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3 years ago

Subtract.Write your answer as a fraction or a whole or mixed number. 10 1/3-3 2/3=

Mathematics

1 answer:

Tanzania [10]3 years ago

7 0

Answer:

as a mixed fraction 6 2/3, improper fraction 20/3

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Can someone please help me with numbers 1, a, b, c, 2, a, b, c

algol [13]

Thank you thank you very much

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3 years ago

Please help with this problem

Andre45 [30]

Answer:

The length of the short side is 14.5 units, the length of the other short side is 18.5 units, and the length of the longest side is 23.5 units.

Step-by-step explanation:

The Pythagorean Theorem

If a and b are the lengths of the legs of a right triangle and c is the length of the hypotenuse, then the sum of the squares of the lengths of the legs is equal to the square of the length of the hypotenuse.

This relationship is represented by the formula:

$a^2+b^2=c^2$

Applying the Pythagorean Theorem to find the lengths of the three sides we get:

$(x)^2+(x+4)^2=(x+9)^2\\\\2x^2+8x+16=x^2+18x+81\\\\2x^2+8x-65=x^2+18x\\\\2x^2-10x-65=x^2\\\\x^2-10x-65=0$

Solve with the quadratic formula

$\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}$

$x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$\mathrm{For\:}\quad a=1,\:b=-10,\:c=-65:\quad x_{1,\:2}=\frac{-\left(-10\right)\pm \sqrt{\left(-10\right)^2-4\cdot \:1\left(-65\right)}}{2\cdot \:1}\\\\x_{1}=\frac{-\left(-10\right)+ \sqrt{\left(-10\right)^2-4\cdot \:1\left(-65\right)}}{2\cdot \:1}=5+3\sqrt{10}\\\\x_{2}=\frac{-\left(-10\right)- \sqrt{\left(-10\right)^2-4\cdot \:1\left(-65\right)}}{2\cdot \:1}=5-3\sqrt{10}$

Because a length can only be positive, the only solution is

$x=5+3\sqrt{10}\approx 14.5$

The length of the short side is 14.5, the length of the other short side is $14.5+4=18.5$ , and the length of the longest side is $14.5+9=23.5$ .

3 0

3 years ago

Will mark brainliest! Questions are in the picture!

alexdok [17]

Answer:

52.9 inches .

Step-by-step explanation:

Given that the triangular indentation has an area of 100 in.² and the base and height of this traingle are represented by expressions 3x and x+3 respectively . We need to find out the perimeter to the nearest tenth.

As we know that the area of triangle is ,

$\rm\longrightarrow Area_{\triangle}=\dfrac{1}{2}(base)(height)$

Substituting the respective values,

$\rm\longrightarrow 100\ =\dfrac{1}{2}(x+3)(3x)\\\\\rm\longrightarrow 2(100) = 3x(x+3)\\\\\rm\longrightarrow200 = 3x^2+9x\\\\\rm\longrightarrow 3x^2+9x-200=0$

On using the Quadratic formula , we have;

$\rm\longrightarrow x =\dfrac{-9\pm\sqrt{9^2-4(3)(-200)}}{2(3)} \\\\\rm\longrightarrow x =\dfrac{-9\pm \sqrt{81+1200}}{6}\\\\\rm\longrightarrow x =\dfrac{-9\pm \sqrt{1281}}{6}$