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klemol [59]
3 years ago
15

Tell whether the angle measures can be those of a triangle.

Mathematics
1 answer:
babunello [35]3 years ago
5 0

No.

The 3 angles of a triangle need to equal 180 degrees.

The sum of the 3 given angles is greater than 180 ( 160 + 20 + 20 = 200) so it cannot form a triangle.

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Complete the identity
spin [16.1K]

Answer: cos(\pi-x)=-cos(x)

Step-by-step explanation:

We need to apply the following identity:

cos(A - B) = cos A*cos B + sinA*sin B

Then, applying this, you know that for cos(\pi-x):

cos(\pi-x)=cos(\pi)*cos(x)+sin(\pi)*sin(x)

We need to remember that:

cos(\pi)=-1 and sin(\pi)=0

Therefore, we need to substitute these values into cos(\pi-x)=cos(\pi)*cos(x)+sin(\pi)*sin(x).

Then, you get:

cos(\pi-x)=(-1)*cos(x)+0*sin(x)

cos(\pi-x)=-1cos(x)+0

cos(\pi-x)=-cos(x)

8 0
3 years ago
F(2)=<br> Please please help me
Whitepunk [10]

Answer:

F(2)=-2

Step-by-step explanation:

Y=F(2) AS SEEN IN GRAPH Y=-2

3 0
3 years ago
Read 2 more answers
Simplify the expression 43 + 2(3 − 2)
Sonja [21]

Answer:

))))))

Step-by-step explanation:

43+2(3-2)=43+2*3-2*2=43+6-4=45

8 0
3 years ago
Read 2 more answers
randy has 128 ounces of dog food. he feeds his dog 8 ounces of food each day. how much days will the dog food last
notsponge [240]
You do 128÷8 and you get 16
5 0
3 years ago
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How do you solve these equations? I don't want you to answer all of them, just tell me how to solve each type of equation on the
miss Akunina [59]

Answer:

8. Identify the common denominator; express each fraction using that denominator; combine the numerators of those rewritten fractions and express the result over the common denominator. Factor out any common factors from numerator and denominator in your result. (It's exactly the same set of instructions that apply for completely numerical fractions.)

9. As with numerical fractions, multiply the numerator by the inverse of the denominator; cancel common factors from numerator and denominator.

10. The method often recommended is to multiply the equation by a common denominator to eliminate the fractions. Then solve in the usual way. Check all answers. If one of the answers makes your multiplier (common denominator) be zero, it is extraneous. (10a cannot have extraneous solutions; 10b might)

Step-by-step explanation:

For a couple of these, it is helpful to remember that (a-b) = -(b-a).

<h3>8d.</h3>

\dfrac{5}{x+2}+\dfrac{25-x}{x^2-3x-10}=\dfrac{5(x-5)}{(x+2)(x-5)}+\dfrac{25-x}{(x+2)(x-5)}\\\\=\dfrac{5x-25+25-x}{(x+2)(x-5)}=\dfrac{4x}{x^2-3x-10}

___

<h3>9b.</h3>

\displaystyle\frac{\left(\frac{x}{x-2}\right)}{\left(\frac{2x}{2-x}\right)}=\frac{x}{x-2}\cdot\frac{-(x-2)}{2x}=\frac{-x(x-2)}{2x(x-2)}=-\frac{1}{2}

___

<h3>10b.</h3>

\dfrac{3}{x-1}+\dfrac{6}{x^2-3x+2}=2\\\\\dfrac{3(x-2)}{(x-1)(x-2)}+\dfrac{6}{(x-1)(x-2)}=\dfrac{2(x-1)(x-2)}{(x-1)(x-2)}\\\\3x-6+6=2(x^2-3x+2) \qquad\text{multiply by the denominator}\\\\2x^2-9x+4=0 \qquad\text{subtract 3x}\\\\(2x-1)(x-4)=0 \qquad\text{factor; x=1/2, x=4}

Neither solution makes any denominator be zero, so both are good solutions.

8 0
3 years ago
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