What positive value of x makes the following equation true: 289? Explain.

aliina [53]

Another way to do it is
2x-y=17
x-y=10, x=10+y
substitute x=10+y for x in 2x-y=17 and get
2(10+y)-y=17
20+2y-y=17
20+y=17
subtract 20 from both sides
y=-3
subsitute y=-3 into equation x-y=10
x-(-3)=10
x+3=10
subtract 3 from both sides
x=7
or subsitute into 2x-y=17
2x-(-3)=17
2x+3=17
subtract 3 from both sides
2x=14
divide both sides by two
x=7

5 0

3 years ago

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Why is a cube root of a negative is always a negative? Need help! Thanks!

Fed [463]

A cube root is representaive of reverse mutiplication. E.g Square root of 4 is 2 (2x2)
Using this, we can see that anytime 3 negatives are multiplied, the answer will always be negative.

7 0

3 years ago

Find the value of y-<br> y^2=169

Blababa [14]

Answer:

13

Step-by-step explanation:

square root y^2 = square root 169

so

y = 13

6 0

4 years ago

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In order to evaluate 7 sec(θ) dθ, multiply the integrand by sec(θ) + tan(θ) sec(θ) + tan(θ) . 7 sec(θ) dθ = 7 sec(θ) sec(θ) + ta

Maurinko [17]

Answer:

$\int {7 \sec(\theta) } \, d\theta = 7\ln(\sec(\theta) + \tan(\theta)) + c$

Step-by-step explanation:

The question is not properly formatted. However, the integral of $\int {7 \sec(\theta) } \, d\theta$ is as follows:

$\int {7 \sec(\theta) } \, d\theta$

Remove constant 7 out of the integrand

$\int {7 \sec(\theta) } \, d\theta = 7\int {\sec(\theta) } \, d\theta$

Multiply by 1

$\int {7 \sec(\theta) } \, d\theta = 7\int {\sec(\theta) * 1} \, d\theta$

Express 1 as: $\frac{\sec(\theta) + \tan(\theta) }{\sec(\theta) + \tan(\theta)}$

$\int {7 \sec(\theta) } \, d\theta = 7\int {\sec(\theta) * \frac{\sec(\theta) + \tan(\theta) }{\sec(\theta) + \tan(\theta)}} \, d\theta$

Expand

$\int {7 \sec(\theta) } \, d\theta = 7\int {\frac{\sec^2(\theta) + \sec(\theta)\tan(\theta) }{\sec(\theta) + \tan(\theta)}} \, d\theta$

Let

$u = \sec(\theta) + \tan(\theta)$

Differentiate

$\frac{du}{d\theta} = \sec(\theta)\tan(\theta) + sec^2(\theta)$

Make $d\theta$ the subject

$d\theta = \frac{du}{\sec(\theta)\tan(\theta) + sec^2(\theta)}$

So, we have:

$\int {7 \sec(\theta) } \, d\theta = 7\int {\frac{\sec^2(\theta) + \sec(\theta)\tan(\theta) }{u}} \,* \frac{du}{\sec(\theta)\tan(\theta) + sec^2(\theta)}$