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2 years ago

3(x+5) +2(2x+3)=0 solveeeeeeeeeeeedddddddddedddddd

Mathematics

1 answer:

morpeh [17]2 years ago

4 0

Step-by-step explanation:

${:}\longrightarrow$ $\sf 3 (x+5)+2 (2x+3)=0$

${:}\longrightarrow$ $\sf 3x+15+4x+6=0$

${:}\longrightarrow$ $\sf 3x+4x+15+6=0$

${:}\longrightarrow$ $\sf 7x+21=0$

${:}\longrightarrow$ $\sf 7x=-21$

${:}\longrightarrow$ $\sf x={\dfrac {-21}{7}}$

${:}\longrightarrow$ $\sf x=-3$

$\maltese{\underline{\boxed{\bf {x=-3}}}}$

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Madison needs to buy enough meat to make 1,000 hamburgers for the company picnic. Each hamburger will weigh 0.25 pound. How many

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2 years ago

Find cos(2*ABC) 100POINTS

juin [17]

Answer:

$-\dfrac{7}{25}$

Step-by-step explanation:

<u>Trigonometric Identities</u>

$\cos(A \pm B)=\cos A \cos B \mp \sin A \sin B$

<u>Trigonometric ratios</u>

$\sf \sin(\theta)=\dfrac{O}{H}\quad\cos(\theta)=\dfrac{A}{H}\quad\tan(\theta)=\dfrac{O}{A}$

where:

$\theta$ is the angle
O is the side opposite the angle
A is the side adjacent the angle
H is the hypotenuse (the side opposite the right angle)

Using the trig ratio formulas for cosine and sine:

$\cos(\angle ABC)=\dfrac{3}{5}$

$\sin(\angle ABC)=\dfrac{4}{5}$

Therefore, using the trig identities and ratios:

$\begin{aligned}\implies \cos(2 \cdot \angle ABC) & = \cos(\angle ABC + \angle ABC)\\\\& = \cos (\angle ABC) \cos (\angle ABC) - \sin(\angle ABC) \sin (\angle ABC)\\\\& = \cos^2(\angle ABC)-\sin^2(\angle ABC)\\\\& = \left(\dfrac{3}{5}\right)^2-\left(\dfrac{4}{5}\right)^2\\\\& = \dfrac{3^2}{5^2}-\dfrac{4^2}{5^2}\\\\& = \dfrac{9}{25}-\dfrac{16}{25}\\\\& = \dfrac{9-16}{25}\\\\& = -\dfrac{7}{25} \end{aligned}$