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4 years ago

Can someone please help me with this ASAP!! (will mark as brainliest if correct)

Mathematics

2 answers:

sattari [20]4 years ago

6 0

Answer:

P = 6x-2

x = 14

Step-by-step explanation:

To find the perimeter, we add the sides

P = 4x-8 + 3-2x + 5x+3

Combine like terms

P = 4x-3x+5x -8+3+3

= 6x -2

Now we know the perimeter

82 = 6x-2

Add 2 to each side

82+2 = 6x-2+2

84 = 6x

Divide each side by 6

84/6 = 6x/6

14=x

Vladimir [108]4 years ago

3 0

Answer:

A) 7x - 2

B) x = 12

Step-by-step explanation:

4x - 8 + 3 - 2x + 5x + 3

7x - 2

7x - 2 = 82

7x = 84

x = 12

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3 years ago

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4 years ago

find all possible value of the given variable

mamaluj [8]

$1.\\ \\ h^2+5h=0 \\ \\h(x+5)=0\\ \\x=0 \ \ \ or \ \ \ x+5 =0\ \ |-5\\ \\x+5-5=0-5\\ \\x=0 \ \ \ or \ \ \ x=-5$

$2.\\ \\ z^2-z=0\\ \\z(x-1)=0\\ \\z=0 \ \ \ or \ \ \ z-1 =0 \ \ | +1\\ \\z-1+1 =0 +1 \\ \\x=0 \ \ \ or \ \ \ z=1$

$3.\\ \\m^2+13m+40=0 \\ \\a=1 ,\ b=13, \ c=40 \\ \\\Delta =b^2-4ac =13^2-4\cdot 1\cdot 40=169 - 1600=-1431 \\ \\and \ we \ know \ when \ \Delta \ is \ negative, \ theres \ no \solution$

$4.\\ \\z^2-3z=0 \\ \\ (z-3)=0\\ \\z=0 \ \ \ or \ \ \ z-3 =0\ \ |+3\\ \\ z-3+3=0+3\\ \\z=0 \ \ \ or \ \ \ z=3$

$5.\\ \\q^2+7q=0 \\ \\q(q+7)=0\\ \\q=0 \ \ \ or \ \ \ q+7 =0\ \ |-7\\ \\q+7-7=0-7\\ \\q=0 \ \ \ or \ \ \ q=-7$

$6.\\ \\k^2+2k=0\\ \\k(k+2)=0\\ \\k=0 \ \ \ or \ \ \ k+2 =0\ \ |-2\\ \\k+2-2=0-2\\ \\k=0 \ \ \ or \ \ \ k=-2$

$7. \\ \\ x^2-3x-70=0 \\ \\a=1,\ b=-3, \ c=-70 \\ \\\Delta =b^2-4ac = (-3)^2-4\cdot 1\cdot (-70)= 9+280=289\\ \\ x_{1}=\frac{-b-\sqrt{\Delta} }{2a}=\frac{3-\sqrt{289}}{2 }=\frac{ 3-17}{2}=\frac{-14}{2}=-7$

$x_{2}=\frac{-b+\sqrt{\Delta} }{2a}=\frac{3+\sqrt{289}}{2 }=\frac{ 3+17}{2}=\frac{20}{2}=10\\ \\(x+7)(x-10)=0$

$8.\\ \\q^2+7q-60=0 \\ \\a=1,\ b=7, \ q=-60 \\ \\\Delta =b^2-4ac = 7^2-4\cdot 1\cdot (-60)=49+240=289 \\ \\ x_{1}=\frac{-b-\sqrt{\Delta} }{2a}=\frac{-7-\sqrt{289}}{2 }=\frac{ -7-17}{2}=\frac{-24}{2}=-12$

$x_{2}=\frac{-b+\sqrt{\Delta} }{2a}=\frac{-7+\sqrt{289}}{2 }=\frac{ -7+17}{2}=\frac{ 10}{2}= 5\\ \\(x+12)(x-5)=0$

$9.\\ \\z^2+9z-36=0 \\ \\a=1,\ b=9, \ q=-36 \\ \\\Delta =b^2-4ac = 9^2-4\cdot 1\cdot (-36)= 81+144=225\\ \\ x_{1}=\frac{-b-\sqrt{\Delta} }{2a}=\frac{-9-\sqrt{225}}{2 }=\frac{ -9-15}{2}=\frac{-24}{2}=-12$

$x_{2}=\frac{-b+\sqrt{\Delta} }{2a}=\frac{-9+\sqrt{225}}{2 }=\frac{ -9+15}{2}=\frac{6}{2}=3\\ \\(x+11)(x-3)=0$

$10.\\ \\d^2-13d+22=0 \\ \\a=1,\ b=-13, \ q=22 \\ \\\Delta =b^2-4ac = (-13)^2-4\cdot 1\cdot 22= 169-88=81\\ \\ d_{1}=\frac{-b-\sqrt{\Delta} }{2a}=\frac{13-\sqrt{81}}{2 }=\frac{ 13-9}{2}=\frac{4}{2}=2$

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7 0

3 years ago

Find the complement of the set given that

iris [78.8K]

Answer:

A ' = {-2,-1,0,1, 4,7}

Step-by-step explanation:

Here, the Universal set U is given as:

U = {x | x is in I and −3 ≤ x ≤ 7}

⇒ U = { -3,-2,-1,0,1,2,3,4,5,6,7}

Now given: A = {−3, 2, 3, 5, 6}

Compliment A set of any set A is the set of all the elements which is in the universal set U but NOT in A.

So, here the set of elements which are in U but not in A is

A ' = {-2,-1,0,1, 4,7}

So here the elements in the compliment of the given set -2,-1,0,1,4,7

4 0

3 years ago

A parallelogram with sides of 6 and 10 has an area of 30. Find the measure of the small angle of the parallelogram.

dmitriy555 [2]

<span>A = 2 * (0.5ab) + b (10 - a) = ab + 10b - ab = 10b

10b = 30√2; b = 3√2

sin α = 3√2 / 6; α = 45 degrees

Small angle: 45°; Large angle: 135°</span><span>
</span>

4 0

4 years ago