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3 years ago

Which expression is equivalent to the expression below? Select all that apply. –2/5(15 – 20d + 5c)

1 answer:

7 0

$Use\ distributive\ property:\\\\-\dfrac{2}{5}(15-20d+5c)=\left(-\dfrac{2}{5}\right)(15)+\left(-\dfrac{2}{5}\right)(-20d)+\left(-\dfrac{2}{5}\right)(5c)\\\\=(-2)(3)+(-2)(-4d)+(-2)(c)\\\\=\boxed{-6+8d-2c}$

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Helpppppppp

wel

The hypotenuse can be solved by this formula. X^2 = 6^2 +6^2

X^2=64
Usually you cut it down to 8 but it wants the squared form so 64.

6 0

3 years ago

A batch of 40 parts contains six defects. If two parts are drawn randomly one at a time without replacement, what is the probabi

adell [148]

Answer:

a) Therefore, the probability is P=1/52.

b) Therefore, the probability is P=9/400.

Step-by-step explanation:

We know that a batch of 40 parts contains six defects.

a) We calculate the probability that the both parts are defective, if two parts are drawn randomly one at a time without replacement.

The probability for the first is 6/40.

The probability for the second is 5/39.

We get:

$P=\frac{6}{40}\cdot \frac{5}{39}\\\\P=\frac{1}{52}$

Therefore, the probability is P=1/52.

b) We calculate the probability that the both parts are defective, if two parts are drawn randomly one at a time with replacement.

The probability for the first is 6/40.

The probability for the second is 6/40.

We get:

$P=\frac{6}{40}\cdot \frac{6}{40}\\\\P=\frac{9}{400}$

Therefore, the probability is P=9/400.

6 0

3 years ago

Given a regular square pyramid with RS = 6 and PX = 4, find the following measure.<br><br> PY =

Dennis_Churaev [7]

Answer:

$PY=5\ units$

Step-by-step explanation:

we know that

Applying the Pythagoras Theorem

$PY^{2}=PX^{2}+XY^{2}$

In this problem we have that

$XY=RS/2=6/2=3\ units$

$PX=4\ units$

substitute

$PY^{2}=4^{2}+3^{2}$

$PY^{2}=25$

$PY=\sqrt{25}\ units$

$PY=5\ units$

5 0

4 years ago

Read 2 more answers

What is 5/8 plus 7/8

Blababa [14]

5/8+7/8=12/8=1 4/8=1 1/2

3 0

4 years ago

Read 2 more answers

Which equation has the solutions X=1 +-SQRT 5?

Vesna [10]

Answer:

D is correct. $x^2-2x-4=0$

Step-by-step explanation:

We are given the root of the equation

$x=1+\pm \sqrt{5}$

If we are given solution of the equation then find equation using formula.

If a and b are the solution of equation then equation would be (x-a)(x-b)=0

Here, $a=1+\sqrt{5} , b=1-\sqrt{5}$

Equation form would be $(x-1-\sqrt{5})(x-1+\sqrt{5})=0$

Now we simplify the above equation to get correct option.

$x^2-x+x\sqrt{5}-x+1-\sqrt{5}-x\sqrt{5}+\sqrt{5}-5=0$

$x^2-2x-4=0$

So, D is correct. $x^2-2x-4=0$

8 0

4 years ago

Read 2 more answers