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babymother [125]
3 years ago
9

Without actually performing the operations, determine mentally the coefficient of the x²-term in the simplified form of the foll

owing expression.
(-5²+3x-4) - (-2x²+x-3) + (-9x²+2x-5).

The coefficient of the x²-term is___.
please i really need help with this one i'm stuck.
Mathematics
1 answer:
zavuch27 [327]3 years ago
5 0

Answer:

-7

Step-by-step explanation:

Because there is no multiplication, it is just the x^2 terms. There is a -(-2x^2) and a -9x^2, so it is 2x^2 and -9x^2 or -7x^2, so the coefficient is -7.

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Answer:

A person in a group where 5 people are sharing 3 small bags of snickers

Step-by-step explanation:

Let the number of calories in one bag of snickers =x

Number of calories in 3 small bags of snickers =3x

  • If 5 people share 3 small bags of snickers, 1 person's share  =\dfrac{3x}{5}

Number of calories in 2 small bags of snickers =2x

  • If 7 people share 2 small bags of snickers, 1 person's share  =\dfrac{2x}{7}

We then compare the two.

Let x=1

\dfrac{3}{5}=0.6$ calories per person\\\dfrac{2}{7}=0.29$ calories per person

Therefore, a person in a group where 5 people are sharing 3 small bags of snickers gets more calories.

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Which sphere has a volume of 0.288Pi in.3? A sphere with radius 0.4 inches. A sphere with radius 0.5 inches. A sphere with radiu
worty [1.4K]

<em>C. A sphere with radius 0.6 inches.</em>

<u><em>Here is why:</em></u>

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V=\frac{4}{3} \pi r^3

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V=\frac{4}{3}*\pi *0.6*0.6*0.6

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Find sin(a)&amp;cos(B), tan(a)&amp;cot(B), and sec(a)&amp;csc(B).​
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Answer:

Part A) sin(\alpha)=\frac{4}{7},\ cos(\beta)=\frac{4}{7}

Part B) tan(\alpha)=\frac{4}{\sqrt{33}},\ tan(\beta)=\frac{4}{\sqrt{33}}

Part C) sec(\alpha)=\frac{7}{\sqrt{33}},\ csc(\beta)=\frac{7}{\sqrt{33}}

Step-by-step explanation:

Part A) Find sin(\alpha)\ and\ cos(\beta)

we know that

If two angles are complementary, then the value of sine of one angle is equal to the cosine of the other angle

In this problem

\alpha+\beta=90^o ---> by complementary angles

so

sin(\alpha)=cos(\beta)

Find the value of sin(\alpha) in the right triangle of the figure

sin(\alpha)=\frac{8}{14} ---> opposite side divided by the hypotenuse

simplify

sin(\alpha)=\frac{4}{7}

therefore

sin(\alpha)=\frac{4}{7}

cos(\beta)=\frac{4}{7}

Part B) Find tan(\alpha)\ and\ cot(\beta)

we know that

If two angles are complementary, then the value of tangent of one angle is equal to the cotangent of the other angle

In this problem

\alpha+\beta=90^o ---> by complementary angles

so

tan(\alpha)=cot(\beta)

<em>Find the value of the length side adjacent to the angle alpha</em>

Applying the Pythagorean Theorem

Let

x ----> length side adjacent to angle alpha

14^2=x^2+8^2\\x^2=14^2-8^2\\x^2=132

x=\sqrt{132}\ units

simplify

x=2\sqrt{33}\ units

Find the value of tan(\alpha) in the right triangle of the figure

tan(\alpha)=\frac{8}{2\sqrt{33}} ---> opposite side divided by the adjacent side angle alpha

simplify

tan(\alpha)=\frac{4}{\sqrt{33}}

therefore

tan(\alpha)=\frac{4}{\sqrt{33}}

tan(\beta)=\frac{4}{\sqrt{33}}

Part C) Find sec(\alpha)\ and\ csc(\beta)

we know that

If two angles are complementary, then the value of secant of one angle is equal to the cosecant of the other angle

In this problem

\alpha+\beta=90^o ---> by complementary angles

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sec(\alpha)=csc(\beta)

Find the value of sec(\alpha) in the right triangle of the figure

sec(\alpha)=\frac{1}{cos(\alpha)}

Find the value of cos(\alpha)

cos(\alpha)=\frac{2\sqrt{33}}{14} ---> adjacent side divided by the hypotenuse

simplify

cos(\alpha)=\frac{\sqrt{33}}{7}

therefore

sec(\alpha)=\frac{7}{\sqrt{33}}

csc(\beta)=\frac{7}{\sqrt{33}}

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