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3 years ago

Create a linear equation in slope intercept form that contains the points (1,4) and (4,7).

Mathematics

1 answer:

S_A_V [24]3 years ago

4 0

Answer:

y = x + 3

Step-by-step explanation:

The equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y- intercept )

Calculate m using the slope formula

m = (y₂ - y₁ ) / (x₂ - x₁ )

with (x₁, y₁ ) = (1, 4) and (x₂, y₂ ) = (4, 7)

m = $\frac{7-4}{4-1}$ = $\frac{3}{3}$ = 1, hence

y = x + c ← is the partial equation

To find c substitute either of the 2 points into the partial equation.

Using (1, 4), then

4 = 1 + c ⇒ c = 4 - 1 = 3

y = x + 3 ← equation in slope- intercept form

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6. Find the function with x-intercepts (-3,0) and (5,0), which also goes through the point (1,8)

Yakvenalex [24]

Answer:

f(x) = (-1/2)(x^2 + 8x - 15)

Step-by-step explanation:

This function has two roots: -3 and 5. Most likely it is a quadratic (all of which have two roots).

Then f(x) = a(x + 3)(x - 5)

The graph goes through (1. 8): Therefore, y = 8 when x = 1:

f(1) = a(1 + 3)(1 - 5) = 8, or

a(4)(-4) = 8, or

-16a = 8, which leads to a = -1/2.

Thus the quadratic in question is f(x) = (-1/2)(x + 3)(x - 5), or

f(x) = (-1/2)(x^2 + 8x - 15)

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2 years ago

Add the two expressions.<br><br> −2.4n−3 and −7.8n+2

Sphinxa [80]

Answer:

-10.2n - 1

Step-by-step explanation:

We have two expressions in variable n and we have to add the two expressions.

An important thing to note is that only like terms can be added. i.e. the term with "n" can only be added or subtracted to the term with "n". Similarly a constant can only be added or subtracted to a constant.

$(-2.4n-3)+(-7.8n+2)\\=-2.4n-3-7.8n+2\\=-2.4n-7.8n-3+2\\=-10.2n-1$

Thus, the two given expressions add up to -10.2n - 1

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2 years ago

A jar contains n nickels and d dimes. There are 28 coins in the jar,and total value of the coins is $2.20. How many nickels and

Harman [31]

N + d = 20 (there are a total of 20 nickels and dimes).05n + .1d = 1.4 (total change equals $1.40)

7 0

3 years ago

Prove the identity secxcscx(tanx+cotx)=2+tan^2x+cot^2x

svetlana [45]

Hello,

$sec(x)= \dfrac{1}{cos(x)} \\

cosec(x)= \dfrac{1}{sin(x)} \\

sec(x)*cosec(x)*(tg(x)+cotg(x))=\dfrac{1}{cos(x)}* \dfrac{1}{sin(x)}*( \frac{sin(x)}{cos(x)} +\frac{cos(x)}{sin(x)})\\

= \dfrac{sin^2(x)+cos^2(x)}{sin^2x*cos^2x} \\

= \dfrac{1}{sin^2x*cos^2x} \\

$
==============================================================
$2+tg^2(x)+cotg^2(x)=2+ \dfrac{sin^2x}{cos^2x} + \dfrac{cos^2x}{sin^2x} \\

=2+ \dfrac{sin^4x+cos^4x}{sin^2x*cos^2x} \\

=\dfrac{2*sin^2x*cos^2x+sin^4x+cos^4x}{sin^2x*cos^2x} \\

= \dfrac{(sin^2x+cos^2x)^2}{sin^2x*cos^2x}} \\

= \dfrac{1}{sin^2x*cos^2x}} 
$

8 0

3 years ago

Read 2 more answers

a student must have an average on five test that is greater than it equal to 80% but less than 90% to receive a final grade of B

Bad White [126]

Answer:

89–100

Step-by-step explanation:

Let x represent the grade on the fifth test. The average for 5 tests wants to be between 80 and 90, so we have ...

80 ≤ (78 +62 +91 +80 +x)/5 ≤ 90

Multiplying by 5 gives

400 ≤ 311 +x ≤ 450

Subtracting 311 gives the range for the fifth test grade.

89 ≤ x ≤ 139

Since the maximum available score is 100, the range on the 5th test for an overall B grade is ...

89 ≤ x ≤ 100 . . . . where x is the grade on the 5th test

6 0

3 years ago

Create a linear equation in slope intercept form that contains the points (1,4) and (4,7).​

Create a linear equation in slope intercept form that contains the points (1,4) and (4,7).