f(x)=4x+6 g(x)=4x-6 if g(x)=f(x)+k what is the value of k

Answer:

60

would be right

Step-by-step explanation:

Total length of cable required is 466.18 m.

Pythagoras theorem states that In a right-angled triangle, the square of the hypotenuse side is equal to the sum of squares of the other two sides.

According to the question

A 100m mast is supported by six cables in two sets of three cables.

They are anchored to the ground at an equal distance from the mast.

The top set of three cables is attached at a point 20m below the top of the mast.

Each cable in the lower set of three cables is 60m long and is attached at a height of 30m above the ground.

One set of guy wires (1 upper and one lower have two right triangles which can solve

The lower right triangle

a = distance from the pole to guy tie point

b = 30 m

c = 60 m the lower guy wire (the hypotenuse)

Using Pythagoras theorem

[tex]a^{2} +30^{2} =60^{2}[/tex]

⇒ [tex]a^{2} = 3600 - 900[/tex]

⇒ [tex]a^{2} =2700[/tex]

⇒ [tex]a = \sqrt{2700}[/tex]

⇒ [tex]a = 51.96[/tex] m from the pole to the tie point.

The other triangle using upper guy wire as the hypotenuse (h) tied to a point on the pole 80 m from the ground (20 m from the top) and also 50m from the tie point.

Using Pythagoras theorem

[tex]h^{2} =(51.96)^{2} +80^{2}[/tex]

⇒ [tex]h^{2}=2700+6400[/tex]

⇒ [tex]h^{2}=9100[/tex]

⇒ [tex]h = \sqrt{9100}[/tex]

⇒ [tex]h = 95.39[/tex] m , the length of the upper guy wire

For one set of guy wires (1 upper & 1 lower):

= 95.39 + 60

= 155.39 m

For 3 sets of guy wires :

= 3(155.39)

= 466.18 m of wire required

Hence, total length of cable required is 466.18 m.

Find out more information about Pythagoras theorem here

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Answer:

4x^3

Step-by-step explanation:

20 x^5 + 28 x^4 + 12x^3

4*5*x^5  + 4*7*x^4 + 4*3*x^3

Factor out 4x^3

4x^3 ( 5x^2 + 7x +3)

Answer:

3rd option

Step-by-step explanation:

Using the identity

sin²x + cos²x = 1  ( subtract sin²x from both sides )

cos²x = 1 - sin²x ( take the square root of both sides )

cosx = ± [tex]\sqrt{1-sin^2x}[/tex]

Given

sinθ = [tex]\frac{21}{29}[/tex] and 0 < θ < 90 , then

cosθ

= [tex]\sqrt{1-(\frac{21}{29})^2 }[/tex]

= [tex]\sqrt{1-\frac{441}{841} }[/tex]

= [tex]\sqrt{\frac{400}{841} }[/tex] = [tex]\frac{\sqrt{400} }{841}[/tex] = [tex]\frac{20}{29}[/tex]

Answer:

Both Ava and Finn are correct

Step-by-step explanation:

There are 4 tables and 9 pencils at each table

4 ( 9) = 36

9 * 4 = 36

Both Ava and Finn are correct

Answer:

Step-by-step explanation:

Answer:

Hence the answer is Letter B.

Step-by-step explanation:

° ° °

Answer:

b = 7/2

x = 11/2

y = 21/2

z = -4

Step-by-step explanation:

2x + 2y + 2z = 24

x + y + z = 12

b + 2x + y + z = 21

2b + 2y = 28

b + y = 14

3x + y + z = 23

we can start anywhere by transforming these equations in a way that always one variable is excised by others.

so, e.g.

b = 14 - y

14 - y + 2x + y + z = 21

2x + z = 7

z = 7 - 2x

3x + y + 7 - 2x = 23

x + y = 16

16 + z = 12

z = -4

-4 = 7 - 2x

-11 = -2x

11 = 2x

x = 11/2

11/2 + y = 16

y = 16 - 11/2 = 32/2 - 11/2 = 21/2

b = 14 - 21/2 = 28/2 - 21/2 = 7/2

Answer:

1320 Rs.

Step-by-step explanation:

let cost price=x

loss=10 %

S.P.=x×90/100=9x/10

9x/10=1080

x=1080×10/9=1200 Rs

so C.P.=1200 Rs

gain=10%

S.P.=110/100×1200=1320 Rs

Answer:

-6/-2<-2x/-2

divide both sides by-2 to remain with the x variable on one side

I hope this helps

Answer:

D) 4x +3y = -6

Step-by-step explanation:

paralell lines so m1 and m2 are equal

m = (3 +1 )/ (0 - 3 )

m = -4/ 3

y -2 = -4/3 (x +3)

y =-4x/3 -2

3y = -4x -6

4x +3y = -6

Answer:

A'(-3,0), B'(0,-3) and C'(4,7)

Step-by-step explanation:

We are given that the vertices of triangle are A(0,-3), B(3,0) and C(-7,4).

We have to find the coordinates of the image of triangle under a rotation of 90° clockwise about the origin.

90° clockwise about the origin

Rule:[tex](x,y)\rightarrow (y,-x)[/tex]

Using the rule

The coordinates of A'

[tex]A(0,-3)\rightarrow A'(-3,0)[/tex]

The coordinates of B'

[tex]B(3,0)\rightarrow B'(0,-3)[/tex]

The coordinates of C'

[tex]C(-7,4)\rightarrow C'(4,7)[/tex]

Hence, the vertices of image of triangle is given by

A'(-3,0), B'(0,-3) and C'(4,7)

Answer:

x = 27°

Step-by-step explanation:

180° - 72° = 108°

108° / 4 = 27°

72° + (27° * 4) = 180°

Hope it helps!

Answer:

x - 3y = 24

B. y = -3x - 1

Step-by-step explanation:

Hope it helped.

° ° °

Answer:

UWUNOE BRO HUUZEM HIHUDHS

Step-by-step explanation:

Answer:

Step-by-step explanation:

When we take an average of something, we have to add up all the data on the somethings and then divide by the number of somethings we have. Ed takes 5 tests, and we have scores for them; we also have his current average. What we don't know for sure are 2 of the 5 test scores, but we have enough to determine what they are.

If one test has a score of x, and the other test is 3 points less than that, the score on that last test is x - 3. Putting all of that together into an average problem:

[tex]\frac{87+85+87+x+(x-3)}{5}=90[/tex] and simplfiying a bit:

[tex]\frac{256+2x}{5}=90[/tex]. Multiply both sides by 5 to get

256 + 2x = 450; subtract 256 from both sides to get

2x = 194 and divide by 2:

x = 97

The one test score was a 97 and the other one, which was 3 less than that, was a 94.

Answer:

x = 37.3

Step-by-step explanation:

Since this is a right triangle, we can use trig functions

tan theta = opp/ adj

tan 63 = x /19

19 tan 63 =x

x=37.28959

To the nearest tenth

x = 37.3

Answer:

b

Step-by-step explanation:

area of triangle = 1/2 x c x d =cd/2

area of semicircle = 1/2 x π x r^2 = 1/2 x π x (a/2)^2 = 1/2 x π x a^2/4 =πa^2/8

area of shape = area of triangle + area of semicircle

===========================================================

Explanation:

Your teacher shows that minor arc BD is 129 degrees. Recall that any minor arc is always less than 180 degrees.

Let's say that we had point A at the center of the circle. This would mean angle DAB is 129 degrees. This central angle subtends or cuts off minor arc BD.

Now focus entirely on quadrilateral DABC. The goal is to find the unknown angle C. The angle A was found earlier at 129 degrees. The angles B and D are 90 degrees each since tangents are perpendicular to the radius at the point of tangency.

We have this so far

Adding all four angles of any quadrilateral will always get us 360 degrees. It's similar to how adding the angles of a triangle gets us 180 degrees. In fact, any quadrilateral can be cut into two triangles (aka the process of triangulation). So adding two triangles gets us 180+180 = 360 more or less.

--------------

Anyways, the four angles A,B,C,D must add to 360. So,

A+B+C+D = 360

129+90+C+90 = 360

C+309 = 360

C = 360-309

C = 51

--------------

As a shortcut, note how B+D = 90+90 = 180

Also note that A+C = 360-(B+D) which becomes A+C = 180

So, C = 180-A = 180-129 = 51

Answer:

f(x) -5x

g(x) -85x²

f(x) . g(x)

-90x

Answer:

36x³ + 25x² + 12x + 2

Step-by-step explanation:

f(x) × g(x)

= (- 4x - 1)(- 9x² - 4x - 2)

Each term in the second factor is multiplied by each term in the first factor

- 4x(- 9x² - 4x - 2) - 1 (- 9x² - 4x - 2) ← distribute parenthesis

= 36x³ + 16x² + 8x + 9x² + 4x + 2 ← collect like terms

= 36x³ + 25x² + 12x + 2

Answers:

===========================================================

Explanation:

I recommend drawing the two triangles as shown below. The diagram isn't to scale, but all that matters is that we get the general idea down.

Your teacher gave you this info:

These two pieces of info are enough to prove that triangle ABC is similar to triangle EDF. Use the angle angle (AA) similarity theorem to do this proof. The order of the lettering matters because it helps us see how the letters pair up.

Since the triangles are similar, the scale factor to go from ABC to EDF is EF/AC = 2/6 = 1/3

In other words, the side lengths of triangle EDF are exactly 1/3 that of the corresponding lengths of ABC. Or we can reverse things to say the sides of ABC are 3 times longer than the sides of EDF.

----------------------

Use this scale factor to help determine the missing length ED.

(ED)/(AB) = scale factor

(ED)/(AB) = 1/3

(ED)/(3.3) = 1/3

ED = (1/3)*(3.3)

ED = 1.1

Note how ED/AB = (1.1)/(3.3) = 11/33 = 1/3

----------------------

While we don't know how long sides BC or DF are, we are told that DF is 3.2 units shorter compared to BC.

This means

DF = BC - 3.2

DF = x - 3.2

Then we can say,

(DF)/(BC) = scale factor

(x-3.2)/(x) = 1/3

3(x - 3.2) = x*1

3x - 9.6 = x

3x - x = 9.6

2x = 9.6

x = (9.6)/2

x = 4.8

So BC is x = 4.8 units long and DF is x-3.2 = 4.8-3.2 = 1.6 units long

Answer:

Step-by-step explanation:

This is simple, but the extra numbers given in the form of the specific times might throw you off.

If Garrett leaves at noon and at 7 Ben is some distance ahead of him, that means that Garrett has been driving for 7 hours. Ben left at 2, so at 7 pm he has been driving for 5 hours. That's part of what's confusing. We'll put that in a table to hopefully make things easier:

              d      =       r      *        t

G                                             7

B                                             5

We also know that Garrett is driving at 10 km/h, so:

              d        =        r        *        t

G                               10       *        7

B                                 r                 5

The r is because Ben's rate is our unknown. Look at the top of the table. That is the formula we are going to use to solve this problem: d = rt.

If Garrett drives for 7 hours at 10 km/hr, then the distance he has traveled is 70 km (that's found by multiplying the rate of 10 km/h by the time of 7 hours). Ben's rate, along those same lines of reasoning, is 5r. Fill that in:

            d        =        r        *        t

G         70       =       10       *       7

B         5r        =        r        *       5

Ok now the table is filled out. Let's look at the rest of the problem. It says that at 7 pm Ben's distance is 15 km more than Garrett's distance. The words "more than" indicate addition. In words that is

"Ben's distance is Garrett's distance plus 15 km" which translates to, mathematically speaking:

5r = 70 + 15 and

5r = 85 so

r = 17

Ben's rate is 17 km/h, choice a.

Answer:

Yes it can be right angle triangle

Answer:

V = 8 m/s

Step-by-step explanation:

 Assuming that the object was at rest, so μ = 0

Equations

F=ma - Newtons 2nd law

[tex]v^{2} = u^{2} + 2a[/tex]Δx - 4th kinematic equatioin

Step 1 - find "a"

F=m/a

a=F/m

a=80/10

a=8m/s^2

Step 2 - find "v"  

v^2 = 0 + 2 * 8 * 4

v^2 = 64    

v=8m/s

Answer:

71°

Step-by-step explanation:

12/36 = 1/3

inverse cos(1/3) = 70.53 = 71

2^2 + 2^2 = h^2

8 = h^2

√8 = h

h = 2.82mm

Answered by Gauthmath must click thanks and mark brainliest

Answer:

7

Step-by-step explanation:

Concepts:

Solving:

1. First, let's set up the equation:

7/8 ÷ 1/8

2. We can get the reciprocal of 1/8 by doing the opposite of division; multiplication. 1/8 becomes 1 · 8, and that's equal to 8.

3. Now, since we know dividing a fraction by another one is exactly the same as multiplying the fraction by the reciprocal of the other, let's multiply 7/8 by the inverse of 1/8, which is 8.

7/8 · 8 = 56/8 = 7

Therefore, 7/8 divided by 1/8 is equal to 7.

Answer:

here,

3x-3x=

[tex] {x}^{2} - {x}^{2} [/tex]

x=0

Answer:

25%

Step-by-step explanation:

1 ÷ 4 = 0.25

0.25 x 100 = 25 = 25%

Answer:

Step-by-step explanation:

the scale factor is 2. one length is 10, and the other is 5. pythagorean therom approved.

Answer:

[tex]\angle m + \angle t + \angle n = 180[/tex]

Step-by-step explanation:

Required

Show that:

[tex]\angle m + \angle t + \angle n = 180^o[/tex]

To make the proof easier, I've added a screenshot of the triangle.

We make use of alternate angles to complete the proof.

In the attached triangle, the two angles beside [tex]\angle m[/tex] are alternate to [tex]\angle t[/tex] and [tex]\angle n[/tex]

i.e.

[tex]\angle 1 = \angle t[/tex]

[tex]\angle 2 = \angle n[/tex]

Using angle on a straight line theorem, we have:

[tex]\angle 1 + \angle m + \angle 2 = 180[/tex]

Substitute values for (1) and (2)

[tex]\angle t + \angle m + \angle n = 180[/tex]

Rewrite as:

[tex]\angle m + \angle t + \angle n = 180[/tex] -- proved