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Snowcat [4.5K]
3 years ago
9

PLEASE HELP IF YOU WANT BRIANLEIST TY!! ^ ^

Mathematics
1 answer:
Sergio039 [100]3 years ago
8 0
I think it’s rhombus and isosceles trapezoid. i’m so sorry if it’s wrong!
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Which of the following is a property of set?
NARA [144]

Answer:

C might be the answer, I guesss

8 0
3 years ago
Which number(s) below belong to the solution set of the equation? Check all
Stells [14]

Answer:

11

Step-by-step explanation:

198 ÷ 18 = 11

x = 11

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4 0
3 years ago
Linnea's company's revenue in 2017 is 36/25 of its revenue in 2016. What is Linnea's company's revenue in 2017 as a percent of i
nekit [7.7K]

Let the 2016 revenue = x

Let the 2017 revenue = y

36/25 * 100 = the percent relationship

% relationship = 144%

y = 144% x

Answer 144% <<<<

6 0
4 years ago
Read 2 more answers
PLEASE HELP ASAP!!!!!
padilas [110]

Answer:

A'''(1,1), B'''(-2,-1) and C'''(-4,0).

Step-by-step explanation:

Triangle ABC has vertices with coordinates A(-6,-1), B(-3,-3) and C(-1,-2) (from given diagram).

1. Translation the triangle ABC 5 units to the right and 2 units up has a rule

(x,y)→(x+5,y+2).

Then the image triangle A'B'C' has vertices A'(-1,1), B'(2,-1) and C'(4,0).

2. Reflection the triangle A'B'C' across the line y=-x has a rule

(x,y)→(-y,-x).

Then the image triangle A"B"C" has vertices A"(-1,1) (this point lie on the line), B"(1,-2) and C"(0,-4).

3. Rotation the triangle A"B"C" counterclockwise about the origin by 270° has  a rule

(x,y)→(y,-x).

Then the image triangle A'''B'''C''' has vertices A'''(1,1), B'''(-2,-1) and C'''(-4,0).

3 0
3 years ago
Read 2 more answers
Find the area of the shaded region.f(x)=4x+3x2−x3,g(x)=0<br> The area is _____.
Tasya [4]

Answer:

The answer is " \bold{\frac{7}{2}}"

Step-by-step explanation:

Given value:

\to f(x)=4x+3x^2-x^3\\\\\to g(x)=0

Find:

area=?

calculation:

\ Area = \int_{-1}^{0} -(4x+3x^2-x^3) dx  + \int_{0}^{1} (4x+3x^2-x^3) dx  \\\\

        =  -(2x^2+x^3- \frac{x^4}{4})_{-1}^{0} + (2x^2+x^3- \frac{x^4}{4})_{0}^{1}\\\\\\=  -( 2x^2+x^3- \frac{x^4}{4})_{-1}^{0} + (2x^2+x^3- \frac{x^4}{4})_{0}^{1}\\\\\\= ( 2- 1 -\frac{1}{4}) + (2+1- \frac{1}{4})\\\\\\=  ( \frac{8-4-1}{4}) + (\frac{8+4-1}{4})\\\\=  ( \frac{3}{4}) + (\frac{11}{4})\\\\=   \frac{3}{4} + \frac{11}{4}\\\\= \frac{11+3}{4}\\\\= \frac{14}{4}\\\\= \frac{7}{2}\\\\

7 0
3 years ago
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