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3 years ago

When six is subtracted from five times a number, the result is 9

Mathematics

1 answer:

lozanna [386]3 years ago

7 0

You do 9=5y-6. And then solve for y. It’s is 3.

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A recipe for giant soap bubbles uses 1/2 cup dishwashing liquid and 4 1/2 cups of water. How many cups of water will be used wit

GREYUIT [131]

Answer: D. 13 1/2.

Step-by-step explanation:

1/2*3= 1 1/2 cups of dishwashing liquid.

4 1/2*3= 13 1/2 cups of water will be used with 1 1/2 cups of dishwashing liquid.

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2 years ago

What is the radius of the following circle

stepladder [879]

I got 11.5 lol I don’t really know why

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3 years ago

Foot pieces you can cut from 5 feet of

Rzqust [24]

Answer:

1p = r

Step-by-step explanation:

You are asking how many 1-foot pieces you can make out of a 5 foot piece of ribbon. You are asking for an equation.

p = the # of pieces

r = the amount of ribbon

So lets try the equation:

1p = r

1(4) = r

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7 0

3 years ago

Suppose that the length of a side of a cube X is uniformly distributed in the interval 9

Nastasia [14]

Answer:

$f(v) = \left \{ {{\frac{1}{3}v^{-\frac{2}{3}}\ 9^3 \le v \le 10^3} \atop {0, elsewhere}} \right.$

Step-by-step explanation:

Given

$9 < x < 10$ --- interval

Required

The probability density of the volume of the cube

The volume of a cube is:

$v = x^3$

For a uniform distribution, we have:

$x \to U(a,b)$

and

$f(x) = \left \{ {{\frac{1}{b-a}\ a \le x \le b} \atop {0\ elsewhere}} \right.$

$9 < x < 10$ implies that:

$(a,b) = (9,10)$

So, we have:

$f(x) = \left \{ {{\frac{1}{10-9}\ 9 \le x \le 10} \atop {0\ elsewhere}} \right.$

Solve

$f(x) = \left \{ {{\frac{1}{1}\ 9 \le x \le 10} \atop {0\ elsewhere}} \right.$

$f(x) = \left \{ {{1\ 9 \le x \le 10} \atop {0\ elsewhere}} \right.$

Recall that:

$v = x^3$

Make x the subject

$x = v^\frac{1}{3}$

So, the cumulative density is:

$F(x) = P(x < v^\frac{1}{3})$

$f(x) = \left \{ {{1\ 9 \le x \le 10} \atop {0\ elsewhere}} \right.$ becomes

$f(x) = \left \{ {{1\ 9 \le x \le v^\frac{1}{3} - 9} \atop {0\ elsewhere}} \right.$

The CDF is:

$F(x) = \int\limits^{v^\frac{1}{3}}_9 1\ dx$

Integrate

$F(x) = [v]\limits^{v^\frac{1}{3}}_9$

Expand

$F(x) = v^\frac{1}{3} - 9$

The density function of the volume F(v) is:

$F(v) = F'(x)$

Differentiate F(x) to give:

$F(x) = v^\frac{1}{3} - 9$

$F'(x) = \frac{1}{3}v^{\frac{1}{3}-1}$

$F'(x) = \frac{1}{3}v^{-\frac{2}{3}}$

$F(v) = \frac{1}{3}v^{-\frac{2}{3}}$

So:

$f(v) = \left \{ {{\frac{1}{3}v^{-\frac{2}{3}}\ 9^3 \le v \le 10^3} \atop {0, elsewhere}} \right.$

8 0

2 years ago

The question is number 6 i don't know if I got it right I really need to know and this is solving quadratics by completing the s

zhenek [66]

Answer:

Step-by-step explanation:

If u = 6 then substitute so

6(6+6)=6 know solve but this isn't going to be correct because 6 plus 6 is 12 times 6 isn't 3

So your answer isn't correct. I think it would be a negative number or decimal because 6 is larger than 3. Hope this helps!

it is u(u+6)=3

I got u=6

4 0

2 years ago