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2 years ago

Please, i need help :(

Mathematics

1 answer:

ollegr [7]2 years ago

7 0

Answer:

First of all, find x from the Pythagorean theorem $a^2+b^2=c^2$ , which c is the hypotenuse.

$7^2+24^2=c^2$

$49+576=c^2$

$c^2=625$

c=25 or x=25

Next, we have to figure out the value of y and z. So, you need to use inverse trigonometric functions.

arctan $(\frac{7}{24})=y=16.3$ degrees. arctan $(\frac{24}{7})=z=73.7$ degrees.

All in all, the values are

x=25

y=16.3

z=73.7

Hope this helps and have a nice day!

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Milton spilled some ink on his homework paper. He can't read the coefficient of $x$, but he knows that the equation has two dist

Lera25 [3.4K]

Answer:

$Sum = -81$

Step-by-step explanation:

See the comment for complete question.

Given

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No coefficient of x^2

Required:

Determine the sum of all distinct positive integers of the coefficient of x

Reading through the complete question, we can see that the question has 3 terms which are:

x^2 ---- with no coefficient

x ---- with an unknown coefficient

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$x^2 + ax + 36 = 0$

Where a is the unknown coefficient

From the question, we understand that the equation has two negative integer solution. This can be represented as:

$x = -\alpha$ and $x = -\beta$

Using the above roots, the equation can be represented as:

$(x + \alpha)(x + \beta) = 0$

Open brackets

$x^2 + (\alpha + \beta)x + \alpha \beta = 0$

To compare the above equation to $x^2 + ax + 36 = 0$ , we have:

$a = \alpha + \beta$

$\alpha \beta = 36$

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The values of $\alpha$ and $\beta$ that satisfy $\alpha \beta = 36$ are:

$\alpha = -1$ and $\beta = -36$

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$a = -1 - 36 = -37$

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$a = -2 - 18 = -20$

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2 years ago

a(0.3−y)+1.1+2.4x (y−1.2) =0 =−1.2(x−0.5) Consider the system of equations above, where aaa is a constant. For which value o

Veseljchak [2.6K]

Answer:

For a = 1.22 there is one solution where y = 1.3

Step-by-step explanation:

Hi there!

Let´s write the system of equations:

a(0.3 - y) + 1.1 +2.4x(y-1.2) = 0

-1.2(x-0.5) = 0

Let´s solve the second equation for x:

-1.2(x-0.5) = 0

x- 0.5 = 0

x = 0.5

Now let´s repalce x = 0.5 and y = 1.3 in the first equation and solve it for a:

a(0.3 - y) + 1.1 +2.4x(y-1.2) = 0

a(0.3 - 1.3) + 1.1 + 2.4(0.5)(1.3 -1.2) = 0

a(-1) + 1.1 + 1.2(0.1) = 0

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Let´s check the solution and solve the system of equations with a = 1.22. Let´s solve the first equation for y:

1.22(0.3 - y) + 1.1 +2.4(0.5)(y-1.2) = 0

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Then, the answer is correct.

Have a nice day!

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