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Kay [80]
3 years ago
5

Solve this quadratic equation 2x^2+x-5 = 0

Mathematics
1 answer:
DIA [1.3K]3 years ago
3 0
2x^2+x-5=0
           +5 +5
2/2x^2+x=5/2
x^2+x=2.5
square root both sides
x+x=1.58
2/2x=1.58/2
x=0.79
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39,012 in expanded form
Helen [10]
30,000 + 9,000 + 10 + 2
6 0
3 years ago
What is the answer to this somebody!!!
Fed [463]

Use the geometric mean for right triangles as the sides are related to the hypotenuse. 30 - 6 = 24. We need that for our proportion. \frac{24}{x}=\frac{x}{30}. We cross-multiply to get x^2=720. Taking the square root of both sides gives us that x=12\sqrt{5}, the second choice above.

7 0
3 years ago
please answer this with an explanation since my homework requires it thanks and I really appreciate it
cricket20 [7]

Answer:

<h2>D. 10</h2>

Step-by-step explanation:

There are 5 students in each scenario, so we just add 5 and 5 together. The answer is 10.

I'm always happy to help :)

8 0
3 years ago
Read 2 more answers
I really need help on this
kvasek [131]

Answer:

i think its c i could be wrong but just go with your gut felling

Step-by-step explanation:

4 0
2 years ago
A boat traveled north for 28 miles, then turned x° southwest and traveled for 25 miles before stopping. When it stopped, the boa
zavuch27 [327]

Law of cosine is applicable to all the triangles. The value of the angle by which the boat turns is 39.195°.

<h3>What is the law of cosine?</h3>

Law of cosine helps us to find the third side of the triangle when 2 sides and an angle are known. It is formulated as,

c=\sqrt{a^{2}+b^{2}-2 a b \cdot \cos \gamma}

a, b, c is the sides of the triangle and,

\gamma = angle opposite c

A.)

Given to us

A boat traveled north for 28 miles, then turned x° southwest and traveled for 25 miles before stopping.

When it stopped, the boat was 18 miles from its starting point.

According to the given statements, sides and angles can be written,

a = 25 miles

b = 28 miles

c = 18 miles

\theta = x^o

Substitute the values,

c=\sqrt{a^{2}+b^{2}-2 a b \cdot \cos \gamma}

18=\sqrt{25^{2}+28^{2}-2(25)(28) \cdot \cos \theta}\\\\\theta = 39.195^o

Hence, the value of the angle by which the boat turns is 39.195°.

B.)

Given to us

Starting at the dock, it travels 18 miles to the left, then 25 miles up and to the right, and then 28 miles down and back to the dock.

The angle between 25 miles and 28 miles is x degrees.

According to the given statements, sides and angles can be written,

a = 25 miles

b = 28 miles

c = 18 miles

\theta = x^o

Substitute the values,

c=\sqrt{a^{2}+b^{2}-2 a b \cdot \cos \gamma}

18=\sqrt{25^{2}+28^{2}-2(25)(28) \cdot \cos \theta}\\\\\theta = 39.195^o

Hence, the value of the angle by which the boat turns is 39.195°.

Learn more about the Law of cosine:

brainly.com/question/17289163

5 0
2 years ago
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