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3 years ago

Which expression shows the first step in dividing 5x + 3 by x?

Mathematics

1 answer:

prisoha [69]3 years ago

4 0

First you need to make table value
Second y=5x+3
Y is your output and x is your input

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Please help me with this math problem

goldfiish [28.3K]

If you look at the graph you can tell the graph is increasing before x = -2.

From x = -2 to x = 0, it's decreasing.

Then it's increasing again from x = 0 to x = 2, then decreasing after x = 2

So answer is the last one
It is increasing before x = -2 and from x = 0 to x = 2

3 0

3 years ago

Question 13

kicyunya [14]

Answer:

The c intercept is 42

The t intercepts are: 6, -1 and 7

Step-by-step explanation:

Given

$c(t) = (t - 6)(t +1)(t-7)$

Solving (a): The c intercept

Simply set t to 0

$c(t) = (t - 6)(t +1)(t-7)$

$c(0) = (0 - 6)(0 +1)(0-7)$

$c(0) = (- 6)(1)(-7)$

$c(0) = 42$

Solving (b): The t intercept

Simply set c(t) to 0

$c(t) = (t - 6)(t +1)(t-7)$

$(t - 6)(t +1)(t-7) = 0$

Split

$t - 6= 0,\ \ t +1= 0,\ \ t-7 = 0$

Solve for t

$t = 6,\ \ t =-1,\ \ t=7$

4 0

2 years ago

Darlene wants to buy a sweater that originally cost 32$ it is on sale for 20% off. Sales tax is 6.75%. What is the final price o

AURORKA [14]

Answer:

Step-by-step explanation:

32 x 0.20 = $6.40

32 - 6.4 = $25.60

25.6 x .0675 ≈ 1.738

25.6 + 1.738 = 27.338 ≈$27.34

6 0

3 years ago

How do I find the domain of a function

Aliun [14]

It is how much the x values can expand on a graph

6 0

3 years ago

Suppose you are investigating the relationship between two variables, traffic flow and expected lead content, where traffic flow

prisoha [69]

Answer:

The answers is " Option B".

Step-by-step explanation:

$CI=\hat{Y}\pm t_{Critical}\times S_{e}$

Where,

$\hat{Y}=$ predicted value of lead content when traffic flow is 15.

$\to df=n-1=8-1=7$

$95\% \ CI\ is\ (463.5, 596.3) \\\\\hat{Y}=\frac{(463.5+596.3)}{2}\\\\$

$=\frac{1059.8}{2}\\\\=529.9$

Calculating thet-critical value $t_{ \{\frac{\alpha}{2},\ df \}}=-2.365$

The lower predicted value $=529.9-2.365(Se)$

$463.5=529.9-2.365(Se)\\\\529.9-463.5=2.365(Se)\\\\66.4=2.365(Se)\\\\Se=\frac{66.4}{2.365} \\\\Se=28.076$

When $99\%$ of CI use as the expected lead content: $\to 529.9\pm t_{0.005,7}\times 28.076 \\\\=(529.9 \pm 3.499 \times 28.076)\\\\=(529.9 \pm 98.238)\\\\=(529.9-98.238, 529.9+98.238)\\\\=(431.662, 628.138)\\\\=(431.6, 628.1)$

8 0

2 years ago