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yKpoI14uk [10]
2 years ago
14

Find the ratio of a to b if 2a=3c and 12c=7b​

Mathematics
1 answer:
postnew [5]2 years ago
6 0

Answer:

I think the answer is 1a : 14b

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HELP! GIVING BRAINLIEST IF ANSWERED CORRECTLY AND FOR EACH SECTION!!!
Anon25 [30]

Answer:

[14] 139

[15] 90

[16] 41

[17] 49

[18] 139

[19] 131

Step-by-step explanation:

[14] Since m ∠ CEB is vertical with m ∠ AED = 139

[15] Knowing that m ∠FBD = 90 degree then m ∠ CEF = 90 degree

[16] Since m ∠ DEB is vertical to m ∠ADC Thus answer = 41

Note: vertical angles are angles opposite each other where two lines cross.

[17] Since m ∠ FED = 90 degree and m ∠ DEB = 41 then 90 - 41 = 49

[18] To Find m ∠CEB we have to subtract from m ∠ DEB

180 - 41 = 139

[19] Knowing that m ∠ FED = 90 degree and m ∠DEB = 41

Then m ∠ AEF =  90 +41=131

<u><em>~Lenvy~</em></u>

4 0
2 years ago
Christian is rewriting an expression of the form y = ax2 + bx + c in the form y = a(x – h)2 + k. Which of the following must be
notsponge [240]
"h and k cannot both equal zero" -- yes, it can. if the vertex of a parabola is at (0, 0), there's nothing incorrect/invalid about that!!

"k and c have the same value" -- k and c do not have the same value. "k" is the y-value of the vertex and c is the constant in your quadratic equation, and the constant is not necessarily the y-value.

"the value of a remains the same" -- this is true. the a's in your equations are the same values, because the a-value is the coefficient of the x-variable in both equations. y = a(x - h)^2 and y = ax^2 -- both of these have a applying to your x-variables.

"h is equal to one half -b" -- this isn't true. the formula for calculating the x value of the vertex (h is the x-value of the vertex) is h = (-b/2a). -b/2a is not the same as one half -b because this answer choice doesn't involve the a-value.
3 0
3 years ago
Read 2 more answers
Evaluate 10.2x+9.4y when x=2 and y=3​
STALIN [3.7K]

Steps to solve:

10.2x + 9.4y when x = 2 and y = 3 ​

= 10.2(2) + 9.4(3)

= 20.4 + 28.2

= 48.6

______

Best Regards,

Wolfyy :)

3 0
3 years ago
Read 2 more answers
Inverse this matrix
SIZIF [17.4K]
In order for the inverse to exist, the matrix cannot be singular, so we need to first examine the conditions for existence of the inverse.

Compute the determinant. The easiest way might be a cofactor expansion along either the first row or third column; I'll do the first.

\begin{vmatrix}x&0&0\\0&1&0\\w&0&x\end{vmatrix}=x\begin{vmatrix}1&0\\0&x\end{vmatrix}=x^2

The matrix is then singular whenever x\neq0.

With this in mind, compute the inverse.

\mathbf A=\begin{bmatrix}x&0&0\\0&1&0\\w&0&x\end{bmatrix}\implies\mathbf A^{-1}=\dfrac1{x^2}\begin{bmatrix}a&0&-w\\0&-x^2&0\\0&0&x\end{bmatrix}=\begin{bmatrix}\frac a{x^2}&0&-\frac w{x^2}\\0&-1&0\\0&0&\frac1x\end{bmatrix}
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3 years ago
What is fig(-1)) for f(x) = 2x - 3 and g(x) = x² +11?​
iVinArrow [24]

Answer:

retype the question or send a picture please

5 0
3 years ago
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