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valina [46]
3 years ago
13

ANSWER PLS ITS URGENT NEED DONE IN ONE MIN BUS IS COMING!!!! pls dont let me fail!!!!! (will mark as brainliest)

Mathematics
2 answers:
nalin [4]3 years ago
7 0

Answer:

8.) x = 5, x=-12

9.)x = 5, x = -8

10.)x = -1/5, x = -2

Hope I helped :)

jenyasd209 [6]3 years ago
4 0

Answer:

10. -2, -⅕ = x

9. -8, 5 = x

8. -12, 5 = x

7. -5, 6 = x

Step-by-step explanation:

10. 5{x}^{2} + 11x + 2 = 0 \\  \\ (5{x}^{2} + x) + (10x + 2) = 0 \\ x(5x + 1) + 2(5x + 1) \\  \\ (x + 2)(5x + 1) = 0 \\  \\  -2, \:  - \frac{1}{5} = x

9.

2{x}^{2} + 6x - 80 = 0 \\  \\ (2{x}^{2} + 16x) - (10x - 80) = 0 \\ 2x(x + 8) -10(x + 8) = 0 \\  \\ (x + 8)(2x - 10) = 0 \\  \\  -8, \: 5 = x

8.

{x}^{2} + 8x - 65 = x - 5 \:→ {x}^{2} + 7x - 60 \\  \\ ({x}^{2} - 5x)  + (12x - 60) \\ x(x - 5) + 12(x - 5) \\  \\ (x + 12)(x - 5) = 0 \\  \\ 5, \:  -12 = x

7.

{x}^{2} = x + 30 →  - {x}^{2} + x + 30 \\  \\  (-{x}^{2} - 5x) + (6x + 30) \\  -x(x + 5) + 6(x + 5) \\  \\  -(x - 6)(x + 5) = 0 \\  \\  -5, \: 6 = x

I am joyous to assist you anytime.

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Hi!
This is a fun one, as it delves into basic trigonometry.

We're going to use the Pythagorean theorem here, which says that for right triangles where "c" is the hypotenuse,

a² + b² = c²

We have to split this large triangle into two parts, both of which are right triangles. (This is why they drew a line in the middle to tell you that the larger triangle is composed of two right triangles.)

Let's do the one on the right first.

We know that the length of the hypotenuse is 10, and that the length of one of the legs is 6.5. If we plug this into our equation, we'll get the length of the other leg. I'm choosing "b" to be 6.5, but it really doesn't matter if you pick "a" or "b", so long as you reserve "c" for the hypotenuse (longest side).

a² + 6.5² = 10²
a² + 42.25 = 100
a² = 57.75
√a² = √57.75
a ≈ 7.6

Therefore, the length of DC is about 7.6.

Find the length of AD using the same method (7.5 is the hypotenuse "c", and 6.5 is one of the legs "a" or "b"). Then, once you have AD, add the lengths of AD and DC to get AC.

Have a great one!
7 0
2 years ago
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Eduardwww [97]

Answer:

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4 0
2 years ago
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Tpy6a [65]

Given:

The fractions are:

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The each fraction as the sum of a whole number and a fraction less than 1.

Solution:

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\dfrac{6}{5}=\dfrac{5+1}{5}

\dfrac{6}{5}=\dfrac{5}{5}+\dfrac{1}{5}

\dfrac{6}{5}=1+\dfrac{1}{5}

Therefore, the given fraction \dfrac{6}{5} can be written as 1+\dfrac{1}{5}.

The given fraction is \dfrac{11}{7}.

\dfrac{11}{7}=\dfrac{7+4}{7}

\dfrac{11}{7}=\dfrac{7}{7}+\dfrac{4}{7}

\dfrac{11}{7}=1+\dfrac{4}{7}

Therefore, the given fraction \dfrac{11}{7} can be written as 1+\dfrac{4}{7}.

The given fraction is \dfrac{21}{4}.

\dfrac{21}{4}=\dfrac{20+1}{4}

\dfrac{21}{4}=\dfrac{20}{4}+\dfrac{1}{4}

\dfrac{21}{4}=5+\dfrac{1}{4}

Therefore, the given fraction \dfrac{21}{4} can be written as 5+\dfrac{1}{4}.

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nlexa [21]

Answer:

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288/2 = 144

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