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3 years ago

Trig help ...............

Mathematics

2 answers:

Flauer [41]3 years ago

4 0

Letter D

A = 135°
B = 30°

cosAcosB - sinAsinB = cos(A+B)

cos(135°+30°)
cos(165°)

masya89 [10]3 years ago

4 0

Answer: D

Step-by-step explanation:

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Michael's team is trying to determine how they can process shipments more quickly. He has convened a large group to come up with

xenn [34]

Step-by-step explanation:

d is correct answer as i think.

4 0

3 years ago

Use deductive reasoning to show that the following procedure always produces the number 6. Procedure: Pick a number. Add 4 to th

Flauer [41]

Answer:

procedure always produces 6

Step-by-step explanation:

Let 'n' be the unknown number

Add 4 to the number : n+4

multiply the sum by 3.

multiply the sum n+4 by 3

$3(n+4) is 3n+12$

Now subtract 6, so we subtract 6 from 3n+12

$3n+12-6=3n+6$

finally decrease the difference by the tripe of the original number

triple of original number is 3n

$3n+6-3n= 6$

so the procedure always produces 6

5 0

3 years ago

Write the equation -4x^2+9y^2+32x+36y-64=0 in standard form. Please show me each step of the process!

IgorC [24]

Hey there, hope I can help!

$-4x^2+9y^2+32x+36y-64=0$

$\mathrm{Add\:}64\mathrm{\:to\:both\:sides} \ \textgreater \ 9y^2+32x+36y-4x^2=64$

$\mathrm{Factor\:out\:coefficient\:of\:square\:terms} \ \textgreater \ -4\left(x^2-8x\right)+9\left(y^2+4y\right)=64$

$\mathrm{Divide\:by\:coefficient\:of\:square\:terms:\:}4$
$-\left(x^2-8x\right)+\frac{9}{4}\left(y^2+4y\right)=16$

$\mathrm{Divide\:by\:coefficient\:of\:square\:terms:\:}9$
$-\frac{1}{9}\left(x^2-8x\right)+\frac{1}{4}\left(y^2+4y\right)=\frac{16}{9}$

$\mathrm{Convert}\:x\:\mathrm{to\:square\:form}$
$-\frac{1}{9}\left(x^2-8x+16\right)+\frac{1}{4}\left(y^2+4y\right)=\frac{16}{9}-\frac{1}{9}\left(16\right)$

$\mathrm{Convert\:to\:square\:form}$
$-\frac{1}{9}\left(x-4\right)^2+\frac{1}{4}\left(y^2+4y\right)=\frac{16}{9}-\frac{1}{9}\left(16\right)$

$\mathrm{Convert}\:y\:\mathrm{to\:square\:form}$
$-\frac{1}{9}\left(x-4\right)^2+\frac{1}{4}\left(y^2+4y+4\right)=\frac{16}{9}-\frac{1}{9}\left(16\right)+\frac{1}{4}\left(4\right)$

$\mathrm{Convert\:to\:square\:form}$
$-\frac{1}{9}\left(x-4\right)^2+\frac{1}{4}\left(y+2\right)^2=\frac{16}{9}-\frac{1}{9}\left(16\right)+\frac{1}{4}\left(4\right)$

$\mathrm{Refine\:}\frac{16}{9}-\frac{1}{9}\left(16\right)+\frac{1}{4}\left(4\right) \ \textgreater \ -\frac{1}{9}\left(x-4\right)^2+\frac{1}{4}\left(y+2\right)^2=1$

$Refine\;once\;more\;-\frac{\left(x-4\right)^2}{9}+\frac{\left(y+2\right)^2}{4}=1$

For me I used
$\frac{\left(y-k\right)^2}{a^2}-\frac{\left(x-h\right)^2}{b^2}= 1$
$As\;\mathrm{it\;\:is\:the\:standard\:equation\:for\:an\:up-down\:facing\:hyperbola}$

I know yours is an equation which is why I did not go any further because this is the standard form you are looking for. I would rewrite mine to get my hyperbola standard form. However the one I have provided is the form you need where mine would be.
$\frac{\left(y-\left(-2\right)\right)^2}{2^2}-\frac{\left(x-4\right)^2}{3^2}=1$

Hope this helps!

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3 years ago

Lauren jogs the rate at a 2 mile every 1/2 hour. what is her until rate?

Fantom [35]

I 4 mph

Step-by-step explanation: I assume you’re writing Asking about the mph it would be be 4

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1 year ago

A) 0,2 x 5/7 + 11/14 = ?

NemiM [27]

I hope this helps you