

Answer:

f(3) = 38

Step-by-step explanation:

Given the quadratic function

f\left(x\right)=\:5x^2\:-\:4x\:+\:9

Evaluating the quadratic function

f\left(x\right)=\:5x^2\:-\:4x\:+\:9

substitute x = 3

f\left(3\right)=\:5\left(3\right)^2\:-\:4\left(4\right)\:+\:9

f(3) = 45-16+9

f(3) = 38

Therefore,

  • f(3) = 38
6 0
2 years ago
A man wants to paint his room with an area of 515 square feet. He mixses the following paints to make a shade of green
larisa [96]
About 3 gallons of blue paint will be needed because if you add 175+175 it equals 350 and then add another 175 it equals 525. 
6 0
3 years ago
Answer ASAP. Please give correct answer. (Involves a bunch of circles and trig identities)
iren [92.7K]

Answer:

45.40

Step-by-step explanation:

First of all, the shape of rope is not a parabola but a catenary, and all catenaries are similar, defined by:

y=acoshxa

You just have to figure out where the origin is (see picture). The hight of the lowest point on the rope is 20 and the pole is 50 meters high. So the end point must be a+(50−20) above the x-axis. In other words (d/2,a+30) must be a point on the catenary:

a+30=acoshd2a(1)

The lenght of the catenary is given by the following formula (which can be proved easily):

s=asinhx2a−asinhx1a

where x1,x2 are x-cooridanates of ending points. In our case:

80=2asinhd2a

40=asinhd2a(2)

You have to solve the system of two equations, (1) and (2), with two unknowns (a,d). It's fairly straightforward.

Square (1) and (2) and subtract. You will get:

(a+30)2−402=a2

Calculate a from this equation, replace that value into (1) or (2) to evaluate d.

My calculation:

a=353≈11.67

d=703arccosh257≈45.40

7 0
3 years ago
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