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3 years ago

Please help! Will mark Brainly!

Mathematics

1 answer:

Ulleksa [173]3 years ago

4 0

10 hot dogs and 5 hamburgers your graph would be 2y:1x

$\frac{2y}{1x}$

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What is the vertex of the function f (x) = x2 - 10x? (5,-25) (5.-75) (-5,75) (-5,25)

JulsSmile [24]

Option A

The vertex is (h, k) = (5, -25)

Solution:

Given function is:

$f(x) = x^2-10x$

The vertex form is given as:

$y = a(x-h)^2+k$

<h3>where (h, k) is the vertex</h3>

Rewrite the equation in vertex form

$f(x) = x^2-10x$

Complete the square for $x^2-10x$

Use the form $ax^2+bx+c$ to find the values of a, b, c

a = 1 , b = -10, c = 0

Consider the vertex form of a parabola

$a(x+d)^2+e$

Substitute the values of a and b into the following formula to find "d" :

$d = \frac{b}{2a}\\\\d = \frac{-10}{2 \times 1}\\\\d = -5$

Find the value of "e" using the formula,

$e = c - \frac{b^2}{4a}\\\\e = 0 - \frac{(-10)^2}{4 \times 1}\\\\e = -25$

Substitute the value of a, d, e into vertex form

$a(x+d)^2+e\\\\1(x-5)^2-25\\\\(x-5)^2-25$

Set y equal to above equation

$y = (x-5)^2-25$

Compare the above equation with vertex form

$y = a(x-h)^2+k$

$y = (x-5)^2-25$

We find, h = 5 and k = -25

Thus the vertex is (h, k) = (5, -25)

7 0

3 years ago

A robot can complete 8 tasks in 5/6 or an hour.each tasks takes the same amount .how long does it take the robot to complete one

lyudmila [28]

So one hour is 60 minutes,
5/6 of an hour would be, 50 minutes. Basically 60/6=10 and 10*5=50
so if the robot does 8 tasks in 50 minutes and each tasks takes the same amount, the equation would be 50/8
50/8= 6.25
Each task is 6 minutes and 25 seconds.

8 0

3 years ago

What is the solution to this inequality?

Mumz [18]

Answer:

x> 1

Step-by-step explanation:

X-7>-6

Add 7 to each side

X-7+7>-6+7

x> 1

3 0

3 years ago

Drag the tiles to the correct boxes to complete the pairs. Match each ratio with its simplest form. 3:1 2:3 1:3 3:4 6:7 9:7 21:2

Sauron [17]

Answer:

Reduce them

8 0

3 years ago

Pleeease open the image and hellllp me

Verdich [7]

1. Rewrite the expression in terms of logarithms:

$y=x^x=e^{\ln x^x}=e^{x\ln x}$

Then differentiate with the chain rule (I'll use prime notation to save space; that is, the derivative of y is denoted y' )

$y'=e^{x\ln x}(x\ln x)'=x^x(x\ln x)'$

$y'=x^x(x'\ln x+x(\ln x)')$

$y'=x^x\left(\ln x+\dfrac xx\right)$

$y'=x^x(\ln x+1)$

2. Chain rule:

$y=\ln(\csc(3x))$

$y'=\dfrac1{\csc(3x)}(\csc(3x))'$

$y'=\sin(3x)\left(-\cot^2(3x)(3x)'\right)$

$y'=-3\sin(3x)\cot^2(3x)$

Since $\cot x=\frac{\cos x}{\sin x}$ , we can cancel one factor of sine:

$y'=-3\dfrac{\cos^2(3x)}{\sin(3x)}=-3\cos(3x)\cot(3x)$

3. Chain rule:

$y=e^{e^{\sin x}}$

$y'=e^{e^{\sin x}}\left(e^{\sin x}\right)'$

$y'=e^{e^{\sin x}}e^{\sin x}(\sin x)'$

$y'=e^{e^{\sin x}+\sin x}\cos x$

4. If you're like me and don't remember the rule for differentiating logarithms of bases not equal to e, you can use the change-of-base formula first:

$\log_2x=\dfrac{\ln x}{\ln2}$