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Y_Kistochka [10]

2 years ago

10

Please help someone..

1 answer:

snow_tiger [21]2 years ago

4 0

Answer: $\bold{\dfrac{1}{5}}$

<u>Step-by-step explanation:</u>

$\dfrac{DF}{AK}=\dfrac{4}{20}=\large\boxed{\dfrac{1}{5}}$

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Simplify 4√16 + 8√16

Mumz [18]

$4\sqrt{16} + 8\sqrt{16}$ :

Since $\sqrt{1} =1$ , we get:

$4\cdot \:1\cdot \:6+8\cdot \:1\cdot \:6=24+48=72$

-----------------------------------------------------------------------------------

$7\sqrt{4} *3\sqrt{8}$

We have $\sqrt{4}=2$

simplify $\sqrt{8} = \sqrt{2^3} = \sqrt{2^2*2}$

then the radical rule is used: $\quad \sqrt[n]{ab}=\sqrt[n]{a}\sqrt[n]{b}$

⇒ $\sqrt{2^2*2} = \sqrt{2^2} \sqrt{2}= \sqrt{4}\sqrt{2} =2\sqrt{2}$

Now we have $\sqrt{4}=2$ and $\sqrt{8}=2 \sqrt{2}$

$7\sqrt{4} *3\sqrt{8} = 7*2*3*2\sqrt{2} = 84 \sqrt{2}$

-----------------------------------------------------------------------------------

$2\sqrt{4}+5 \sqrt{9}$

$2\sqrt{4} =2*2= 4$ and $5\sqrt{9} =5*3= 15$

$2\sqrt{4}+5 \sqrt{9} = 4+15 = 19$

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$4\sqrt{5} -2\sqrt{5} = 2\sqrt{5}$

since $2\sqrt{5}$ is the half of $4\sqrt{5}$

-----------------------------------------------------------------------------------

$-4\sqrt{2}\:+\:5\sqrt{2} =1 \sqrt{2} = \sqrt{2}\\$ <em>(-4+5=1)</em>

4 0

2 years ago

What is the square root of 67 times 15? I get lost in the numbers.

koban [17]

67 times 15 will be 1005

3 0

2 years ago

A bag holds 12 red marbles, 11 green marbles, 17 blue marbles, and 5 yellow marbles. What is the probability that you will NOT c

grigory [225]

The probability that you will NOT choose a blue marble is 28/45.
45 because there are 45 marbles in total 12+11+17+5=45
And then 28 because you add all the marbles that are NOT blue 12+11+5=28

3 0

2 years ago

<img src="https://tex.z-dn.net/?f=%20%5Cunderline%7B%20%5Cunderline%7B%20%5Ctext%7BQuestion%7D%7D%7D%20%3A%20" id="TexFormula1"

Ghella [55]

Answer:

See Below.

Step-by-step explanation:

Statements: Reasons:

$1)\text{ } AD\text{ bisects } \angle BAC$ Given

$2)\text{ } \angle BAD \cong \angle CAD$ Definition of Bisector

$\displaystyle 3) \text{ } A D \parallel E C$ Given

$\displaystyle 4)\text{ } \angle CAD\cong \angle ACE$ Alternate Interior Angles Theorem*

$\displaystyle 5) \text{ } \angle BAD\cong \angle BEC$ Corresponding Angles Theorem*

$6)\text{ } \angle CAD\cong \angle BEC$ Substitution

$7)\text{ } \angle ACE\cong\angle BEC$ Substitution

$\displaystyle 8)\text{ } AC = A E$ Isosceles Theorem Converse

*Please refer to the attachment below.

Let me know if you have any questions!

3 0

2 years ago

Read 2 more answers

Lines / and m are perpendicular. The slope of / is 3/5 what is the slope of m

DIA [1.3K]

$\bf \stackrel{\textit{perpendicular lines have \underline{negative reciprocal} slopes}} {\stackrel{slope}{\cfrac{3}{5}}\qquad \qquad \qquad \stackrel{reciprocal}{\cfrac{5}{3}}\qquad \stackrel{negative~reciprocal}{-\cfrac{5}{3}}}$

5 0

2 years ago