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3 years ago

When trying to distribute

Mathematics

1 answer:

marshall27 [118]3 years ago

3 0

You always do the exponent first. so its:
- ( x^2+2x+2) +4
-x^2-2x-2+4
-x^2-2x+2

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A number line that represents k > 5

schepotkina [342]

Answer:

any number from 6 and above

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3 years ago

A quadrilateral has no pairs of parallel sides. Which name best describes the figure?

Alecsey [184]

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4 years ago

Read 2 more answers

The total surface area of a cuboid is 112cm^2. find the value of x. values given: 10cm and 2cm.

Tatiana [17]

Answer:

x= 5.6

Step-by-step explanation:

we will need to put the information we currently have into an equation:

10•2•x= 112

then multiply 10•2 which equals 20...then the equation wil turn out like this:

20•x=112

now you will isolate the x... but how?? dividing 20 by itslef.. cancelling it out, and if you do one thing to the other side of the equation, you have to do it to the other...

20/20•x=112/20

you will end up with x= 122/20

you can use a calculator for this step, but to practice long division... (check the screen shot thingie)

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6 0

4 years ago

If I varies inversely with R and I = 0.25 when R = 18, which equation should be used to show this relationship?

pashok25 [27]

Inverse relationships are of the form xy=k or if you prefer y=k/x so to find the constant we can say:

18(0.25)=k=4.5

So you could say: IR=4.5 or I=4.5/R

7 0

4 years ago

HELP! Find the value of sin 0 if tan 0 = 4; 180 < 0< 270

BabaBlast [244]

Hi there! Use the following identities below to help with your problem.

$\large \boxed{sin \theta = tan \theta cos \theta} \\ \large \boxed{tan^{2} \theta + 1 = {sec}^{2} \theta}$

What we know is our tangent value. We are going to use the tan²θ+1 = sec²θ to find the value of cosθ. Substitute tanθ = 4 in the second identity.

$\large{ {4}^{2} + 1 = {sec}^{2} \theta } \\ \large{16 + 1 = {sec}^{2} \theta } \\ \large{ {sec}^{2} \theta = 17}$

As we know, sec²θ = 1/cos²θ.

$\large \boxed{sec \theta = \frac{1}{cos \theta} } \\ \large \boxed{ {sec}^{2} \theta = \frac{1}{ {cos}^{2} \theta} }$

And thus,

$\large{ {cos}^{2} \theta = \frac{1}{17}} \\ \large{cos \theta = \frac{ \sqrt{1} }{ \sqrt{17} } } \\ \large{cos \theta = \frac{1}{ \sqrt{17} } \longrightarrow \frac{ \sqrt{17} }{17} }$

Since the given domain is 180° < θ < 360°. Thus, the cosθ < 0.

$\large{cos \theta = \cancel\frac{ \sqrt{17} }{17} \longrightarrow cos \theta = - \frac{ \sqrt{17} }{17}}$

Then use the Identity of sinθ = tanθcosθ to find the sinθ.

$\large{sin \theta = 4 \times ( - \frac{ \sqrt{17} }{17}) } \\ \large{sin \theta = - \frac{4 \sqrt{17} }{17} }$

Answer

sinθ = -4sqrt(17)/17 or A choice.

4 0

3 years ago