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densk [106]
4 years ago
5

Help..

Mathematics
1 answer:
Leona [35]4 years ago
7 0

Hey there!

In order for you to do this equation you have to do <u>PEMDAS</u>

  1. \bold{Parentheses, Exponents,Multiplication,Division,Addition,Subtraction}
  • \bold{\frac{1}{2}(7)(4)+6(5)}
  • \bold{\frac{1}{2}(7)=\frac{7}{2}}
  • \bold{\frac{7}{2}(4)=14}
  • \bold{6(5)=30}
  • \bold{14+30=44}
  • \boxed{\boxed{\bold{Answer\#1:44}}}

  • \bold{12(6)+\frac{1}{4}\times2^2}
  • \bold{12(6)=72}
  • \bold{2^2=4}
  • \bold{\frac{1}{4}\times4=1}
  • \bold{72+1=73}
  • \boxed{\boxed{\bold{Answer\#2:72}}}

Good luck on your assignment and enjoy  your day!

~\frak{LoveYourselfFirst:)}

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which of the following is the most likely order of magnitude for the population of a large U.S city such as chicago
Mariulka [41]

Answer:6

Step-by-step explanation:

4 0
3 years ago
Points $M$, $N$, and $O$ are the midpoints of sides $\overline{KL}$, $\overline{LJ}$, and $\overline{JK}$, respectively, of tria
Ivan

The midpoint theorem states that the line joining the mid points of two sides of a triangle is parallel to the third and facing side and equal to half of the length of the third side

Based on the midpoint theorem, the area of triangle ΔLPQ is 63 square units

The reason the value of the area of triangle ΔLPQ as given above is correct is as follows:

The given parameters;

The midpoint of \overline {KL} = M; The midpoint of \overline {LJ} = N; The midpoint of \overline {JK} = O

The midpoint of \overline {NO} = P; The midpoint of \overline {OM} = Q; The midpoint of \overline {MN} = R

The area of triangle ΔPQR = 21

The required parameter:

Calculate the area of triangle ΔLPQ

Method:

The definition of midpoint, area ratio, and area of a triangle formula can be used to find the area of triangle ΔLPQ

Solution:

According to the midpoint theorem, we have;

\overline {QR} = (1/2) × \overline {NO}

\overline {PR} = (1/2) × \overline {OM}

\overline {PQ} = (1/2) × \overline {MN}

Given that \overline {QR} is parallel to \overline {ON}, and \overline {PR} is parallel to, we have;

∠MON = ∠PRQ

Similarly, we have, ∠MNO = ∠PQR

Therefore, ΔPQR is similar to triangle ΔMON, which is also similar to ΔJKS

The area of triangle ΔPQR = 21, by area ratio = (Side ratio)², we have;

The sides of ΔMON = 2 × The side length of ΔPQR

The area of triangle ΔMON = 2² × The area of ΔPQR

∴ The area of triangle ΔMON = 4 × 21

Similarly the area of ΔJKS = 4 × 4 × 21

PQ = JK/4

The area of LPQ = (1/2) × PQ × h

h = (3/4×JL) × sin(x°)

∴ The area of LPQ = (1/2)×JK/4×(3/4×JL) × sin(x°)

However; (1/2)×JK×JL× sin(x°) = Area of ΔJKS = 4 × 4 × 21

Therefore;

The area of ΔLPQ = (Area of ΔJKS)/4×(3/4) = (4 × 4 × 21)/4×(3/4) = 63

The area of triangle ΔLPQ = 63 square units

Learn more about the midpoint theorem here:

brainly.com/question/15227899

8 0
3 years ago
In the figure, ∆ABC~∆XYZ. What is the perimeter of ∆ABC? Show your work.
Elina [12.6K]

Answer:

49 units

Step-by-step explanation:

Given the triangles are similar then the ratios of corresponding sides are equal, that is

\frac{BC}{YZ} = \frac{AB}{XY}, substitute values

\frac{BC}{44} = \frac{11}{22} ( cross- multiply )

22BC = 484 ( divide both sides by 22 )

BC = 22

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

Similarly

\frac{AC}{XZ} = \frac{AB}{XY}, substitute values

\frac{AC}{32} = \frac{11}{22} ( cross- multiply )

22AC = 352 ( divide both sides by 22 )

AC = 16

Thus perimeter of Δ ABC is

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5 0
4 years ago
If p is a polynomial show that lim x→ap(x)=p(a
Lostsunrise [7]

Let p(x) be a polynomial, and suppose that a is any real number. Prove that

lim x→a p(x) = p(a) .

 

Solution. Notice that

 

2(−1)4 − 3(−1)3 − 4(−1)2 − (−1) − 1 = 1 .

 

So x − (−1) must divide 2x^4 − 3x^3 − 4x^2 − x − 2. Do polynomial long division to get 2x^4 − 3x^3 − 4x^2 – x – 2 / (x − (−1)) = 2x^3 − 5x^2 + x – 2.

 

Let ε > 0. Set δ = min{ ε/40 , 1}. Let x be a real number such that 0 < |x−(−1)| < δ. Then |x + 1| < ε/40 . Also, |x + 1| < 1, so −2 < x < 0. In particular |x| < 2. So

 

|2x^3 − 5x^2 + x − 2| ≤ |2x^3 | + | − 5x^2 | + |x| + | − 2|

= 2|x|^3 + 5|x|^2 + |x| + 2

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= 40

 

Thus, |2x^4 − 3x^3 − 4x^2 − x − 2| = |x + 1| · |2x^3 − 5x^2 + x − 2| < ε/40 · 40 = ε.

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3 years ago
Which of the following statements best describes the effect of replacing the graph of f(x) with the graph of f(x) − 5? (1 point)
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The entire graph is transformed 5 units down
3 0
4 years ago
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