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3 years ago

What are both of these?

Mathematics

1 answer:

allochka39001 [22]3 years ago

3 0

Pshhhh she said math negatives I’m pretty sure.

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I need help its my homework and it's due today Find the base angles.

MrRa [10]

Answer:

Each base angle would be 70°.

Step-by-step explanation:

*Base angles means the two equal angles formed on the base are called base angles.

Base means the third unequal side of the triangle. *

Given triangle is isosceles triangle.

The angle of vertex which is given is 40°.

Let each base angle be x.

We know that,

There are two base angles in a isosceles triangle.

And, Sum of angles of triangle is 180° .

Therefore,

$\tt \implies \: x + x + 40 {}^{ \circ} = 180 {}^\circ$

Now Solve For x. That would be the solution.

Steps:

Combine like terms:

$\tt \implies2x + 40{}^{ \circ} = 180{}^{ \circ}$

Subtract 40 from both sides:

$\tt \implies2x + 40{}^{ \circ} - 40 = 180{}^{ \circ} - 40{}^{ \circ}$

Simplify the LHS and RHS:

$\tt \implies2x + 0 = 140{}^{ \circ}$

$\tt \implies2x = 140 {}^{ \circ}$

Divide both sides by 2:

$\tt \implies \cfrac{2x}{2} = \cfrac{140{}^{ \circ} }{2{}^{ \circ} }$

Use cancellation method and cancel LHS and RHS:

$\tt \implies \cfrac{ \cancel2x}{ \cancel2} = \cfrac{ \cancel{140^{ \circ}} }{ \cancel{2{}^{ \circ} }}$

$\tt \implies \cfrac{1x}{1} = \cfrac{70{}^{ \circ} }{1{}^{ \circ} }$

$\tt \implies{1x} = {70}^{ \circ}$

$\tt \implies{x} = 70{}^{ \circ}$

Hence, each base angle would be 70°.

Verification:

As we know that sum of angles of triangle is 180°.

So,

$\tt \implies \: Base \: angle + other \: base \: angle + vertex \: angle= 180{}^{ \circ}$

We got that one base angles of the given isosceles triangle is 70° and other is 70°, and the vertex angle is 40°[Given].

$\tt \implies70{}^{ \circ} + 70{}^{ \circ} + 40{}^{ \circ} = 180{}^{ \circ}$

Solve it.

$\tt \implies140{}^{ \circ} + 40{}^{ \circ} = 180{}^{ \circ}$

$\tt \implies180{}^{ \circ} = 180{}^{ \circ}$

$\tt \: LHS = RHS$

$\star \sf \: Hence, Verified.$

$\rule{225pt}{2pt}$

I hope this helps!

Let me know if you have any questions.

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Remove all grouping symbols and combine like terms.

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$(3x+x^2)-\{4x^2-3x+[2x^2-x-(5x^2+4x)]\}\\\\=3x+x^2-\{4x^2-3x+[2x^2-x-5x^2-4x]\}\\\\=3x+x^2-\{4x^2-3x+[-3x^2-5x]\}\\\\=3x+x^2-\{4x^2-3x-3x^2-5x\}\\\\=3x+x^2-\{x^2-8x\}\\\\=3x+x^2-x^2+8x\\\\=11x$

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Claire has two $20 bills, two $5 bills, and three

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Answer:

153

Step-by-step explanation: 20 x 2 + 5 x 2 + 3 x 1+ 100

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The answer is false

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Solve for x 6(2x - 5) = -(x + 4)

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Answer:

x=2

Step-by-step explanation:

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