All categories

3 years ago

Multiplication to find ratios that are equivalent to 4:5

Mathematics

1 answer:

Sindrei [870]3 years ago

8 0

Answer:

You can multiply 4 and 5 by a number, and it will be equivalent.

Let's use the number 3.

4 x 3=12

5 x 3=15

So, it will be 12:15

You might be interested in

What is 10% of 80 anwser soon

Elena-2011 [213]

The correct answer is 8
10% of 80 is 8

Hope this helped :)

5 0

3 years ago

Read 2 more answers

Please helpppp i don’t understand

atroni [7]

Answer:

Think the answer will just be 5

Step-by-step explanation:

I don't use step by step so yeah

5 0

3 years ago

Does 2.5, 6, 6.5 form a right triangle

yawa3891 [41]

Yes I think it’s s right triangle

5 0

3 years ago

Read 2 more answers

If line a is perpendicular to line b and the slope of line a is 7 whats the slope of line b

amm1812

Answer:

If a line is perpendicular to another line, that means that the slope is completely opposite that of the original line. The first thing that we do to the slope is we negate the number which means that if we have a slope of $-7$ our slope because $7$ in this step. In our case our slope is $7$ so in this step it becomes $-7$ .

Moving onto the second part which is to get the reciprocal of the number which means that if we have $\frac{1}{2}$ then we would switch it to $2$ . In our case our number is $-7$ so we would make that into a fraction like this $-\frac{1}{7}$ .

In conclusion, our final slope of the perpendicular line is $-\frac{1}{7}$ .

Hope this helps! Let me know if you have any questions

8 0

2 years ago

Read 2 more answers

Please answer correctly !!!!!!!!! Will Mark brainliest !!!!!!!!!!!!!

il63 [147K]

Answer:

$x=-\frac{-20+\sqrt{-20w+3600}}{10},\:x=-\frac{-20-\sqrt{-20w+3600}}{10}$

Step-by-step explanation:

$w=-5\left(x-8\right)\left(x+4\right)\\\mathrm{Expand\:}-5\left(x-8\right)\left(x+4\right):\quad -5x^2+20x+160\\w=-5x^2+20x+160\\Switch\:sides\\-5x^2+20x+160=w\\\mathrm{Subtract\:}w\mathrm{\:from\:both\:sides}\\-5x^2+20x+160-w=w-w\\Simplify\\-5x^2+20x+160-w=0\\Solve\:with\:the\:quadratic\:formula\\\mathrm{Quadratic\:Equation\:Formula:}\\\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}\\x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$\mathrm{For\:}\quad a=-5,\:b=20,\:c=160-w:\quad x_{1,\:2}=\frac{-20\pm \sqrt{20^2-4\left(-5\right)\left(160-w\right)}}{2\left(-5\right)}\\x=\frac{-20+\sqrt{20^2-4\left(-5\right)\left(160-w\right)}}{2\left(-5\right)}:\quad -\frac{-20+\sqrt{-20w+3600}}{10}\\x=\frac{-20-\sqrt{20^2-4\left(-5\right)\left(160-w\right)}}{2\left(-5\right)}:\quad -\frac{-20-\sqrt{-20w+3600}}{10}\\The\:solutions\:to\:the\:quadratic\:equation\:are\\x=-\frac{-20+\sqrt{-20w+3600}}{10},\:x=-\frac{-20-\sqrt{-20w+3600}}{10}$

6 0

3 years ago