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2 years ago

What is the range of the function shown in the graph below?

Mathematics

1 answer:

alexandr1967 [171]2 years ago

4 0

Answer:

y= -3/2X + 2

Step-by-step explanation:

we get the gradient by picking any two points from the graph( it a linear graph)

using the point:

(x,y)= (-2,5) , (4,-4)

using M= (y2-y1)/x2-x1)

we get our gradient to be -9/6 which is

-3/2

using the equation of a straight line graph which is

y-y1=m(x-x1) and the point (4,-4)

we get

y-(-4)=-3/2(x-4)

----> y+4= -3/2x + 6

make y subject of the formula to take the form

y=mx+c

we get :

y=-3/2x + 2

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What type of number is this?<br>0.707070...

AfilCa [17]

Answer:

A repeating decimal...

Step-by-step explanation:

Because the numbers after the decimal points keep repeating.

4 0

2 years ago

Steve is buying apples for the fifth grade each bag holds $12 if there are 75 students total how many bags of apples will Steve

schepotkina [342]

Steve will have to buy "6.25" bags of apples if he wants to give one apple to each student.
Explanation:
75 divided by 12 = 6.25

8 0

2 years ago

(3x5)-1 expression word

pshichka [43]

Answer:

Subtract the result of three times five with 1.

Step-by-step explanation:

$(3 \times 5)-1$

$15-1$

$14$

Subtract the result of three times five with 1.

4 0

3 years ago

3. The curve C with equation y=f(x) is such that, dy/dx = 3x^2 + 4x +k

Andreas93 [3]

a. Given that y = f(x) and f(0) = -2, by the fundamental theorem of calculus we have

$\displaystyle \frac{dy}{dx} = 3x^2 + 4x + k \implies y = f(0) + \int_0^x (3t^2+4t+k) \, dt$

Evaluate the integral to solve for y :

$\displaystyle y = -2 + \int_0^x (3t^2+4t+k) \, dt$

$\displaystyle y = -2 + (t^3+2t^2+kt)\bigg|_0^x$

$\displaystyle y = x^3+2x^2+kx - 2$

Use the other known value, f(2) = 18, to solve for k :

$18 = 2^3 + 2\times2^2+2k - 2 \implies \boxed{k = 2}$

Then the curve C has equation

$\boxed{y = x^3 + 2x^2 + 2x - 2}$

b. Any tangent to the curve C at a point (a, f(a)) has slope equal to the derivative of y at that point:

$\dfrac{dy}{dx}\bigg|_{x=a} = 3a^2 + 4a + 2$

The slope of the given tangent line $y=x-2$ is 1. Solve for a :

$3a^2 + 4a + 2 = 1 \implies 3a^2 + 4a + 1 = (3a+1)(a+1)=0 \implies a = -\dfrac13 \text{ or }a = -1$

so we know there exists a tangent to C with slope 1. When x = -1/3, we have y = f(-1/3) = -67/27; when x = -1, we have y = f(-1) = -3. This means the tangent line must meet C at either (-1/3, -67/27) or (-1, -3).

Decide which of these points is correct:

$x - 2 = x^3 + 2x^2 + 2x - 2 \implies x^3 + 2x^2 + x = x(x+1)^2=0 \implies x=0 \text{ or } x = -1$

So, the point of contact between the tangent line and C is (-1, -3).

7 0

2 years ago

Helppppp

sasho [114]

Answer:

186.5 + 2x

Step-by-step explanation:

First, radify 25. Luckily it's a perfect square so it will be 5.

Second, divide within the parentheses. You're expression should look like this.

(93-5+x+42/8)2

Divide 42/8. The decimal form would be 5.25, while the fraction should be 5 1/4.

Third, solve within the parentheses from left to right. Right now you're expression should look like this:

(93-5+x+5.25)2

Add all like terms

93-5+5.25=93.25

Fourth, multiply. Right now the expression should look like this:

(93.25 + x) 2

186.5 + 2x

8 0

2 years ago