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Softa [21]
3 years ago
11

The cost of the studios internet service is given by the formula c=90+30x, where x is the number of months of usage. What is the

studios cost for 11 moths of internet?
Mathematics
1 answer:
blondinia [14]3 years ago
6 0

Answer:

$420

Step-by-step explanation:

C=90+30(11)

11x30=330

330+90=420

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Two rigid transformations are used to map δhjk to δlmn. The first is a translation of vertex h to vertex l. What is the second t
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B a rotation about point H

Step-by-step explanation:

Two rigid transformations are used to map ΔHJK to ΔLMN. The first is a translation of vertex H to vertex L. What is the second transformation?

a reflection across the line containing HK

a rotation about point H

a reflection across the line containing HJ

a rotation about point K

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G the population of a town today is 48000 people. if the population decreases with a constant rate of change and the decrease is
tiny-mole [99]
The population in ten years will be 208,000
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3 years ago
In triangle KLM, if K is congruent to L, KL = 9x - 40, LM = 7x - 37, & KM = 3x + 23, find x & the measure of each angle.
Dennis_Churaev [7]

Answer: x = 15, ∠K = 45.7°, ∠L = 45.7°, ∠M = 88.6°

<u>Step-by-step explanation:</u>

Since ∠K ≅ ∠L, then ΔKLM is an isoceles triangle with base KL

 KM ≅ LM

3x + 23 = 7x - 37

       23 = 4x - 37

       60 = 4x

        15 = x

KM = LM = 3x + 23

               = 3(15) + 23

               = 45 + 23

               = 68

KL = 9x - 40

    = 9(15) - 40

    = 135 - 40

    = 95

Next, draw a perpendicular bisector KN from K to KL. Thus, N is the midpoint of KL and ΔMNL is a right triangle.

  • Since N is the midpoint of KL and KL = 95, then NL = 47.5
  • Since ∠N is 90°, then NL is adjacent to ∠l and ML is the hypotenuse

Use trig to solve for ∠L (which equals ∠K):

cos ∠L = \frac{adjacent}{hypotenuse}

cos ∠L = \frac{47.5}{68}

      ∠L = cos⁻¹ (\frac{47.5}{68})

      ∠L = 45.7  

Triangle sum Theorem:

∠K + ∠L + ∠M = 180°    

45.7 + 45.7 + ∠M = 180

       91.4     + ∠M = 180

                      ∠M = 88.6

       

7 0
3 years ago
If ​f(x)=x^2- 4 and ​g(x)=x^2 plus 2 x​,
gregori [183]

Answer:

\huge\boxed{(f-g)\left(-\dfrac{1}{3}\right)=-3\dfrac{1}{3}}

Step-by-step explanation:

\text{We have}:\\\\f(x)=x^2-4\ \text{and}\ g(x)=x^2+2x\\\\(f-g)(x)=f(x)-g(x)\\\\\text{therefore}\\\\(f-g)(x)=(x^2-4)-(x^2+2x)=x^2-4-x^2-2x\\\\=(x^2-x^2)-2x-4=-2x-4\\\\(f-g)\left(-\dfrac{1}{3}\right)\to\text{put}\ x=-\dfrac{1}{3}\ \text{to the equation of the function}\ (f-g)(x):\\\\(f-g)\left(-\dfrac{1}{3}\right)=-2\left(-\dfrac{1}{3}\right)-4=\dfrac{2}{3}-4=-3\dfrac{1}{3}

6 0
3 years ago
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