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2 years ago

5

Nut B fits on Bolt B. Nut A fits on Bolt C, but does not fit on Bolt A. Bolts B and A are exactly the same. Given the scenario,

will nut A fit on Bolt B?

1 answer:

Elena L [17]2 years ago

6 0

Nut B -> Bolt B
Nut A -> Bolt C, but NOT Bolt A
Bolt B is also the same as Bolt A, so Bolt B = Bolt A
Then, Nut B -> Bolt A.
Remember that Nut A didn't fit on Bolt A, and Bolt A is the same as Bolt B.

So, Nut A will NOT fit on Bolt B.

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Andy drove 840 miles in 12 hours. How much miles can he drive per hour?

Sergio039 [100]

Answer:

70

Step-by-step explanation:

840 / 12 = 70

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2 years ago

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20 plus 2 x divided by 5 subtract x

irina [24]

20+2x/5-x
Start by writing it as a fraction

20 + 2/5x (fraction) - x
Next you calculate the difference

20 - 3/5x (fraction) is the answer

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2 years ago

For all values of x,<br>f(x) = x^2 + 3x and g(x) = x - 4<br>Show that fg(x) = x^2 - 5x + 4

Westkost [7]

Answer:

f(g(x)) = x² - 5x +4

Step-by-step explanation:

<u><em>Explanation</em></u>

Given that the functions

f(x) = x² + 3 x and g(x) = x-4

f(g(x)) = f(x -4) (∵ g(x) = x-4)

= (x-4)² + 3(x-4)

= x² - 2(4)x + 4² + 3x -12

= x² - 8 x + 16 + 3x -12

= x² - 5x +4

∴ f(g(x)) = x² - 5x +4

4 0

2 years ago

Find the vertices and foci of the hyperbola. 9x2 − y2 − 36x − 4y + 23 = 0

Xelga [282]

Hey there, hope I can help!

NOTE: Look at the image/images for useful tips
$\left(h+c,\:k\right),\:\left(h-c,\:k\right)$

$\frac{\left(x-h\right)^2}{a^2}-\frac{\left(y-k\right)^2}{b^2}=1\:\mathrm{\:is\:the\:standard\:equation\:for\:a\:right-left\:facing:H}$
with the center of (h, k), semi-axis a and semi-conjugate - axis b.
NOTE: H = hyperbola

$9x^2-y^2-36x-4y+23=0 \ \textgreater \ \mathrm{Subtract\:}23\mathrm{\:from\:both\:sides}$
$9x^2-36x-4y-y^2=-23$

$\mathrm{Factor\:out\:coefficient\:of\:square\:terms}$
$9\left(x^2-4x\right)-\left(y^2+4y\right)=-23$

$\mathrm{Divide\:by\:coefficient\:of\:square\:terms:\:}9$
$\left(x^2-4x\right)-\frac{1}{9}\left(y^2+4y\right)=-\frac{23}{9}$

$\mathrm{Divide\:by\:coefficient\:of\:square\:terms:\:}1$
$\frac{1}{1}\left(x^2-4x\right)-\frac{1}{9}\left(y^2+4y\right)=-\frac{23}{9}$

$\mathrm{Convert}\:x\:\mathrm{to\:square\:form}$
$\frac{1}{1}\left(x^2-4x+4\right)-\frac{1}{9}\left(y^2+4y\right)=-\frac{23}{9}+\frac{1}{1}\left(4\right)$

$\mathrm{Convert\:to\:square\:form}$
$\frac{1}{1}\left(x-2\right)^2-\frac{1}{9}\left(y^2+4y\right)=-\frac{23}{9}+\frac{1}{1}\left(4\right)$

$\mathrm{Convert}\:y\:\mathrm{to\:square\:form}$
$\frac{1}{1}\left(x-2\right)^2-\frac{1}{9}\left(y^2+4y+4\right)=-\frac{23}{9}+\frac{1}{1}\left(4\right)-\frac{1}{9}\left(4\right)$

$\mathrm{Convert\:to\:square\:form}$
$\frac{1}{1}\left(x-2\right)^2-\frac{1}{9}\left(y+2\right)^2=-\frac{23}{9}+\frac{1}{1}\left(4\right)-\frac{1}{9}\left(4\right)$

$\mathrm{Refine\:}-\frac{23}{9}+\frac{1}{1}\left(4\right)-\frac{1}{9}\left(4\right) \ \textgreater \ \frac{1}{1}\left(x-2\right)^2-\frac{1}{9}\left(y+2\right)^2=1 \ \textgreater \ Refine$
$\frac{\left(x-2\right)^2}{1}-\frac{\left(y+2\right)^2}{9}=1$

Now rewrite in hyperbola standardform
$\frac{\left(x-2\right)^2}{1^2}-\frac{\left(y-\left(-2\right)\right)^2}{3^2}=1$

$\mathrm{Therefore\:Hyperbola\:properties\:are:}\left(h,\:k\right)=\left(2,\:-2\right),\:a=1,\:b=3$
$\left(2+c,\:-2\right),\:\left(2-c,\:-2\right)$

Now we must compute c
$\sqrt{1^2+3^2} \ \textgreater \ \mathrm{Apply\:rule}\:1^a=1 \ \textgreater \ 1^2 = 1 \ \textgreater \ \sqrt{1+3^2}$

$3^2 = 9 \ \textgreater \ \sqrt{1+9} \ \textgreater \ \sqrt{10}$

Therefore the hyperbola foci is at $\left(2+\sqrt{10},\:-2\right),\:\left(2-\sqrt{10},\:-2\right)$

For the vertices we have $\left(2+1,\:-2\right),\:\left(2-1,\:-2\right)$

Simply refine it
$\left(3,\:-2\right),\:\left(1,\:-2\right)$
Therefore the listed coordinates above are our vertices

Hope this helps!

8 0

3 years ago

Expression is equivalent to 486 – 9 + 6 + 33 × 2

mojhsa [17]

Answer:

549

Step-by-step explanation:

A calculator can give you an equivalent expression.

I might work from the middle outward:

= 486 -3 +33×2

= 483 +66

= (480 +60) + (3 +6)

= 540 + 9

= 549

7 0

3 years ago