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3 years ago

Please heLp :’) thank you

Mathematics

1 answer:

galben [10]3 years ago

5 0

Answer:

Step-by-step explanation:

Just like in the last problem, we have a right triangle but in this case, x is a leg, 10 / 2 = 5 is the other leg, and 13 is the hypotenuse. Using the 5 - 12 - 13 Pythagorean Triple, we know that x = 12.

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Please help me solve for x

kkurt [141]

$\qquad \qquad\huge \underline{\boxed{\sf Answer}}$

The shown pair of angles are Co interior angle pair, therefore the sum of those angles is 180°

$\qquad \sf \dashrightarrow \: 130 \degree + 7x + 1 \degree = 180 \degree$

$\qquad \sf \dashrightarrow \: 131 \degree + 7x = 180 \degree$

$\qquad \sf \dashrightarrow \: 7x= 180 \degree - 131 \degree$

$\qquad \sf \dashrightarrow \: 7x= 49 \degree$

$\qquad \sf \dashrightarrow \: x=49 \degree \div 7$

$\qquad \sf \dashrightarrow \: x=7 \degree$

The required value of x is 7°

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3 years ago

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Brrunno [24]

Answer:

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Step-by-step explanation:

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4 years ago

What is the range of f(x)=43*2.4^x

Aliun [14]

This is an exponential function.
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7 0

3 years ago

26. Julie says that sin A/cos A = tan A and sin B / cos B = tan B. Use the triangle shown to prove

PSYCHO15rus [73]

Julie is correct because:

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1 year ago

What is the quadratic regression equation for the data set? <br><br> PLEASEE HELP ME

Soloha48 [4]

You're trying to find constants $a_0,a_1,a_2$ such that $\hat y=a_0+a_1\hat x+a_2{\hat x}^2$ . Equivalently, you're looking for the least-square solution to the following matrix equation.

$\underbrace{\begin{bmatrix}1&6&6^2\\1&3&3^2\\\vdots&\vdots&\vdots\\1&9&9^2\end{bmatrix}}_{\mathbf A}\underbrace{\begin{bmatrix}a_0\\a_1\\a_2\end{bmatrix}}_{\mathbf x}=\underbrace{\begin{bmatrix}100\\110\\\vdots\\70\end{bmatrix}}_{\mathbf b}$

To solve $\mathbf{Ax}=\mathbf b$ , multiply both sides by the transpose of $\mathbf A$ , which introduces an invertible square matrix on the LHS.

$\mathbf{Ax}=\mathbf b\implies\mathbf A^\top\mathbf{Ax}=\mathbf A^\top\mathbf b\implies\mathbf x=(\mathbf A^\top\mathbf A)^{-1}\mathbf A^\top\mathbf b$

Computing this, you'd find that

$\mathbf x=\begin{bmatrix}a_0\\a_1\\a_2\end{bmatrix}\approx\begin{bmatrix}121.119\\-3.786\\-0.175\end{bmatrix}$

which means the first choice is correct.

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4 years ago