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3 years ago

What is the gradient of the graph shown? Give your answer in its simplest form.

Mathematics

1 answer:

Minchanka [31]3 years ago

6 0

Answer:

m=3/2

Step-by-step explanation:

Two points are shown

point 1: (0;-3). 4;3. -2-6

point 2: (2;0)

$m = \frac{y2 - y1}{x2 - x1}$

$m = \frac{0 - ( - 3)}{2 - 0}$

$m = \frac{3}{2}$

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At the county fair, Darius can purchase 3 candy apples and 4 bags of peanuts for $11.33. He can purchase 9 candy apples and 5 ba

Ksenya-84 [330]

Let c represents the cost of a candy apple and b represents the cost of a bag of peanuts.

Darius can purchase 3 candy apples and 4 bags of peanuts. So his total cost would be 3c + 4b. Darius can buy 3 candy apples and 4 bags of peanuts in $11.33,so we can write the equation as:

3c + 4b = 11.33 (1)

Darius can purchase 9 candy apples and 5 bags of peanuts. So his total cost would be 9c + 5b. Darius can buy 9 candy apples and 5 bags of peanuts in $23.56,so we can write the equation as:

9c + 5b = 23.56 (2)

<span>Darius decides to purchase 2 candy apples and 3 bags of peanuts. The total cost in this case will be 2c + 3b. To find this first we need to find the cost of each candy apple and bag of peanuts by solving the above two equations.

Multiplying equation 1 by three and subtracting equation 2 from it, we get:

3(3c + 4b) - (9c + 5b) = 3(11.33) - 23.56

9c + 12b - 9c - 5b = 10.43

7b = 10.43

b = $1.49

Using the value of b in equation 1, we get:

3c + 4(1.49) = 11.33

3c = 5.37

c = $ 1.79

Thus, cost of one candy apple is $1.79 and cost of one bag of peanuts is $1.49.

So, 2c + 3b = 2(1.79) + 3(1.49) = $ 8.05

Therefore, Darius can buy 2 candy apples and 3 bags of peanuts in $8.05</span>

3 0

3 years ago

I need help , I don’t understand this

marta [7]

#2. First, we factor each polynomial. Then, if any terms on both the top and the bottom of the fraction match, they cancel out. So... we do just that. You end up with:

$\frac{x(x-4)}{(x+9)(x-4)}$

Notice there's an (x-4) on both top and bottom. So they cancel out. That leaves us with your answer of $\frac{x}{(x+9)}$

#3. We do the same thing as above then multiply and simplify. In the interest of space, I'll cut straight to some simplification.

$\frac{2(x+2)^{3} }{6x(x+2)} ( \frac{5}{(x-2)^{2} } )$

Now we start cancelling. For the first fraction, there are 3 (x+2)'s on top and 1 on the bottom so we will cancel out the one on the bottom and leave 2 (x+2)'s on top. There are no more polynomials to cancel out so now we multiply across:

$\frac{10(x+2)^{2} }{6x(x-2)^{2} }$

10 and 6 share a GCF of 2 so we divide both of those by 2. This leaves us with the final answer of:

$\frac{5(x+2)^{2} }{3x(x-2)^{2} }$

#4. This equation introduces division and because of it, we must flip the second fraction to make the division sign into a multiplication symbol. Again for space, I'll flip the fraction and simplify in one step.

$\frac{3(x+2)(x-2)}{(x+4)(x-2)} ( \frac{x+4}{6(x+3)})$

Now we do our cancelling. First fraction has (x - 2) in the top and bottom. They're gone. The first fraction has a (x + 4) on the bottom and the second fraction has one on the top. Those will also cancel. This leaves you with:

$\frac{3(x+2)}{6(x+3)}$

3 and 6 share a GCF of 3 so we divide both numbers by this. This leaves you with your final answer:

$\frac{x+2}{2(x+3)}$

#5. We are adding so we first factor both fractions and see what we need to multiply by to make the denominators the same. I'll do the former first. (10 - x) and (x - 10) are not the same so we multiply the first equation (top and bottom) by (x - 10) and the second equation by (10 - x). Because they will now have the same denominator we can combine them already. This gives us:

$\frac{(3+2x)(x-10)+(13+x)(10-x)}{(10-x)(x-10)}$

Now we FOIL each to expand and then simplify by combining like terms. Again for space, I'm just showing the result of this; you end up with:

$\frac{x^{2}-20x+100}{(10-x)(x-10)}$

Now we factor the top. This gives you 2 (x - 10)'s on top and one on bottom. So we just leave one on the top and cancel the bottom one out. This leaves you with your answer:

$\frac{x+10}{10-x}$

#6. Same process for this one so I won't repeat. I'll just show the work.

$\frac{3}{(x-3)(x+2)} + \frac{2}{(x-3)(x-2)}$ becomes

$\frac{3(x-2) + 2(x+2)}{(x-3)(x+2)(x-2)}$ which equals

$\frac{3x - 6 + 2x + 4}{(x-3)(x+2)(x-2)}$ giving you the final answer

$\frac{5x - 2}{(x-3)(x+2)(x-2)}$

#7. For this question we find the least common denominator to make the denominators match. For 5, x, and 2x, the LCD is 10x. So we multiply top and bottom of each fraction by what would make the bottom equal 10x. This rewrites the fraction as:

$\frac{3x}{5} ( \frac{2x}{2x}) * ( \frac{5}{x}( \frac{10}{10}) - \frac{5}{2x} ( \frac{5}{5}))$

Simplify to get:

$\frac{3x}{5} * ( \frac{25}{10x})$

After simplifying again, you end up with your final answer:

$\frac{3}{2}$

8 0

3 years ago

Let $f(x) = x^2$ and $g(x) = \sqrt{x}$. Find the area bounded by $f(x)$ and $g(x).$

Anna [14]

Answer:

$\large\boxed{1\dfrac{1}{3}\ u^2}$

Step-by-step explanation:

Let's sketch graphs of functions f(x) and g(x) on one coordinate system (attachment).

Let's calculate the common points:

$x^2=\sqrt{x}\qquad\text{square of both sides}\\\\(x^2)^2=\left(\sqrt{x}\right)^2\\\\x^4=x\qquad\text{subtract}\ x\ \text{from both sides}\\\\x^4-x=0\qquad\text{distribute}\\\\x(x^3-1)=0\iff x=0\ \vee\ x^3-1=0\\\\x^3-1=0\qquad\text{add 1 to both sides}\\\\x^3=1\to x=\sqrt[3]1\to x=1$

The area to be calculated is the area in the interval [0, 1] bounded by the graph g(x) and the axis x minus the area bounded by the graph f(x) and the axis x.

We have integrals:

$\int\limits_{0}^1(\sqrt{x})dx-\int\limits_{0}^1(x^2)dx=(*)\\\\\int(\sqrt{x})dx=\int\left(x^\frac{1}{2}\right)dx=\dfrac{2}{3}x^\frac{3}{2}=\dfrac{2x\sqrt{x}}{3}\\\\\int(x^2)dx=\dfrac{1}{3}x^3\\\\(*)=\left(\dfrac{2x\sqrt{x}}{2}\right]^1_0-\left(\dfrac{1}{3}x^3\right]^1_0=\dfrac{2(1)\sqrt{1}}{2}-\dfrac{2(0)\sqrt{0}}{2}-\left(\dfrac{1}{3}(1)^3-\dfrac{1}{3}(0)^3\right)\\\\=\dfrac{2(1)(1)}{2}-\dfrac{2(0)(0)}{2}-\dfrac{1}{3}(1)}+\dfrac{1}{3}(0)=2-0-\dfrac{1}{3}+0=1\dfrac{1}{3}$