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3 years ago

What is a possible equation for a function with an x-intercept of (-5,0) and a y-intercept of (0,3)?

Mathematics

1 answer:

Feliz [49]3 years ago

5 0

Answer:

The equation of a straight line is

-3 x+5 y=15

Step-by-step explanation:

Given x - intercept of (-5,0) and y- intercept is (0,3)

here (-5,0 ) point lie on x- axis and (0,3) this point lie on y- axis

we know that the x- intercept 'a' and y- intercept 'b' formula is

$\frac{x}{a} +\frac{y}{b} = 1$

so given x - intercept a =-5 and y- intercept is b= 3

now the straight line equation is $\frac{x}{-5} +\frac{y}{3} =1$

now simplify $\frac{-3 x+5 y}{15} =1$

$-3 x+5 y=15$

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G(x) = x3 + 6x2 + 12x + 8<br><br><br><br> Determine the function’s value when x = −1.

salantis [7]

Answer:

g(-1) = 1

Step-by-step explanation:

Synthetic division is by far the fastest way to evaluate this function at x = -1. Set up synth. div. as follows:

-1 ) 1 6 12 8

-1 -5 -7

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

1 5 7 1

since the remainder is 1, g(-1) = 1

5 0

3 years ago

A coin is to be tossed as many times as necessary to turn up one head. Thus the elements c of the sample space C are H, TH, TTH,

slavikrds [6]

Answer:

Step-by-step explanation:

As stated in the question, the probability to toss a coin and turn up heads in the first try is $\frac{1}{2}$ , in the second is $\frac{1}{4}$ , in the third is $\frac{1}{8}$ and so on. Then, P(C) is given by the next sum:

$P(C)=\sum^{\infty}_{n=1}(\frac{1}{2} )^{n}=1$

This is a geometric series with factor $\frac{1}{2}$ . Then is convergent to $\frac{1}{1-\frac{1}{2}}-1=1.$ . With this we have proved that P(C)=1.

Now, observe that

$P(H)=\frac{1}{2}, P(TH)=\frac{1}{4},P(TTH)=\frac{1}{8},P(TTTH)=\frac{1}{16},P(TTTTH)=\frac{1}{32},P(TTTTTH)=\frac{1}{64}.$

Then

$P(C1)=P(H)+P(TH)+P(TTH)+P(TTTH)+P(TTTTH)=\frac{1}{2} +\frac{1}{4} +\frac{1}{8} +\frac{1}{16} +\frac{1}{32} =\frac{31}{32}$

$P(C2)=P(TTTTH)+P(TTTTTH)=\frac{1}{32}+\frac{1}{64} =\frac{3}{64}$

$P(C1\cap C2)=P(TTTTH)=\frac{1}{32}$

and

$P(C1\cup C2)=P(H)+P(TH)+P(TTH)+P(TTTH)+P(TTTTH)+P(TTTTTH)=\frac{1}{2} +\frac{1}{4} +\frac{1}{8} +\frac{1}{16} +\frac{1}{32} +\frac{1}{64}=\frac{63}{64}$