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4 years ago

HELP PLEASE..answer fast please and show all work and explain how to solve

Mathematics

1 answer:

zhenek [66]4 years ago

7 0

$\cfrac{a}{sinA}= \cfrac{b}{sinB} \ \ \to \\ sinB= \cfrac{b*sinA}{a} = \cfrac{10*sin15^o}{9}= 0.2875 \ \ \to \ m \angle B=16.7^o \\ \\ \\ m \angle C=180- m \angle A- m \angle B=180-15-16.7=148.3^o \\ \\ \\ \cfrac{a}{sinA}= \cfrac{c}{sinC} \ \ \to \ c= \cfrac{a*sinC}{sinA}= \cfrac{9*sin148.3^o}{sin15^o} \approx 18.3$

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Factor x^2-6x+9<br><br> A.) (x-3)<br> B.) (x-3)(x+3)<br> C.) (x-3)^2<br> D.) (x+3)^2

ra1l [238]

There must be two binomial factors. Thus, we eliminate Answer A, which contains only one such factor.

(x-3)(x+3) has the form (a-b)(a+b), which equals a^2 - 3^2. This is a special product. Eliminate Answer B, because a^2 - 3^2 does not equal x^2 - 6x + 9.

The middle term of x^2 - 6x + 9 is negative. This quadratic has the form x^2 - y^2, which is special product and whose factors are (x-y) and (x-y).

Thus, x^2-6x+9 = (x-3)(x-3) (Answer C)

8 0

3 years ago

Help needed please im stuck

defon

Answer:

a) b^3

b) x^3

Step-by-step explanation:

It can be helpful to think of the exponent as specifying the number of times the base is a factor in the product. Of course, common factors in the numerator and denominator cancel.

$\dfrac{b^7}{b^4}=\dfrac{b\cdot b\cdot b\cdot b\cdot b\cdot b\cdot b}{b\cdot b\cdot b\cdot b}=b\cdot b\cdot b=b^{7-4}=\boxed{b^3}$

$\dfrac{x\cdot x^5}{x^2\cdot x}=\dfrac{x\cdot x\cdot x\cdot x\cdot x\cdot x}{x\cdot x\cdot x}=x\cdot x\cdot x=x^{6-3}=\boxed{x^3}$

4 0

2 years ago

QUICKLY<br> A sphere has a radius of 4 in. Which equation finds the volume of the sphere?

MissTica

Volume of sphere formula: V = 4/3πr³

We are given the radius of 4, so, input that with r in the formula.

V = 4/3π(4)³

As we can see, option B has what we wrote above. So, that is the correct answer.

The answer is Option B.

Best of Luck!

6 0

3 years ago

Find an equation for the plane which passes through the point p(1,3,3) and contains the line: l:x(t)=3t,y(t)=2t,z(t)=4+3t

BARSIC [14]

Answer:

Step-by-step explanation:

Given that the plane passes through (1,3,3)

Also the plane contains the line x(t)=3t,y(t)=2t,z(t)=4+3t.

This means all points lying in this line will also lie in the plane.

Find out two points on this line.

First point: Let t =0. Point is (0,0,4)

Next point : Let t = 1: Point is (3,2,7)

Now we have 3 non collinear points (0,0,4) (3,2,7) and (1,3,3) lying on the plane.

Equation of the plane is

$\left[\begin{array}{ccc}x-0&y-0&z-4\\3&2&3\\1&3&-1\end{array}\right] =0$

Simplify to get

x(-2-9)-y(-3-3)+(z-4)(9-2)=0

i.e -11x+6y+7z-28 =0

11x-6y-7z+28 =0 is the equation of the plane.

6 0

4 years ago

Answer in simplest fractional form. negative 1 fifth + negative 3 fourths =

kramer

-1/5 + (-3/4)
= -1/5 - 3/4
= -4/20 - 15/20
= -19/20

5 0

4 years ago