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2 years ago

Explain how to use Distributive Property when simplifying an expression

Mathematics

1 answer:

dybincka [34]2 years ago

6 0

Answer:

Expand the equation.

Multiply (distribute) the first numbers of each set, outer numbers of each set, inner numbers of each set, and the last numbers of each set.

Combine like terms.

Solve the equation and simplify, if needed.

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Which of the following is a correct conversion of 50 kg

wariber [46]

Answer:

c. 50kg = 50,000g

Step-by-step explanation:

We have to solve these following rule of three to see which of the following is a correct conversion of 50kg.

Each kg has 1000g

1kg - 1000g

50kg - xg

x = 50*1000

x = 50,000g

Conversion a) is wrong

Each kg has 1000000mg

1kg - 1000000g

50kg - x mg

x = 50*1000000

x = 50,000,000 mg

Conversion b) is wrong

In a), we have already converted 50kg to g. We have verified that 50kg = 50,000g. So c. is correct

In b), we have already converrted 50kg to mg. We have verified that 50kg = 50,000,000g. So d. is wrong

4 0

2 years ago

Fill in the blank. In the triangle below, y=

arsen [322]

Answer:

47.10

Step-by-step explanation:

The mnemonic SOH CAH TOA reminds you that ...

... Cos = Adjacent/Hypotenuse

The value y is the length of the hypotenuse, and the given length 35 is the side adjacent to the given angle. Thus, the cosine relationship will be helpful.

Filling in the given values, we have ...

... cos(42°) = 35/y

Multiplying by y/cos(42°), we can find y to be ...

... y = 35/cos(42°) ≈ 35/0.7431

... y ≈ 47.10

Y will be approximately 47.10 in length

Step-by-step explanation:

we will use this form that we memorized in our schools ............. SOH CAH TOA

CAH means , Cos ∅= Adjacent/Hypotenuse

Given values

∅ =42°

hypotenuse= y

adjacent side length = 35

now putting in our values into

Cos ∅= Adjacent/Hypotenuse

cos(42°) = 35/y

by multiplying y on both sides we get

y= 35/cos( 42°)

cos(42°)= 0.743

so y= 35/0.743

y = 47.10632

y ≈ 47.10

3 0

2 years ago

Write an expression for the 12th partial sum of the series 3/2+7/3+19/6+... using summation notation

lapo4ka [179]

Answer:

$S_{12}=\sum_{i=1}^{12} [\frac{3}{2}+(i-1)\times \frac{5}{6}]$

$S_{12}=73$

Step-by-step explanation:

$First\ term\ of\ the\ series(a_1)=\frac{3}{2}\\\\Second\ term\ of\ the\ series(a_2)=\frac{7}{3}\\\\Third\ term\ of\ the\ series(a_3)=\frac{19}{6}\\\\a_2-a_1=\frac{7}{3}-\frac{3}{2}=\frac{5}{6}\\\\a_3-a_2=\frac{19}{6}-\frac{7}{3}=\frac{5}{6}\\\\Hence\ it\ is\ an\ Arithmetic\ Series\ with\ first\ term=\frac{3}{2}\ and\ constant\ difference=\frac{5}{6}$

$a_1=\frac{3}{2}+0\times \frac{5}{6}\\\\a_2=\frac{3}{2}+1\times \frac{5}{6}\\\\a_3=\frac{3}{2}+2\times \frac{5}{6}\\\\.\\.\\.\\a_n=\frac{3}{2}+(n-1)\times \frac{5}{6}\\\\S_n=a_1+a_2+a_3+......+a_n\\\\S_n=(\frac{3}{2}+0\times \frac{5}{6})+(\frac{3}{2}+1\times \frac{5}{6})+(\frac{3}{2}+2\times \frac{5}{6})+....+(\frac{3}{2}+[n-1]\times \frac{5}{6})\\\\S_n=\sum_{i=1}^n [\frac{3}{2}+(i-1)\times \frac{5}{6}]\\\\S_n=(\frac{3}{2}+\frac{3}{2}+\frac{3}{2}+...n\ times)+\frac{5}{6}(1+2+3+4+...+(n-1))\\\\$