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3 years ago

Answer this fast and correct and you will get a thanks, a brainliest and a 5-star review along with 10 points!

Mathematics

2 answers:

hammer [34]3 years ago

7 0

<h3>Answer: 50</h3>

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Work Shown:

55% = 55/100 = 0.55

55% of 200 = 0.55*200 = 110 people prefer dogs

30% of 200 = 0.30*200 = 60 people prefer cats

Subtract the results: 110-60 = 50

victus00 [196]3 years ago

6 0

The answer is 25% if you subtract 55 by 30 you will get 25

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Please help thank you!❤️

aniked [119]

Answer:

Step-by-step explanation:

because 1 dozen cost $1

8 0

2 years ago

ABCD rhombus, AC:BD = 4:3 Perimeter ABCD = 20 cm Find: Area of ABCD

AlladinOne [14]

Given that the perimeter of rhombus ABCD is 20 cm, the length of the sides will be:
length=20/4=5 cm
the ratio of the diagonals is 4:3, hence suppose the common factor on the diagonals is x such that:
AC=4x and BD=3x
using Pythagorean theorem, the length of one side of the rhombus will be:
c^2=a^2+b^2
substituting our values we get:
5²=(2x)²+(1.5x)²
25=4x²+2.25x²
25=6.25x²
x²=4
x=2
hence the length of the diagonals will be:
AC=4x=4×2=8 cm
BD=3x=3×2=6 cm
Hence the area of the rhombus wll be:
Area=1/2(AC×BD)
=1/2×8×6
=24 cm²

6 0

3 years ago

I meedddddddde helppp plsss pls

Marina86 [1]

Your answer is 300 miles

3 0

3 years ago

Rounded to the nearest hundredth 0.96

Andru [333]

0.963=0.96 rounded to the nearest hundredth
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8 0

3 years ago

Read 2 more answers

◆ Quadratic Equations ◆<br>Please help !

Mila [183]

I'm sure there's an easier way of solving it than the way I did, but I'm not sure what it could be. Never dealt with a problem like this before.

Anyway, I just plugged in and tested. Chose random values for a, b, c, and d, which follow the rule 0 < a < b < c < d:

a = 1
b = 2
c = 3
d = 4

$\sf ax^2+(1-a(b+c))x+abc-d)$

$\sf 1x^2+(1-1(2+3))x+(1)(2)(3)-(4))$

Simplify into standard form:

$\sf x^2+(1-1(5))x+6-4$

$\sf x^2+(1-5)x+2$

$\sf x^2-4x+2$

Use the quadratic formula to solve:

$\sf x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$

For functions in the form of $\sf ax^2+bx+c$ . So in this case:

a = 1
b = -4
c = 2

Plug them in:

$\sf x=\dfrac{4\pm\sqrt{(-4)^2-4(1)(2)}}{2(1)}$

Solve for 'x':

$\sf x=\dfrac{4\pm\sqrt{16-8}}{2}$

$\sf x=\dfrac{4\pm\sqrt{8}}{2}$

$\sf x\approx\dfrac{4\pm 2.83}{2}$

$\sf x\approx 0.59,3.41$

So the answer would be A.

3 0

3 years ago