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2 years ago

What goes in the boxes?

Mathematics

1 answer:

Anna [14]2 years ago

5 0

Answer: machine 1: 3/5 machine 2: 3/4

machine 1 is working at a faster rate

Step-by-step explanation:

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What is the answer to this equation?

RoseWind [281]

This might be right?

7 0

2 years ago

Read 2 more answers

Given that α and β are the roots of the quadratic equation <img src="https://tex.z-dn.net/?f=2x%5E%7B2%7D%20%2B6x-7%3Dp" id="Tex

siniylev [52]

Answer:

$\large \boxed{\sf \ \ \ p=-11 \ \ \ }$

Step-by-step explanation:

Hello,

$\alpha \text{ and } \beta \text{ are the roots of the following equation}$

$2x^2+6x-7=p$

It means that

$2\alpha^2+6\alpha-7=p \\\\2\beta ^2+6\beta -7=p \\\\$

And we know that

$\alpha= 2\cdot \beta$

So we got two equations

$2(2\beta)^2+6\cdot 2 \cdot \beta -7=p \\\\8\beta^2+12\beta -7=p\\\\ and \ 2\beta ^2+6\beta -7=p \ So \\\\\\8\beta^2+12\beta -7 = 2\beta ^2+6\beta -7\\\\6\beta^2+6\beta =0\\\\\beta(\beta+1)=0\\\\ \beta =0 \ or \ \beta=-1$

For $\beta =0, \ \ \alpha =0, \ \ p = -7$

For $\beta =-1, \ \ \alpha =-2, \ \ p= 2-6-7=-11, \ p=2*4-12-7=-11$

I assume that we are after two different roots so the solution for p is p=-11

b) $\alpha +2 =-2+2=0 \ and \ \beta+2=-1+2=1$

So a quadratic equation with the expected roots is

$x(x-1)=x^2-x$

Hope this helps.

Do not hesitate if you need further explanation.

Thank you

3 0

2 years ago

A bookstore is having a sale in which customers can purchase 6 books for a price of b dollars. Which expression represents the p

Evgen [1.6K]

That would be 6/b <====

5 0

2 years ago

Multiply out 5x (x + 3y)

GrogVix [38]

Answer:

5x²+15xy

Step-by-step explanation:

5x(x+3y)=

5x²+15xy

If you need more steps or more explanation, reply to this answer.

7 0

2 years ago

Read 2 more answers

Find the x-intercept(s) of y = 2x2 - 3x - 5.<br> A) -5 <br> B) 2 1/2, -1<br> C) 3, -5 <br> D) 5, -1

Alexus [3.1K]

Answer:

$\large\boxed{B)\ 2\frac{1}{2},\ -1}$

Step-by-step explanation:

x-intercepts are for y = 0.

We have the equation:

$y=2x^2-3x-5$

Therefore

$2x^2-3x-5=0\\\\2x^2+2x-5x-5=0\\\\2x(x+1)-5(x+1)=0\\\\(x+1)(2x-5)+0\iff x+1=0\ \vee\ 2x-5=0\\\\x+1=0\qquad\text{subtract 1 from both sides}\\\boxed{x=-1}\\\\2x-5=0\qquad\text{add 5 to both sides}\\2x=5\qquad\text{divide both sides by 2}\\\boxed{x=2\frac{1}{2}}$

7 0

3 years ago