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3 years ago

2.) Write an equation of the line with a slope of —3 and y-intercept of 0. y=

Mathematics

1 answer:

vovangra [49]3 years ago

8 0

Answer:

y= -3x

Step-by-step explanation:

Put the given numbers into the formula y= mx + b

m= the slope, which is -3

x= variable

b= y-int/constant

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Help solving this problem, the second problem not the third

yawa3891 [41]

Given:

1 Van and 4 buses filled with 195 students.

1 Van and 5 buses filled with 241 students.

Let x and y be the number of students in Van and number of students in bus.

$\begin{gathered} x+4y=195\ldots\text{ (1)} \\ x+5y=241\ldots\text{ (2)} \end{gathered}$

Subtract equation (1) from equation (2)

$\begin{gathered} x+5y-x-4y=241-195 \\ y=46 \end{gathered}$

Substitute y=46 in equation (1)

$\begin{gathered} x+4(46)=195 \\ x+184=195 \\ x=195-184 \\ x=11 \end{gathered}$

Number of students that van can carry is 11.

Number of students that bus can carry is 46

7 0

1 year ago

The answers are, s=0.20p

larisa86 [58]

Answer:

s=0.20p

Step-by-step explanation:

7 0

3 years ago

What is the sequence for g(1)=-29 g(n)=g(n-1)x(-4)

kupik [55]

Answer:

g(n) = -29×(-4)ⁿ⁻¹

Step-by-step explanation:

This the recursive formula of a sequence g :

g(1)=-29

g(n)=g(n-1)x(-4)

now ,we want to deduce the standard explicit formula of that sequence

It’s clear from the recursive formula (above) that :

_The common ratio = -4

_the first term = -29

Then

the standard explicit formula is : g(n) = -29×(-4)ⁿ⁻¹

8 0

3 years ago

Determine whether the given description corresponds to an experiment or observational study.a. A stock analyst selects a stock f

ICE Princess25 [194]

Answer:

A: experiment

B: observational study

C: observational study

D: experiment

Step-by-step explanation:

3 0

4 years ago

Use Definition 7.1.1. DEFINITION 7.1.1 Laplace Transform Let f be a function defined for t ≥ 0. Then the integral ℒ{f(t)} = [inf

Alla [95]

$f(t)=\begin{cases}-1&\text{for }0\le t$

Write $f(t)$ in terms of the unit step function $u(t)$ :

$f(t)=-1(u(t)-u(t-1))+1u(t-1)=2u(t-1)-u(t)$

where

$u(t)=\begin{cases}1&\text{for }t\ge0\\0&\text{for }t$

Then the Laplace transform of $f(t)$ , denoted $F(s)$ , is

$F(s)=\displaystyle\int_0^\infty f(t)e^{-st}\,\mathrm dt$

$F(s)=\displaystyle2\int_1^\infty e^{-st}\,\mathrm dt-\int_0^\infty e^{-st}\,\mathrm dt$

$\implies F(s)=\dfrac{2e^{-s}}s-\dfrac1s=\dfrac{2e^{-s}-1}s$

3 0

3 years ago