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3 years ago

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Mental Math Situation: You exercised 24 hours each month for a year. How many hours did you exercise by the end of the year? You

may be able to do the math mentally thanks to expanded notation and the Distributive Property.

2 answers:

Aloiza [94]3 years ago

8 0

Answer:

So 24 hours each month times 12 would be 288 hours of working out of one year.

Hope this helps!

Step-by-step explanation:

WINSTONCH [101]3 years ago

5 0

Answer: 288h

Step-by-step explanation:

There are 12 months in a year if you exercise 24h per month you have to multiply 12 x 24 = 288h

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X^+5x+8=0<br> What is the answer to this question

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Step-by-step explanation:

X^+5x+8=0

x^2+5x=-8 kajvdvdhxjxjjxjxjxjxhx

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4 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
$\frac{\left(x-2\right)^2}{1^2}-\frac{\left(y-\left(-2\right)\right)^2}{3^2}=1$

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$\left(2+c,\:-2\right),\:\left(2-c,\:-2\right)$

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
$\left(3,\:-2\right),\:\left(1,\:-2\right)$
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4 years ago

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X=-39/5 and y=10 :) i hope this helps you out !

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4 years ago

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Nady [450]

Answer:

Side a = 35ft

Side b = 35ft

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Step-by-step explanation:

The formula for the perimeter of a triangle = Side a + Side b + Side c

In an isosceles triangle, 2 sides are equal to each other.

So, Side a = Side b

In the question, we are told that:

The two equal sides are each 5 ft less than twice the length of the third side.

Hence,

a = 2c - 5

b = 2c - 5

P = 90ft

P = a + b + c

90 = 2c - 5 + 2c - 5 + c

Collect like terms

90 = 5c - 10

90 + 10 = 5c

100 = 5c

c = 100/5

c = 20

The length of the third side = 20ft

a = 2c - 5

= 2 × 20 - 5

= 40 - 5

= 35 ft

b =2c - 5

= 2 × 20 - 5

= 40 - 5

= 35 ft

Therefore,

Side a = 35ft

Side b = 35ft

Side c = 20ft

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The converse of the statement “If a quadrilateral has four right angles, then it is a rectangle” is "______."

Nataly [62]

A quadrilateral because you just reverse the question.

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