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andrew-mc [135]
3 years ago
13

H(x)=x-4 Determine for each 2-value whether it is in the domain of h or not.

Mathematics
1 answer:
MrRissso [65]3 years ago
3 0

Answer:

  1. in domain
  2. Not I'm domain
  3. Not in domain
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Find the area ratio of a cube with volume 125 m to a cube with volume 64 m.
Viefleur [7K]
V=s^3
SA=6s^2
so

125=v
125=s^3
cube root both sides
5=s
A=6(5)^2=6(25)=150 m^2


64=v
64=s^3
cube root both sides
4=s

A=6(4)^2=6(16)=96 m^2

ratio is

150:96=25:16
5 0
3 years ago
Read 2 more answers
Simplify the expression -3 ÷ (-2/5 ​
Harlamova29_29 [7]

Answer:

7.5

Step-by-step explanation:

To solve this expression, we can use PEMDAS or order of operations.

<u>Parenthesis</u>

<u>Exponents</u>

<u>Multiplication>Division (Depends on which comes first from left to right)</u>

<u>Addition>Subtraction (See Multiplication>Division)</u>

So First, lets solve the <u>Parenthesis</u>. -2/5=-.4

So now, our equation is -3/-.4

There is no exponents or multiplication, so lets just <u>Divide</u>.

-3/-.4=7.5

The solution is 7.5

Hope this helps!

6 0
3 years ago
It takes Benjamin 28 minutes to mow 2 lawns. Assuming the lawns are the same size and Benjamin works at the same speed, about ho
podryga [215]

Answer:

70

Step-by-step explanation:

28 b y 2 14 times 5

3 0
3 years ago
Pleaseee helppp with thissss asapppp
Natasha2012 [34]

As we know that the standard equation of circle is {\bf{(x-h)^{2}+(y-k)^{2}=r^{2}}} , where <em>r</em> is the radius of circle and centre at <em>(h,k) </em>

Now , as the circle passes through <em>(2,9)</em> so it must satisfy the above equation after putting the values of <em>h</em> and <em>k</em> respectively

{:\implies \quad \sf \{2-(-1)\}^{2}+(9-5)^{2}=r^{2}}

{:\implies \quad \sf (3)^{2}+(4)^{2}=r^{2}}

{:\implies \quad \sf 9+16=r^{2}}

After raising ½ power to both sides , we will get <em>r = +5 , -5</em> , but as radius can never be -<em>ve</em> . So <em>r = +</em><em>5</em><em> </em>

Now , putting values in our standard equation ;

{:\implies \quad \sf \{x-(-1)\}^{2}+(y-5)^{2}=(5)^{2}}

{:\implies \quad \bf \therefore \quad \underline{\underline{(x+1)^{2}+(y-5)^{2}=25}}}

<em>This is the required equation of </em><em>Circle</em>

Refer to the attachment as well !

8 0
2 years ago
Which one is it?????
photoshop1234 [79]
It is right 1 down 3
8 0
3 years ago
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