Your "<span>h (x)=x 2 squared +6x-3" is ambiguous. If you meant "x squared," then you need only write x^2 OR "x squared," but NOT "x 2 squared."
I will assume that by "</span><span>h (x)=x 2 squared +6x-3" you actually meant:
</span><span>h(x)=x^2 +6x-3. To find h(3), subst. 3 for x: h(3) = (3)^2 + 6(3) - 3, or
h(3) = 9 + 18 - 3, or h(3) = 24.</span>
Answer:
63 and 53
Step-by-step explanation:
you could use trail and error. So:
13+23=9
33+43=91
63+73=559
53+63=341
Over the x-axis: (x, y) (x, –y)
Over the y-axis: (x, y) (–x, y)
Over the line y = x: (x, y) (y, x)
Through the origin: (x, y) (–x, –y)
Answer:
Approximately 3.5 feet - Option B
Step-by-step explanation:
Imagine that you have this walkway around the garden, with dimensions 30 by 20 feet. This walkway has a difference of x feet between it's length, and say the dimension 30 feet. In fact it has a difference of x along both dimensions - on either ends. Therefore, the increases length and width should be 30 + 2x, and 20 + 2x, which is with respect to an increases area of 1,000 square feet.
( 30 + 2x )
( 20 + 2x ) = 1000 - Expand "( 30 + 2x )
( 20 + 2x )"
600 + 100x + 4
= 1000 - Subtract 1000 on either side, making on side = 0
4
+ 100x - 400 = 0 - Take the "quadratic equation formula"
( Quadratic Equation is as follows ) -
,
,

There can't be a negative width of the walkway, hence our solution should be ( in exact terms )
. The approximated value however is 3.5081...or approximately 3.5 feet.
Answer:
Shift the graph of y = x2 left 13 units and then up 6 units.
Step-by-step explanation: