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Find the value of x. A. 35 B. 10 C. 55 D. 25

dolphi86 [110]

Answer:

A.35

Step-by-step explanation:

trust bro

5 0

3 years ago

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Find an equation of the line in the form ax+by=c whose x-intercept is 12 and y-intercept is 4, where a, b, and c are integer

Lostsunrise [7]

Answer:

The equation of the line in $ax+by=c$ form where $a,b,c$ are integers with no factor common to all three, and greater than or equal 0 is:

$x+3y=12$

Step-by-step explanation:

Given:

x-intercept = 12

y-intercept = 4

To find the equation of line in the form $ax+by=c$ where $a,b,c$ are integers with no factor common to all three, and greater than or equal 0.

Solution:

The x-intercept is the point at which the line cuts the x-axis. The point of x-intercept is given as $(x,0)$

Thus, on point of the line is (12,0)

The y-intercept is the point at which the line cuts the y-axis. The point of y-intercept is given as $(0,y)$ .

Thus, on point of the line is (0,4)

Using the two points we can find the slope of the line using the slope formula.

$m=\frac{y_2-y_1}{x_2-x_1}$

So, slope of the line can b given as:

$m=\frac{4-0}{0-12}$

$m=\frac{4}{-12}$

$m=-\frac{1}{3}$

The slope intercept form of the equation of he line is given as:

$y=mx+b$

where $m$ represents slope of the line and $b$ represents the y-intercept.

So, the equation of the line can be given as:

$y=-\frac{1}{3}x+4$

Multiplying each term by 3 to remove fraction.

$3y=3.-\frac{1}{3}x+4(3)$

$3y=-x+12$

Adding $x$ both sides.

$3y+x=-x+x+12$

$3y+x=12$

Thus, the equation of the line in $ax+by=c$ form where $a,b,c$ are integers with no factor common to all three, and greater than or equal 0 is:

$x+3y=12$

8 0

3 years ago

If x = -6, which inequality is true?

igomit [66]

Answer:

C. 1 - 2x > 13

Step-by-step explanation:

you just substitute

8 0

3 years ago

Given the function f(x) = x^4 + 3x^3 - 2x^2 - 6x - 1, use intermediate theorem to decide which of the following intervals contai

marta [7]

$f(x) = x^4 + 3x^3 - 2x^2 - 6x - 1$

Lets check with every option

(a) [-4,-3]

We plug in -4 for x and -3 for x

$f(-4) = (-4)^4 + 3(-4)^3 - 2(-4)^2 - 6(-4) - 1= 55$

$f(-3) = (-3)^4 + 3(-3)^3 - 2(-3)^2 - 6(-3) - 1= -1$

f(-4) is positive and f(-3) is negative. there is some value at x=c on the interval [-4,-3] where f(c)=0. so there exists atleast one zero on this interval.

(b) [-3,-2]

We plug in -3 for x and -2 for x

$f(-3) = (-3)^4 + 3(-3)^3 - 2(-3)^2 - 6(-3) - 1= -1$

$f(-2) = (-2)^4 + 3(-2)^3 - 2(-2)^2 - 6(-2) - 1= -5$

f(-2) is negative and f(-3) is negative. there is no value at x=c on the interval [-3,-2] where f(c)=0.

(c) [-2,-1]

We plug in -2 for x and -1 for x

$f(-2) = (-2)^4 + 3(-2)^3 - 2(-2)^2 - 6(-2) - 1= -5$

$f(-1) = (-1)^4 + 3(-1)^3 - 2(-1)^2 - 6(-1) - 1= 1$

f(-2) is negative and f(-1) is positive. there is some value at x=c on the interval [-2,-1] where f(c)=0. so there exists atleast one zero on this interval.

(d) [-1,0]

We plug in -1 for x and 0 for x

$f(-1) = (-1)^4 + 3(-1)^3 - 2(-1)^2 - 6(-1) - 1= 1$

$f(0) = (0)^4 + 3(0)^3 - 2(0)^2 - 6(0) - 1= -1$

f(-1) is positive and f(0) is negative. there is some value at x=c on the interval [-1,0] where f(c)=0. so there exists atleast one zero on this interval.

(e) [0,1]

We plug in 0 for x and 1 for x

$f(0) = (0)^4 + 3(0)^3 - 2(0)^2 - 6(0) - 1= -1$

$f(1) = (1)^4 + 3(1)^3 - 2(1)^2 - 6(1) - 1= -5$

f(0) is negative and f(1) is negative. there is no value at x=c on the interval [0,1] where f(c)=0.

(f) [1,2]

We plug in 1 for x and 2 for x

$f(1) = (1)^4 + 3(1)^3 - 2(1)^2 - 6(1) - 1= -5$

$f(2) = (2)^4 + 3(2)^3 - 2(2)^2 - 6(2) - 1= 19$

f(-4) is positive and f(-3) is negative. there is some value at x=c on the interval [-4,-3] where f(c)=0. so there exists atleast one zero on this interval.

so answers are (a) [-4,-3], (c) [-2,-1], (d) [-1,0], (f) [1,2]

8 0

3 years ago

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Given that ab = x, evaluate the following: c 2a3b+a2b+ab

Mrrafil [7]

It will be ......
.......

4 0

3 years ago