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barxatty [35]
3 years ago
10

Please help on the production one.

Mathematics
1 answer:
kirza4 [7]3 years ago
5 0
5x + 8000 = 3x + 10000
x = 1000
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Can y’all help me with this plz
Sergeeva-Olga [200]

Answer:

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Step-by-step explanation:

4 0
3 years ago
Which number satisfies the inequality 9h + 15 > 55?
telo118 [61]
9h + 15 > 55
     - 15    - 15
9h > 40
h > 40/9
h > 4.44444.
any number greater than 4.44444
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3 years ago
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Please help I don't under stand how to do this ​
stealth61 [152]
You want to figure out what the variables equal to, all of these are parallelograms meaning opposite sides and angles are equal to each other.

In question 1 start with 3x+10=43, this means that 3x is 10 less than 43 which is 33, 33 divided by 3 is 11 meaning x=11.

Same thing can be done with the sides 124=4(4y-1), start by getting rid of the parentheses with multiplication to get 124=16y-4, this means that 16y is 4 more than 124, so how many times does 16 go into 128? 8 times, so x=11 and y=8

Question 2 can be solved because opposite angles are the same in a parallelogram, so u=66 degrees

You can find the sum of the interial angles with the formula 180(n-2) where n is the number of sides the shape has, a 4 sided shape has a sum of 360 degrees, so if we already have 2 angles that add up to a total of 132 degrees and there are only 2 angles left and both of those 2 angles have to be the same value then it’s as simple as dividing the remainder in half, 360-132=228 so the other 2 angles would each be 114, 114 divided into 3 parts is 38 so u=66 and v=38

Question 3 and 4 can be solved using the same rules used in question 1 and 2, just set the opposite sides equal to each other
6 0
2 years ago
Use the quadratic formula to solve the following equation -3x^2-x-3=0
tigry1 [53]

<u>Answer:</u>

x=-\frac{1}{6}-\frac{\sqrt{35}}{6} i \text { and } x=-\frac{1}{6}+\frac{\sqrt{35}}{6} i are two roots of equation -3 x^{2}-x-3=0

<u>Solution:</u>

Need to solve given equation using quadratic formula.

-3 x^{2}-x-3=0

General form of quadratic equation is a x^{2}+b x+c=0

And quadratic formula for getting roots of quadratic equation is

x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}

In our case b = -1 , a = -3 and c = -3

Calculating roots of the equation we get

\begin{array}{l}{x=\frac{-(-1) \pm \sqrt{(-1)^{2}-4(-3)(-3)}}{2 \times-3}} \\\\ {x=-\frac{1}{6} \pm\left(-\frac{\sqrt{-35}}{6}\right)}\end{array}

Since b^{2}-4 a c is equal to -35, which is less than zero, so given equation will not have real roots and have complex roots.

\begin{array}{l}{\text { Hence } x=-\frac{1}{6}-\frac{\sqrt{35}}{6} i \text { and } x=-\frac{1}{6}+\frac{\sqrt{35}}{6} i \text { are two roots of equation - }} \\ {3 x^{2}-x-3=0}\end{array}

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4 years ago
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