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2 years ago

How would you graph y-intercept is 8 and x-intercept is 9

Mathematics

2 answers:

vitfil [10]2 years ago

7 0

Answer:

all shown below

Step-by-step explanation:

y-intercept means a dot's y coordinate is 8 and on the y axis, x-i intercept means a dot's x coordinate is 9 and on the x axis

ivann1987 [24]2 years ago

4 0

Answer:

see the attached graph

Step-by-step explanation:

The intercepts are the points where a graph crosses the coordinate axes.

<h3>Y-intercept</h3>

The y-intercept is the the point where the graph crosses the y-axis. The value of x is zero there. The coordinates of the y-intercept are (0, 8).

<h3>X-intercept</h3>

The x-intercept is the point where the graph crosses the x-axis. The value of y is zero there. The coordinates of the x-intercept are (9, 0).

You graph these intercepts by plotting the points at these coordinates.

In the attached graph, the intercepts are blue dots, and we have drawn a line through them. The line is not part of the graph of the intercepts. It is merely an instance of a function that has these intercepts.

You might be interested in

Find the y and the x intercepts of the graph of the equation of y=3x+1

BaLLatris [955]

Answer:

x-intercept ⇨ -1/3

y-intercept ⇨ 1

Step-by-step explanation:

⟺ Finding the x-intercept, substitute y = 0

$y=3x+1\\0=3x+1\\3x+1=0\\$

Move 1 to subtract the another side.

$3x=-1$

Then move 3 to divide -1, leaving only x as a subject since we want to find the x-intercept.

$x=-\frac{1}{3}$ #

⟺ Finding the y-intercept, substitute x = 0

$y=3x+1\\y=3(0)+1\\y=0+1$

$y=1$ #

Tips:

Here's the tips about finding the intercepts of the graph.

⟺ For x-intercept, It's like solving an equation to find the x-term.

⟺ For y-intercept, It's like using the constant to answer.

As for y = mx+b where m = slope and b = y-intercept.

For a linear function, It's not necessary to substitute x = 0 just to find y-intercept as we can answer the constant as our y-intercept.

7 0

3 years ago

Need help with this problem.

Galina-37 [17]

Answer:

Step-by-step explanation:

To find this we must divide 45 by 20 which gives the answer of 2.25. As we know this we now must multiply that by 8 resulting in to 18.

Therefore the answer is 18

5 0

3 years ago

Jose received $250 for his birthday from his family. He wishes to buy a motorcycle and decides to use his birthday money towards

BartSMP [9]

Answer:

He will have $276.10 available towards the down payment for his motorcycle

Step-by-step explanation:

The compound interest formula is given by:

$A = P(1 + \frac{r}{n})^{nt}$

Where A is the amount of money, P is the principal(the initial sum of money), r is the interest rate(as a decimal value), n is the number of times that interest is compounded per unit t and t is the time the money is invested or borrowed for.

In this problem, we have that:

$P = 250, t = 2, r = 0.05$

Compounded quarterly, so n = 12/4 = 3.

We have to find A.

$A = P(1 + \frac{r}{n})^{nt}$

$A = 250(1 + \frac{0.05}{3})^{3*2}$

$A = 276.1$

He will have $276.10 available towards the down payment for his motorcycle

8 0

3 years ago

Overview

Alisiya [41]

Answer:

125.4

Step-by-step explanation:

Given

$Number = 125.3546$

Required

Round to 1 decimal place

Up till the first decimal place, the number is:

$Number = 125.3$

The digit after .3 is 5

The conditions for approximation are:

If n > 4, approximate to 1

Else: approximate to 0

In this case: 5 > 4, so we approximate to 1

Add this "1" to the last digit of 125.3. This becomes 125.4

Hence: when the number is approximated to 1 decimal place, the digit is 125.4

7 0

3 years ago

How to solve this trig

n200080 [17]

Hi there!

To find the Trigonometric Equation, we have to isolate sin, cos, tan, etc. We are also given the interval [0,2π).

First Question

What we have to do is to isolate cos first.

$\displaystyle \large{ cos \theta = - \frac{1}{2} }$

Then find the reference angle. As we know cos(π/3) equals 1/2. Therefore π/3 is our reference angle.

Since we know that cos is negative in Q2 and Q3. We will be using π + (ref. angle) for Q3. and π - (ref. angle) for Q2.

Find Q2

$\displaystyle \large{ \pi - \frac{ \pi}{3} = \frac{3 \pi}{3} - \frac{ \pi}{3} } \\ \displaystyle \large \boxed{ \frac{2 \pi}{3} }$

Find Q3

$\displaystyle \large{ \pi + \frac{ \pi}{3} = \frac{3 \pi}{3} + \frac{ \pi}{3} } \\ \displaystyle \large \boxed{ \frac{4 \pi}{3} }$

Both values are apart of the interval. Hence,

$\displaystyle \large \boxed{ \theta = \frac{2 \pi}{3} , \frac{4 \pi}{3} }$

Second Question

Isolate sin(4 theta).

$\displaystyle \large{sin 4 \theta = - \frac{1}{ \sqrt{2} } }$

Rationalize the denominator.

$\displaystyle \large{sin4 \theta = - \frac{ \sqrt{2} }{2} }$

The problem here is 4 beside theta. What we are going to do is to expand the interval.

$\displaystyle \large{0 \leqslant \theta < 2 \pi}$

Multiply whole by 4.

$\displaystyle \large{0 \times 4 \leqslant \theta \times 4 < 2 \pi \times 4} \\ \displaystyle \large \boxed{0 \leqslant 4 \theta < 8 \pi}$

Then find the reference angle.

We know that sin(π/4) = √2/2. Hence π/4 is our reference angle.

sin is negative in Q3 and Q4. We use π + (ref. angle) for Q3 and 2π - (ref. angle for Q4.)

Find Q3

$\displaystyle \large{ \pi + \frac{ \pi}{4} = \frac{ 4 \pi}{4} + \frac{ \pi}{4} } \\ \displaystyle \large \boxed{ \frac{5 \pi}{4} }$

Find Q4

$\displaystyle \large{2 \pi - \frac{ \pi}{4} = \frac{8 \pi}{4} - \frac{ \pi}{4} } \\ \displaystyle \large \boxed{ \frac{7 \pi}{4} }$

Both values are in [0,2π). However, we exceed our interval to < 8π.

We will be using these following:-

$\displaystyle \large{ \theta + 2 \pi k = \theta \: \: \: \: \: \sf{(k \: \: is \: \: integer)}}$

Hence:-

For Q3

$\displaystyle \large{ \frac{5 \pi}{4} + 2 \pi = \frac{13 \pi}{4} } \\ \displaystyle \large{ \frac{5 \pi}{4} + 4\pi = \frac{21 \pi}{4} } \\ \displaystyle \large{ \frac{5 \pi}{4} + 6\pi = \frac{29 \pi}{4} }$

We cannot use any further k-values (or k cannot be 4 or higher) because it'd be +8π and not in the interval.

For Q4

$\displaystyle \large{ \frac{ 7 \pi}{4} + 2 \pi = \frac{15 \pi}{4} } \\ \displaystyle \large{ \frac{ 7 \pi}{4} + 4 \pi = \frac{23\pi}{4} } \\ \displaystyle \large{ \frac{ 7 \pi}{4} + 6 \pi = \frac{31 \pi}{4} }$

Therefore:-

$\displaystyle \large{4 \theta = \frac{5 \pi}{4} , \frac{7 \pi}{4} , \frac{13\pi}{4} , \frac{21\pi}{4} , \frac{29\pi}{4}, \frac{15 \pi}{4} , \frac{23\pi}{4} , \frac{31\pi}{4} }$

Then we divide all these values by 4.

$\displaystyle \large \boxed{\theta = \frac{5 \pi}{16} , \frac{7 \pi}{16} , \frac{13\pi}{16} , \frac{21\pi}{16} , \frac{29\pi}{16}, \frac{15 \pi}{16} , \frac{23\pi}{16} , \frac{31\pi}{16} }$

Let me know if you have any questions!

3 0

3 years ago