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Vlad1618 [11]
3 years ago
5

SOMEONE PLEASE HELP I DONT UNDERSTAND THIS . ILL GIVE YOU BRAINLIST .

Mathematics
1 answer:
GarryVolchara [31]3 years ago
5 0

Answer:

d: 56

e: 68

f: 56

Step-by-step explanation:

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Could you please answer this precisely. Thank you!
OleMash [197]

The mathematical functions (equations) are matched to their graph line respectively as follows:

  • y = 960 - 78x: purple straight line.
  • y = 8·3^x: violet line curve that slopes to the left.
  • y = 960·(1/2)^x: green line curve that slopes to the right.

<h3>What is a linear function?</h3>

A linear function can be defined as a type of function whose equation is graphically represented by a straight line on the cartesian coordinate. Mathematically, a linear function is given by this equation:

y = mx ± c.

<h3>What is an exponential function?</h3>

An exponential function can be defined as a type of mathematical function whose values are generated by a constant that is raised to the power of the argument. Mathematically, an exponential function is represented by this formula:

f(x) = keˣ

Next, we would match each of the mathematical functions (equations) to their graph line respectively as follows:

  • y = 960 - 78x: purple straight line.
  • y = 8·3^x: violet line curve that slopes to the left.
  • y = 960·(1/2)^x: green line curve that slopes to the right.

Read more on graphs here: brainly.com/question/24298987

#SPJ1

3 0
2 years ago
What is the result of substituting for y in the bottom equation<br> y=x+3<br> y=x^2+2x-4
8_murik_8 [283]

The solutions are (2.1925, 5.1925) and (-3.1925, -0.1925)

<em><u>Solution:</u></em>

Given that,

y = x + 3 ------- eqn 1\\\\y = x^2 + 2x - 4  ----- eqn 2

<em><u>We have to substitute eqn 1 in eqn 2</u></em>

x + 3 = x^2 + 2x - 4

\mathrm{Switch\:sides}\\\\x^2+2x-4=x+3\\\\\mathrm{Subtract\:}3\mathrm{\:from\:both\:sides}\\\\x^2+2x-4-3=x+3-3\\\\\mathrm{Simplify}\\\\x^2+2x-7=x\\\\\mathrm{Subtract\:}x\mathrm{\:from\:both\:sides}\\\\x^2+2x-7-x=x-x\\\\\mathrm{Simplify}\\\\x^2+x-7=0

\mathrm{Solve\:with\:the\:quadratic\:formula}\\\\\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}\\\\x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\\\\mathrm{For\:}\quad a=1,\:b=1,\:c=-7\\\\x =\frac{-1\pm \sqrt{1^2-4\cdot \:1\left(-7\right)}}{2\cdot \:1}

x = \frac{-1 \pm \sqrt{ 1 + 28}}{2}\\\\x = \frac{ -1 \pm \sqrt{29}}{2}

x = \frac{ -1 \pm 5.385 }{2}\\\\We\ have\ two\ solutions\\\\x = \frac{ -1 + 5.385 }{2}\\\\x = 2.1925

Also\\\\x = \frac{ -1 - 5.385 }{2}\\\\x = -3.1925

Substitute x = 2.1925 in eqn 1

y = 2.1925 + 3

y = 5.1925

Substitute x = -3.1925 in eqn 1

y = -3.1925 + 3

y = -0.1925

Thus the solutions are (2.1925, 5.1925) and (-3.1925, -0.1925)

6 0
4 years ago
Write with fractional exponents. 6^5 square root x^2 y​
sertanlavr [38]

Answer:

6xy^2/5 what im assuming is right

Step-by-step explanation:

4 0
3 years ago
CAN somebody plz plz help me with raios
Kaylis [27]
I can help what are the problems
5 0
3 years ago
Please tell me 1-3 which is similar and not thanks
gogolik [260]
The second and the third one are similar.
7 0
3 years ago
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