Determine the solution to the system of equations {4x+15y=40

Mathematics

1 answer:

oee [108]3 years ago

8 0

Step-by-step explanation:

4x+15y= 40-5y= 20

= 4x+15y+5y= 40-20

4x+20y= 20

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NO LINKS!! Verify each identity. Show work please

PIT_PIT [208]

Answer:

Trigonometric Identities used:

$\csc \theta=\dfrac{1}{\sin \theta}$

$\sec \theta=\dfrac{1}{\cos \theta}$

$\cot \theta=\dfrac{1}{\tan \theta}$

$\sin^2 \theta + \cos^2 \theta=1$

$\csc^2 \theta = 1 + \cot^2 \theta$

Question 33

$\large \begin{aligned}\dfrac{\sin \theta}{\csc \theta}+\dfrac{\cos \theta}{\sec \theta}& = \sin \theta \cdot \dfrac{1}{\csc \theta}+\cos \theta \cdot \dfrac{1}{\sec \theta}\\\\& = \sin \theta \cdot \dfrac{1}{\frac{1}{\sin \theta}}+\cos \theta \cdot \dfrac{1}{\frac{1}{\cos \theta}}\\\\& = \sin \theta \cdot \sin \theta+\cos \theta \cdot \cos \theta\\\\& = \sin^2 \theta + \cos^2 \theta\\\\& = 1\end{aligned}$

Question 34

$\large \begin{aligned}\tan \theta \csc^2 \theta - \tan \theta & = \tan \theta ( \csc^2 \theta - 1)\\\\& = \tan \theta (1 + \cot^2 \theta -1)\\\\& = \tan \theta \cot^2 \theta \\\\& = \dfrac{1}{\cot \theta} \cdot \cot^2 \theta \\\\& = \dfrac{\cot^2 \theta}{\cot \theta}\\\\& = \cot \theta\end{aligned}$

8 0

2 years ago

Read 2 more answers

If x = 8 and y = -3, evaluate the following expression: 20 - 5y + 2x

solniwko [45]

Answer:

Step-by-step explanation:

Given the following question:

$20-5y+2x$
$x=8$
$y=-3$

Substitute and solve using PEMDAS:

$20-5y+2x$
$20-5(-3)+2(8)$
$5\times-3=-15$
$20--15+2(8)$
$2\times8=16$
$20-(-15)+16$
$20--15=35$
$35+16=51$
$=51$

Your answer is "51."

Hope this helps.