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3 years ago

What measurements could form a triangle

Mathematics

1 answer:

max2010maxim [7]3 years ago

6 0

There are different types of triangles obviously. One specific measurement you could use is a 30°-40°-50°.

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Brock is a plumber. He charges a flat rate of $40 to visit a house to inspect it's plumbing. He charges an additional $20 for ev

rodikova [14]

Y= 40+(20x), where you asking for something like that?

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3 years ago

Jillian is using integer tiles to add 7 + (negative 2). She uses the steps below.Step 17 positive tiles.Step 27 positive tiles a

NemiM [27]

Answer:

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2 years ago

Which are the solutions of the quadratic equation? x2 = –5x – 3 –5, 0 StartFraction negative 5 minus StartRoot 13 EndRoot Over 2

boyakko [2]

<u>Given</u>:

The quadratic equation is $x^{2}=-5 x-3$

We need to determine the solutions of the quadratic equation.

<u>Solution</u>:

Let us solve the equation to determine the value of x.

Adding both sides of the equation by 5x and 3, we get;

$x^{2}+5 x+3=0$

The solution of the equation can be determined using quadratic formula.

Thus, we get;

$x=\frac{-5 \pm \sqrt{5^{2}-4 \cdot 1 \cdot 3}}{2 \cdot 1}$

$x=\frac{-5 \pm \sqrt{25-12}}{2 }$

$x=\frac{-5 \pm \sqrt{13}}{2 }$

Thus, the two roots of the equation are $x=\frac{-5 + \sqrt{13}}{2 }$ and $x=\frac{-5- \sqrt{13}}{2 }$

Hence, the solutions of the equation are $x=\frac{-5 + \sqrt{13}}{2 }$ and $x=\frac{-5- \sqrt{13}}{2 }$

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3 years ago

Read 2 more answers

What are the real and complex solutions of the polynomial equation? x^4-41x^2=-400. *work needed*

pickupchik [31]

<h3>Answer: x = -5, -4, 4 and 5.</h3><h3>All four zeros are real solutions.</h3>

Step-by-step explanation:

Given the polynomial equation $x^4-41x^2=-400$ .

Adding 400 on both sides to get rid 400 from right side and set 0 on right side, we get

$x^4-41x^2+400=-400+400$ .

$x^4-41x^2+400=0$ .

Factoring by product sum rule.

We need product of 400 and sum upto -41.

We can see that 400 = -25 × -16 = 400 and -25-16 = -41.

Therefore,

$x^4-25x^2-16x^2+400=0$

Making it into two groups, we get

$(x^4-25x^2)+(-16x^2+400)=0$

Factoring out GCF of each group, we get

$x^2(x^2-25)-16(x^2-25)=0$

$(x^2-25)(x^2-16) =0$

Factoring out $(x^2-25)$ and $(x^2-16)$ separately by difference of the squares identity $a^2-b^2=(a-b)(a+b)$ , we get

$(x^2-25) = x^2-5^2= (x-5)(x+5)$ and

$x^2-16 = x^2-4^2 =(x-4)(x+4)$ .

Therefore,

$(x-5)(x+5)(x-4)(x+4) =0$

Applying zero product rule,

x-5=0

x+5=0

x-4=0 and

x+4=0.

Therefore,

<h3>x = -5, -4, 4 and 5.</h3><h3>All four zeros are real solutions.</h3>

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3 years ago

Solve the equation 1 + 5x - 10 = 7x - 9 - 2x

Tom [10]

<span>All real numbers are solutions.

</span>

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3 years ago