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LekaFEV [45]
3 years ago
14

sqrt \frac{256}{4\\}" align="absmiddle" class="latex-formula">
Mathematics
1 answer:
Usimov [2.4K]3 years ago
4 0

Answer:

8

Step-by-step explanation:

\sqrt{\frac{256}{4} }

\sqrt{\frac{16*16}{2*2} }

\sqrt{(\frac{16}{2})^2 }

square and root gets cancel . so ,

\frac{16}{2}

8

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Answer:

.0222

The probability of selecting a nickle then a dime without replacement is 2.22%.

(1/10)*(2/9)= .0222

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3 years ago
Determine whether the equation 3(4g+6)=2(6g+9) has one solution, no solution, or infinitely many solutions. The equation has . Q
taurus [48]

So firstly, foil 3(4g + 6) and 2 (6g + 9): 12g+18=12g+18

Now since the quantity is the same on both sides of this equation, this means that <u>this equation has infinite solutions.</u>

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HELP PLZZ will give brainliest &lt;3
melisa1 [442]

Answer:

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1

Step-by-step explanation:

First question:

You are given a side, a, and its opposite angle, A. You are also given side b. Use that in the law of sines and solve for the other angle, B.

\dfrac{a}{\sin A} = \dfrac{b}{\sin B}

\dfrac{10}{\sin 30^\circ} = \dfrac{40}{\sin B}

\dfrac{1}{0.5} = \dfrac{4}{\sin B}

\sin B = 2

The sine function can never equal 2, so there is no triangle in this case.

Answer: no triangle

Second question:

You are given a side, b, and its opposite angle, B. You are also given side c. Use that in the law of sines and solve for the other angle, C.

\dfrac{b}{\sin B} = \dfrac{c}{\sin C}

\dfrac{10}{\sin 63^\circ} = \dfrac{}{\sin C}

\sin C = \dfrac{8.9\sin 63^\circ}{10}

C = \sin^{-1} \dfrac{8.9\sin 63^\circ}{10}

C \approx 52.5^\circ

One triangle exists for sure. Now we see if there is a second one.

Now we look at the supplement of angle C.

m<C = 52.5°

supplement of angle C: m<C' = 180° - 52.5° = 127.5°

We add the measures of angles B and the supplement of angle C:

m<B + m<C' = 63° + 127.5° = 190.5°

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3 0
3 years ago
NO LINKS!! Use the method of substitution to solve the system. (If there's no solution, enter no solution). Part 11z​
Roman55 [17]

Answer:

(x,y)=\left(\; \boxed{-1,-8} \; \right)\quad \textsf{(smaller $x$-value)}

(x,y)=\left(\; \boxed{5,16} \; \right)\quad \textsf{(larger $x$-value)}

Step-by-step explanation:

Given system of equations:

\begin{cases}y=x^2-9\\y=4x-4\end{cases}

To solve by the method of substitution, substitute the first equation into the second equation and rearrange so that the equation equals zero:

\begin{aligned}x^2-9&=4x-4\\x^2-4x-9&=-4\\x^2-4x-5&=0\end{aligned}

Factor the quadratic:

\begin{aligned}x^2-4x-5&=0\\x^2-5x+x-5&=0\\x(x-5)+1(x-5)&=0\\(x+1)(x-5)&=0\end{aligned}

Apply the <u>zero-product property</u> and solve for x:

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\implies x-5=0 \implies x=5

Substitute the found values of x into the <u>second equation</u> and solve for y:

\begin{aligned}x=-1 \implies y&=4(-1)-4\\y&=-4-4\\y&=-8\end{aligned}

\begin{aligned}x=5 \implies y&=4(5)-4\\y&=20-4\\y&=16\end{aligned}

Therefore, the solutions are:

(x,y)=\left(\; \boxed{-1,-8} \; \right)\quad \textsf{(smaller $x$-value)}

(x,y)=\left(\; \boxed{5,16} \; \right)\quad \textsf{(larger $x$-value)}

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