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3 years ago

3. Find which of the following are in proportion: (1) 12, 16, 6,8

Mathematics

1 answer:

IgorLugansk [536]3 years ago

8 0

Answer:

thus, the ratios 8 : 16 and 6 : 12 are equal

Step-by-step explanation:

8 : 16 and 6: 12

8 : 16 = 1/2

6: 12= 1/2

Thus, the ratios 8 : 16 and 6 : 12 are equal

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Is number square root of 2,500 rational or irrational

ddd [48]

Rational

$\sqrt{2500 } = 50$

4 0

3 years ago

Apparently I'm supposed to do one at a time so here we go again

Nat2105 [25]

A) 60mph
B) 209.7 miles
C) 2 hours and 45 minutes

3 0

3 years ago

What is the value of C in the matrix equation below?

lara [203]

Given:

The matrix equation is:

$\begin{bmatrix}-18&3&5\end{bmatrix}-C=\begin{bmatrix}-22&1&12\end{bmatrix}$

To find:

The value of matrix C.

Solution:

Let $A=\begin{bmatrix}-18&3&5\end{bmatrix},B=\begin{bmatrix}-22&1&12\end{bmatrix}$ . Then the given equation can be rewritten as

$A-C=B$

$-C=B-A$

$C=-(B-A)$

$C=A-B$

On substituting the values of the matrices, we get

$C=\begin{bmatrix}-18&3&5\end{bmatrix}-\begin{bmatrix}-22&1&12\end{bmatrix}$

$C=\begin{bmatrix}-18-(-22)&3-1&5-12\end{bmatrix}$

$C=\begin{bmatrix}-18+22&2&-7\end{bmatrix}$

$C=\begin{bmatrix}-4&2&-7\end{bmatrix}$

Therefore, the correct option is C.

6 0

3 years ago

Read 2 more answers

2x + 5y = -3<br> 2x + 2y = 6

oee [108]

Answer:

(6, - 3 )

Step-by-step explanation:

Given the 2 equations

2x + 5y = - 3 → (1)

2x + 2y = 6 → (2)

Subtracting (1) from (2) term by term will eliminate the x- term

(2x - 2x) + (2y - 5y) = 6 - (- 3), that is

- 3y = 9 ( divide both sides by - 3 )

y = - 3

Substitute y = - 3 in either of the 2 equations and solve for x

Substituting y = - 3 in (1)

2x + 5(- 3) = - 3

2x - 15 = - 3 ( add 15 to both sides )

2x = 12 ( divide both sides by 2 )

x = 6

Solution is (6, - 3 )

8 0

3 years ago

Read 2 more answers

Solve the equation 2cosA = 3tanA

zavuch27 [327]

$\bf sin^2(\theta)+cos^2(\theta)=1\implies cos^2(\theta)=1-sin^2(\theta)
\\\\\\
tan(\theta)=\cfrac{sin(\theta)}{cos(\theta)}\\\\
-----------------------------\\\\

2cos(A)=3tan(A)\implies 2cos(A)=3\cfrac{sin(A)}{cos(A)}
\\\\\\
2cos^2(A)=3sin(A)\implies 2[1-sin^2(A)]=3sin(A)
\\\\\\
2-2sin^2(A)=3sin(A)\implies 2sin^2(A)+3sin(A)-2$

$\bf \\\\\\
0=[2sin(A)-1][sin(A)+2]\implies 
\begin{cases}
0=2sin(A)-1\\
1=2sin(A)\\
\frac{1}{2}=sin(A)\\\\
sin^{-1}\left( \frac{1}{2} \right)=\measuredangle A\\\\
\frac{\pi }{6},\frac{5\pi }{6}\\
----------\\
0=sin(A)+2\\
-2=sin(A)
\end{cases}$

now, as far as the second case....well, sine of anything is within the range of -1 or 1, so -1 < sin(A) < 1

now, we have -2 = sin(A), which simply is out of range for a valid sine, so there's no angle with such sine

so, only the first case are the valid angles for A

8 0

3 years ago