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4 years ago

Evaluate the function, f(m) = 13 - 3m, when m is 4 f(4) =

Mathematics

2 answers:

qwelly [4]4 years ago

8 0

Answer:

Step-by-step explanation:

f(4)=13-3m (where m=4)

f(4)= 13-3×4

f(4)= 14-12

hence f(4)= 1

djverab [1.8K]4 years ago

7 0

Answer:

f(4)= 13-3(4)

4f= 13-12

4f=1

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Points B, D, and F are midpoints of the sides of ΔACE.

Ainat [17]

Answer:

Step-by-step explanation:

D to c = 15, DtoB =17 so Cto B= 17

and A to C =34

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3 years ago

F(x) = -7x - 1<br> f() = -8

xxMikexx [17]

Answer:

f(1) = -8

Step-by-step explanation:

Apparently, you want the argument of the function such that the function value is -8. Fill in the given information and solve for x:

-8 = -7x -1

-7 = -7x . . . . add 1

1 = x . . . . . . . divide by -7

Now, you know that ...

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3 years ago

Question is in picture! answer asap! will give brainliest!

nataly862011 [7]

Answer:solve for x

-3x+5y=-15

Add -5y to both sides

-3x+5y+-5y=-15+-5y

-3x=-5y-15

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-3x/-3=-5y-15/-3

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I hope that's help!

Step-by-step explanation: plz give brainlist

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3 years ago

Sin theta+costheta/sintheta -costheta+sintheta-costheta/sintheta+costheta=2sec2/tan2 theta -1

sleet_krkn [62]

$\dfrac{sin\theta + cos\theta}{sin\theta-cos\theta}+\dfrac{sin\theta-cos\theta}{sin\theta+cos\theta}=\dfrac{2sec^2\theta}{tan^2\theta-1}$

From Left side:

$\dfrac{sin\theta + cos\theta}{sin\theta-cos\theta}\bigg(\dfrac{sin\theta+cos\theta}{sin\theta+cos\theta}\bigg)+\dfrac{sin\theta-cos\theta}{sin\theta+cos\theta}\bigg(\dfrac{sin\theta-cos\theta}{sin\theta-cos\theta}\bigg)$

$\dfrac{sin^2\theta+2cos\thetasin\theta+cos^2\theta}{sin^2\theta-cos^2\theta}+\dfrac{sin^2\theta-2cos\thetasin\theta+cos^2\theta}{sin^2\theta-cos^2\theta}$

NOTE: sin²θ + cos²θ = 1

$\dfrac{1 + 2cos\theta sin\theta}{sin^2\theta-cos^2\theta}+\dfrac{1-2cos\theta sin\theta}{sin^2\theta-cos^2\theta}$

$\dfrac{1 + 2cos\theta sin\theta+1-2cos\theta sin\theta}{sin^2\theta-cos^2\theta}$

$\dfrac{2}{sin^2\theta-cos^2\theta}$

$\dfrac{2}{\bigg(sin^2\theta-cos^2\theta\bigg)\bigg(\dfrac{cos^2\theta}{cos^2\theta}\bigg)}$

$\dfrac{2sec^2\theta}{\dfrac{sin^2\theta}{cos^2\theta}-\dfrac{cos^2\theta}{cos^2\theta}}$

$\dfrac{2sec^2\theta}{tan^2\theta-1}$

Left side = Right side <em>so proof is complete</em>

8 0

3 years ago

Read 2 more answers

At a local seaside, a vendor sells single-cone ice-creams for $3 and double-cone ice-creams for $4.50. The vendor stocks a maxim

guapka [62]

Answer:

The answer is below

Step-by-step explanation:

Let x represent the number of single cone ice cream and let y represent the number of double cone ice cream.

Since the vendor stocks a maximum of 70 single cones and a maximum of 45 double cones. hence:

0 < x ≤ 70, 0 < y ≤ 45 (1)

The vendor expects to sell no more than 50 ice creams, hence:

x + y ≤ 50

Plotting the constraint using geogebra online graphing tool, we can see that the solution to the problem is at (5, 45)

Since the vendor sells single-cone ice-creams for $3 and double-cone ice-creams for $4.50, hence:

Revenue = 3x + 4.5y

At the point (5, 45), the revenue is:

Revenue = 3(5) + 4.5(45) = $217.5

5 0

3 years ago

Evaluate the function, f(m) = 13 - 3m, when m is 4 f(4) = ​

Evaluate the function, f(m) = 13 - 3m, when m is 4 f(4) =