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VMariaS [17]
3 years ago
10

Use the properties of operations to simplify this algebraic expression. rewrite the expression by following the directions in ea

ch step.
5(x - 4) + 3x - 9x+7

Mathematics
2 answers:
o-na [289]3 years ago
5 0

Step 3:

5x + 20 + 3x + 9x + 7 becomes

5x + 3x + 9x + 20 + 7

Step 4:

(5x + 3x + 9x) + (20 + 7)

Step 5:

17x + 27

If you have any further questions feel free to ask.

Hope this helps

Nezavi [6.7K]3 years ago
4 0

Answer and Explanation:

Given : Expression 5(x-4)+3x-9x+7

To find : Use the properties of operations to simplify this algebraic expression. Rewrite the expression by following the directions in each step.

Solution :

Step 1 - Write subtraction as addition,

=5(x+4)+3x+9x+7

Step 2 - Use the distributive property, a(b+c)=ab+ac

=5x+20+3x+9x+7

Step 3 - Use the commutative property of addition to reorder terms so that like terms are together,

=5x+3x+9x+20+7

Step 4 - Use the associative property of addition to group like terms,

=(5x+3x+9x)+(20+7)

Step 5 - Simplify

=17x+27

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Use two different methods to find an explain the formula for the area of a trapezoid that has parallel sides of length a and B a
evablogger [386]

Answer:

Formula of Trapezoid:

A = (a + b) × h / 2

The formula can be derived in different ways. for now, we have discussed two ways:

1. By using the formula of a triangle

2. By dividing into different sections

Step-by-step explanation:

1. By using the formula of a triangle

One of the ways to explain a formula for an area of a trapezoid using a formula for a triangle can be as follows.

Assume a trapezoid PQRS with lower base SR and upper base PQ (they are parallel) and sides PS and QR.

The image is attached below.

Connect vertices P and R with a diagonal.

Consider triangle ΔPQR as having a base PQ and an altitude from vertex R down to point M on base PQ (RM⊥PQ).

Its area is

S1=\frac{1}{2} *PQ*RM

Consider triangle ΔPRS as having a base SR and an altitude from vertex P up to point N on-base SR (PN⊥SR).

Its area is

S2=\frac{1}{2} *SR*PN

Altitudes RM and PN are equal and constitute the distance between two parallel bases PQ and SR.

They both are equal to the altitude of the trapezoid h.

Therefore, we can represent areas of our two triangles as

S1=\frac{1}{2}*PQ*h

S2=\frac{1}{2}*SR*h

Adding them together, we get the area of the whole trapezoid:

S=S1+S2=\frac{1}{2} (PQ+SR)h,

which is usually represented in words as "half-sum of the bases times the altitude".

2. By dividing into different sections

Trapezoid PQRS is shown below, with PQ parallel to RS.

Figure 1 - Trapezoid PQRS with PQ parallel to RS(image is attached below.)

We are going to derive the area of a trapezoid by dividing it into different sections.

If we drop another line from Q, then we will have two altitudes namely PT and QU.

Figure 2 - Trapezoid PQRS divided into two triangles and a rectangle. (image is attached below.)

From Figure 2, it is clear that Area of PQRS = Area of PST + Area of PQUT + Area of QRU. We have learned that the area of a triangle is the product of its base and altitude divided by 2, and the area of a rectangle is the product of its length and width. Hence, we can easily compute the area of PQRS. It is clear that

=> A_{PQRS} = (\frac{ah}{2}) + b_{1}h + \frac{ch}{2}

Simplifying, we have

=>A= \frac{ah+2b_{1+C} }{2}

Factoring we have,

=> A_{PQRS} = (a+ 2b_{1} + c)\frac{h}{2}  \\= > {(a+ b_{1} + c) + b_{1} }\frac{h}{2}

 But, a+ b_{1} + c  is equal to b_{2}, the longer base of our trapezoid.

Hence, A_{PQRS}= (b_{1} + b_{2} )\frac{h}{2}

We have discussed two ways by which we can derive area of a trapezoid.

Read to know more about Trapezoid

brainly.com/question/4758162?referrer=searchResults

#SPJ10

5 0
2 years ago
Please help needed its due soon
Brums [2.3K]
25yd=75 . sorry mane had to get these points up you know
6 0
2 years ago
Find a and b if the point p(6,0) and Q(3,2) lie on the graph of ax+ by=12
tiny-mole [99]

to get the equation of any straight line we only need two points off of it, hmmm let's use P and Q here and then let's set the equation in standard form, that is

standard form for a linear equation means

• all coefficients must be integers, no fractions

• only the constant on the right-hand-side

• all variables on the left-hand-side, sorted

• "x" must not have a negative coefficient

(\stackrel{x_1}{6}~,~\stackrel{y_1}{0})\qquad (\stackrel{x_2}{3}~,~\stackrel{y_2}{2}) ~\hfill \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{2}-\stackrel{y1}{0}}}{\underset{run} {\underset{x_2}{3}-\underset{x_1}{6}}}\implies \cfrac{2}{-3}\implies -\cfrac{2}{3} \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{0}=\stackrel{m}{-\cfrac{2}{3}}(x-\stackrel{x_1}{6})

\stackrel{\textit{multiplying both sides by }\stackrel{LCD}{3}}{3(y-0)~~ = ~~3\left( -\cfrac{2}{3}(x-6) \right)}\implies 3y=-2(x-6) \\\\\\ 3y=-2x+12 \implies \stackrel{a}{2} x+\stackrel{b}{3} y=12

5 0
2 years ago
The equation of a line is given below.
Yanka [14]

Answer:

Please mark as Brainliest :)

Step-by-step explanation:

7 0
3 years ago
CAN SOMEONE PLEASE HELP ME I DOUBTER UNDER HALf these questions Please help:(
Vaselesa [24]

Answer: I can only help you with number Im sorry

Step-by-step explanation:

10(-1.2 + 6) = 48

X = 3    Y = 2


-4 x 3 = -12    30 x 2 = 60


-12 + 60

48  


48 = 48

Im in middle school so Im sorry that it took me long to reply I had a hard time finding the answer. The other questions I haven’t been taught yet so maybe search up what the question is about for help if you don’t get any helpful replies :)(:

5 0
2 years ago
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