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3 years ago

Choose the correct problem for the fraction

Mathematics

2 answers:

Karolina [17]3 years ago

6 0

Answer:

Step-by-step explanation:

dont choose mine just yet but i think it is correct

zlopas [31]3 years ago

5 0

7 divide 8 because fractions always you division

You might be interested in

1. Grace was given the equation 2x + 20 = 15. She decides to add 10 to both sides of the equation. a.) Is that an acceptable mov

vlada-n [284]

9514 1404 393

Answer:

yes
2x +30 = 25

Step-by-step explanation:

Adding the same number to both sides of the equation is an acceptable move. The addition property of equality tells Grace that the value of the variable will remain unchanged by such a move.

Her new equation would be ...

2x +20 +10 = 15 +10

2x +30 = 25 . . . . the result of Grace's move

_____

Additional comment

There may be very good reasons why Grace would want to do that. If solving the equation is Grace's intent, that move would be counterproductive. For the purpose of solving the equation, it would be more productive to either subtract 20 from both sides, or divide both sides by 2. These steps would give, respectively, ...

2x = -5

x +10 = 7.5

6 0

3 years ago

Please help me with my homework my homework question

Paladinen [302]

Answer:

(B) Subtract 3x from both sides of the equation, and then divide both sides by 2.

Can't read the second question fully.

(A) 0.53

Step-by-step explanation:

Number 1:

If we have the equation $3x + 2 = 5x$ , our first goal is to get rid of the x term on one side.

To do this we can subtract 3x from both sides. This leaves our equation to $2=2x$ . To find x, we want to divide both sides by 2 since 2x divided by 2 is just x. Our goal is to isolate x. This leaves $x=1$ .

I couldn't read Number 2 fully - I'm sorry :c

Number 3:

Given the equation $3(x+0.2)=2.2$ , we want to isolate x on one side.

To do this, we first apply the distributive property to the left side.

$3x + 0.6 = 2.2$

Now subtract 0.6 from both sides:

$3x=1.6$

And divide both sides by 3.

$x=0.5\overline{3}$

This rounds to 0.53.

Hope this helped!

7 0

3 years ago

Read 2 more answers

Help please tell me what the answer is ASAP

Amiraneli [1.4K]

Answer:

The approximated length of EF is 2.2 units ⇒ A

Step-by-step explanation:

The tangent ratio in the right triangle is the ratio between the opposite side to the adjacent side of one of the acute angle in the triangle

In the given figure

∵ The triangle DFE has a right angle F

∵ The opposite side to angle D is EF

∵ The adjacent side to angle D is DF

→ By using the tangent ratio above

∴ tan(∠D) = $\frac{EF}{DF}$

∵ DF = 6 units

∵ m∠D = 20°

→ Substitute then in the ratio above

∴ tan(20°) = $\frac{EF}{6}$

→ Multiply both sides by 6

∴ 6 tan(20°) = EF

∴ 2.183821406 = EF

→ Approximate it to the nearest tenth

∴ 2.2 = EF

∴ The approximated length of EF is 2.2 units

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

3 years ago

Put the following equation of a line into slope-intercept form, simplifying all

Gnom [1K]

At the moment it’s in standard form. We can easily turn this into slope int form (y= mx + b) tho

3y= -18x -18
y= -6x - 6

7 0

3 years ago