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vladimir1956 [14]
3 years ago
14

The perimeter of the square is the same as the perimeter of the rectangle.

Mathematics
2 answers:
Elina [12.6K]3 years ago
6 0
You would have to find out the perimeter first to get the length
taurus [48]3 years ago
3 0
12 I think.....hope this helps
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Write the equation of the line that passes through (4,-3) and (8,-12)
german

Answer:

Step-by-step explanation:

(4 , -3)   ;  (8 , -12)

Slope =\dfrac{y_{2}-y_{1}}{x_{2}-x_{1}}\\\\=\dfrac{-12-[-3]}{8-4}\\\\=\dfrac{-12+3}{4}\\\\=\dfrac{-9}{4}

m = -9/4 ; (4 , -3)

y - y_{1}= m(x -x_{1}\\\\y - [-3]= \dfrac{-9}{4}(x -4)\\\\\\y + 3 = \dfrac{-9}{4}x -4*\dfrac{-9}{4}\\\\\\y + 3 =\dfrac{-9}{4}x+9\\\\\\y = \dfrac{-9}{4}x+9-3\\\\y=\dfrac{-9}{4}x+6

3 0
3 years ago
I need help I don’t understand
Gelneren [198K]
B is your answer lol
5 0
3 years ago
ABC is reflected over the y-axis. What are the vertices of
finlep [7]

Answer:

The answer is the first option.

Step-by-step explanation:

When you flip it over the y axis. A would still be the same. ANd the first one matches A.

5 0
3 years ago
Use the Pythagorean theorem to find the lengths of the sides of the right triangle. Use a calculator when necessary. A right tri
Doss [256]

Answer:

The length of the sides of the right angle triangle

15 , 25, 20

Step-by-step explanation:

<u>Explanation</u>:-

<u>Step(i):-</u>

we know that by using Pythagorean theorem to find the lengths of the sides of the triangle

AC² = AB² + BC²

Given a  right triangle has legs labeled 3m and 2m+10

let us assume that AB = 3m and BC = 2m + 10

Given a hypotenuse  labeled 5m

let us assume that hypotenuse AC = 5m

<u>Step(ii)</u>:-

Now by using Pythagorean theorem

AC² = AB² + BC²

(5m)² = (3m)² + (2m+10)²

25m² = 9m² + 4m² + 40m + (10)²  ( since (a + b)² = a²+2ab+b²) )

on simplification , we get

25m²-13m² -40m -100 =0

12m² -40m -100 =0

4(3m² -10m -25) =0

3m² -10m -25 =0

3m² - 15m + 5m -25 =0

3m(m-5) + 5(m-5) =0

(3m +5) (m-5) =0

3m +5 =0 and m-5=0

3m = -5 and m =5

m = \frac{-3}{5} and m=5

we can not choose negative value

so the value m=5

<u>Step (iii)</u>:-

The sides of right angle triangle

AB = 3m

AB = 3(5) = 15 and

BC = 2m + 10

BC = 2(5) +10 = 20

The hypotenuse AC =  5m

                           AC = 25

<u>Conclusion:</u>-

The  lengths of the sides of the right triangle

15, 25 ,20

3 0
4 years ago
Problem 3. if f(x) = /2x-5 and g(x) = 5x? -- 3, find (gºf)(x).<br><br> (Step by step please&lt;3)
kramer

(1) <em>f(x)</em> = <em>x</em>² - 4<em>x</em> + 2, <em>g(x)</em> = 3<em>x</em> - 7

<em>(f</em> o<em> g) (x)</em> = <em>f(g(x))</em> = <em>f</em> (3<em>x</em> - 7)

= (3<em>x</em> - 7)² - 4(3<em>x</em> - 7) + 2

= (9<em>x</em>² - 42<em>x</em> + 49) + (-12<em>x</em> + 28) + 2

= 9<em>x</em>² - 57<em>x</em> + 79

(2) <em>g(x)</em> = -6<em>x</em> + 5, <em>h(x)</em> = -9<em>x</em> - 11

<em>(g</em> o <em>h) (x)</em> = <em>g(h(x))</em> = <em>g</em> (-9<em>x</em> - 11)

= -6(-9<em>x</em> - 11) + 5

= 54<em>x</em> + 66 + 5

= 54<em>x</em> + 71

(3) <em>f(x)</em> = √(2<em>x</em> - 5), <em>g(x)</em> = 5<em>x</em>² - 3

<em>(g</em> o <em>f) (x)</em> = <em>g(f(x))</em> = <em>g</em>(√(2<em>x</em> - 5))

= 5 (√(2<em>x</em> - 5))² - 3

= 5 (2<em>x</em> - 5) - 3

= 10<em>x</em> - 25 - 3

= 10<em>x</em> - 28

8 0
3 years ago
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