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1 year ago

Determine the x- and y-intercepts of the graph ofx−4y=8Then plot the intercepts to graph the equation.

Mathematics

1 answer:

Over [174]1 year ago

8 0

The x-intercept is the point where the graph cuts the x-axis, and the y-intercept is the point where the graph cuts the y-axis.

The x-axis is the line y = 0 and the y-axis is the line x = 0. To find the intercept between each axis and our graph, we just need to evaluate our function at x = 0 and y = 0.

Calculating the x-intercept, we have

$\begin{gathered} x-4\cdot0=8 \\ x=8 \end{gathered}$

The x-intercept is (8, 0).

Calculating the y-intercept, we have

$\begin{gathered} 0-4y=8 \\ y=-\frac{8}{4} \\ y=-2 \end{gathered}$

The y-intercept is (0, -2).

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Feliz [49]

Answer:

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7 0

3 years ago

Find the product of the given complex number and its conjugate.<br> 1/5+4i

nasty-shy [4]

Answer:

-15.96

Step-by-step explanation:

A conjugate is a binomial with the sign inside changed. So the conjugate of (1/5 + 4i) is (1/5 - 4i)

Set the original and the conjugate next to each other and F.O.I.L. Multiply the first numbers of each binomial, the 1/5 and the 1/5 to get 1/25. This is the "F."

Multiply the outer members, the 1/5 and the 4i to get - 60i. This is the "O."

Multiply the inner numbers ( the + 4i and the 1/5) to get + 60i. This is the "I."

Multiply the positive 4i and the negative 4i to get 16i squared

The positive 60i and the negative 60i cancel each other out.

The i squared changes into - 1. This makes the 16 negative.

Add 1/25 to - 16 to get - 15.96

4 0

3 years ago

Simplify the expression given below: (18x^3-7)-(14x^3-36)

vovangra [49]

Answer:

2x6 + 4x5 + x4 + 11x3 + 2x2 + 4x + 4

Step-by-step explanation:

6 0

3 years ago

What was the property of equality

loris [4]

Reflexive Property=For all real number X,X=X

A number equals itself

Addition Property=For all real number X,y & Z,. if

X=Y then X+y=y+Z

Subtraction Property=For all real number X,Y And Z if X=y then X-Z=y-Z.

7 0

3 years ago

Let a1, a2, a3, ... be a sequence of positive integers in arithmetic progression with common difference

Bezzdna [24]

Since $a_1,a_2,a_3,\cdots$ are in arithmetic progression,

$a_2 = a_1 + 2$

$a_3 = a_2 + 2 = a_1 + 2\cdot2$

$a_4 = a_3+2 = a_1+3\cdot2$

$\cdots \implies a_n = a_1 + 2(n-1)$

and since $b_1,b_2,b_3,\cdots$ are in geometric progression,

$b_2 = 2b_1$

$b_3=2b_2 = 2^2 b_1$

$b_4=2b_3=2^3b_1$

$\cdots\implies b_n=2^{n-1}b_1$

Recall that

$\displaystyle \sum_{k=1}^n 1 = \underbrace{1+1+1+\cdots+1}_{n\,\rm times} = n$

$\displaystyle \sum_{k=1}^n k = 1 + 2 + 3 + \cdots + n = \frac{n(n+1)}2$

It follows that

$a_1 + a_2 + \cdots + a_n = \displaystyle \sum_{k=1}^n (a_1 + 2(k-1)) \\\\ ~~~~~~~~ = a_1 \sum_{k=1}^n 1 + 2 \sum_{k=1}^n (k-1) \\\\ ~~~~~~~~ = a_1 n + n(n-1)$