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2 years ago

Which equation could be used to solve for the measure of side Q in the triangle QRS shown below given...

Mathematics

1 answer:

Rama09 [41]2 years ago

8 0

The image is missing, so i have attached it

Answer:

Equation is;

q/sin 105 = 15/sin 30

Step-by-step explanation:

We are given;

Angle Q = 105°

Angle S = 30°

s = 15

From sine rule we know that;

a/sin A = b/sin B = c/sin C.

Thus, applying that to this question, we can find side q.

Thus;. q/Sin Q = s/Sin S

Plugging in the relevant values to obtain ;

q/sin 105 = 15/sin 30

You might be interested in

Is 120% less than, more than or equal to 1

kati45 [8]

120% is the same as 120/100

Divide 120 with 100

120/100 = 1.2

1.2 > 1 ∴ 120% is 'more than' 1

hope this helps

6 0

3 years ago

What is 1 +1 pls HELP

ElenaW [278]

Answer:

Step-by-step explanation:

Subject: Re: Need the math proof for 1 + 1 = 2

The proof starts from the Peano Postulates, which define the natural

numbers N. N is the smallest set satisfying these postulates:

P1. 1 is in N.

P2. If x is in N, then its "successor" x' is in N.

P3. There is no x such that x' = 1.

P4. If x isn't 1, then there is a y in N such that y' = x.

P5. If S is a subset of N, 1 is in S, and the implication

(x in S => x' in S) holds, then S = N.

Then you have to define addition recursively:

Def: Let a and b be in N. If b = 1, then define a + b = a'

(using P1 and P2). If b isn't 1, then let c' = b, with c in N

(using P4), and define a + b = (a + c)'.

Then you have to define 2:

Def: 2 = 1'

2 is in N by P1, P2, and the definition of 2.

Theorem: 1 + 1 = 2

Proof: Use the first part of the definition of + with a = b = 1.

Then 1 + 1 = 1' = 2 Q.E.D.

Note: There is an alternate formulation of the Peano Postulates which

replaces 1 with 0 in P1, P3, P4, and P5. Then you have to change the

definition of addition to this:

Def: Let a and b be in N. If b = 0, then define a + b = a.

If b isn't 0, then let c' = b, with c in N, and define

a + b = (a + c)'.

You also have to define 1 = 0', and 2 = 1'. Then the proof of the

Theorem above is a little different:

Proof: Use the second part of the definition of + first:

1 + 1 = (1 + 0)'

Now use the first part of the definition of + on the sum in

parentheses: 1 + 1 = (1)' = 1' = 2 Q.E.D.

3 0

3 years ago

Someone please help me will give BRAILIEST!!!!!!

Misha Larkins [42]

The answer would be 1/5

8 0

3 years ago

Sara had 324 dollars to spend on 9 books. After buying them she had 18 dollars. How much did each book cost?

larisa86 [58]

Each book costs $34 for this problem you need to first subtract 18 from 324 which gets you 306 then divide 306 by 9 and you get 34

7 0

3 years ago

Read 2 more answers

Hi, can someone please help, and possibly explain the answer please.

mr_godi [17]

Answer:

$x=\frac{-(-2)\±\sqrt{(-2)^2-4(3)(0)} }{2(3)}$

Step-by-step explanation:

Quadratic formula: $x=\frac{-b\±\sqrt{b^2-4ac} }{2a}$ when the equation is $0=ax^2+bx+c$

The given equation is $1=-2x+3x^2+1$ . Let's first arrange this so its format looks like $y=ax^2+bx+c$ :

$1=-2x+3x^2+1$

$1=3x^2-2x+1$

Subtract 1 from both sides of the equation

$1-1=3x^2-2x+1-1\\0=3x^2-2x+0$

Now, we can easily identify 3 as a, -2 as b and 0 as c. Plug these into the quadratic formula:

$x=\frac{-b\±\sqrt{b^2-4ac} }{2a}\\x=\frac{-(-2)\±\sqrt{(-2)^2-4(3)(0)} }{2(3)}$

I hope this helps!

8 0

2 years ago