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Juli2301 [7.4K]
3 years ago
10

Please help me with this math

Mathematics
1 answer:
victus00 [196]3 years ago
5 0

Answer:

5

Step-by-step explanation:

These two angles are equal to each other

set the formulas equal to each other and solve

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2664

Step-by-step explanation:

9.25×36+(380÷20)×8

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Square root3(square root3+2square root12) simplify it.Answer should be square root 3.
julsineya [31]

Answer:

sqrt3[sqrt3+2sqrt12]=sqrt3[sqrt3+4sqrt3] = sqrt3[5sqrt3= 3x5 =15

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A student solved the equation 2x+6=12 using algebra tiles. incorrectly says the solution is 9. Solve the equation. What mistake
alexdok [17]
Instead of subtracting 6 from 12 then dividing they added 6 to 12 & ended up with 18. 18 by 2 is 9.
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I need help with my pre-calculus homework, please show me how to solve them step by step if possible. The image of the problem i
Anna [14]

We know that the law of sines states that:

\frac{\sin\alpha}{a}=\frac{\sin\beta}{b}

For simplicity, let:

\beta=m\angle A_1BC

In triangle A1BC this leads to:

\begin{gathered} \frac{\sin31}{5}=\frac{\sin\beta}{6} \\ \sin\beta=\frac{6}{5}\sin31 \\ \beta=\sin^{-1}(\frac{6}{5}\sin31) \\ \beta=38.174 \end{gathered}

Therefore:

m\angle A_1BC=38.174

Now, triangle A2BC is isosceles which means that both the base angles are equal, since angle CA1B and A2A1B are supplementary we have:

m\angle CA_2B=180-38.174=141.826

8 0
1 year ago
H(x) = x2 1 k(x) = x – 2 (h k)(2) = (h – k)(3) = Evaluate 3h(2) 2k(3) =.
natima [27]

Quadratic equation is the equation in which only one variable is unknown. The highest power of the variable is 2.The value of the given functions are,

(h+k)(x)=5

(h-k)(x)=9

3h(2)+2k(3)=17

<h3>Given information-</h3>

The given function is,

h(x)=x^2+1

k(x)=x-2

<h3>Quadratic equation</h3>

Quadratic equation is the equation in which only one variable is unknown. The highest power of the variable is 2.

1) The value of the function (h+k)(2),

(h+k)(x)=h(x)+k(x)

(h+k)(x)=x^2+1+x-2

(h+k)(2)=2^2+1+2-2

(h+k)(x)=5

2)The value of the function (h-k)(3),

(h-k)(x)=h(x)-k(x)

(h-k)(x)=x^2+1-x+2

(h-k)(3)=3^2+1-3+2

(h-k)(x)=9

3) The value of the function 3h(2)+2k(3)

3h(x)+2k(x)=3x^2+3+2x-2\times 2

3h(2)+2k(3)=3\times2^2+3+2\times2-2\times 2

3h(2)+2k(3)=17

Hence the value of the given functions are,

(h+k)(x)=5

(h-k)(x)=9

3h(2)+2k(3)=17

Learn more about the quadratic equation here;

brainly.com/question/2263981

4 0
3 years ago
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