Determine which function has a range of {y∣y≤3}. f f(x)=-2x−3 g f(x)=-2|x+3|−3 h f(x)=-2(x−4)2+3 j f(x)=-2x3+3x2+x−1
1 answer:
See the attached figure.
<span>f(x)=-2x−3 ⇒⇒ black graph ⇒⇒ range = R
</span>
<span>g(x)=-2|x+3|−3 ⇒⇒ red graph ⇒⇒ range = </span><span>{y∣y ≤ -3}
</span>
<span>h(x)=-2(x−4)²+3 ⇒⇒ blue graph ⇒⇒ </span><span>range = {y∣y ≤ 3}
</span>
<span>j(x)=-2x³+3x²+x−1 ⇒⇒ green graph </span>⇒⇒ range = R
∴ <span>The function which has a range of {y∣y ≤ 3} is </span><span>h(x)=-2(x−4)²+3
</span>
<span></span> Which is the third choice
<span></span>
<span>f(x)=-2x−3 ⇒⇒ black graph ⇒⇒ range = R
</span>
<span>g(x)=-2|x+3|−3 ⇒⇒ red graph ⇒⇒ range = </span><span>{y∣y ≤ -3}
</span>
<span>h(x)=-2(x−4)²+3 ⇒⇒ blue graph ⇒⇒ </span><span>range = {y∣y ≤ 3}
</span>
<span>j(x)=-2x³+3x²+x−1 ⇒⇒ green graph </span>⇒⇒ range = R
∴ <span>The function which has a range of {y∣y ≤ 3} is </span><span>h(x)=-2(x−4)²+3
</span>
<span></span> Which is the third choice
<span></span>
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Answer:
The largest total area that can be enclosed will be a square of length 272 yards.
Step-by-step explanation:
First we get the perimeter of the large rectangular enclosure.
Perimeter of a rectangle =2(l + w)
Perimeter of the large rectangular enclosure= 1088 yard
Therefore:
2(L+W)=1088
The region inside the fence is the area
Area: A = LW
We need to solve the perimeter formula for either the length or width.
2L+ 2W= 1088 yd
2W= 1088– 2L
W =
W = 544–L
Now substitute W = 544–L into the area formula
A = LW
A = L(544 – L)
A = 544L–L²
Since A is a quadratic expression, we re-write the expression with the exponents in descending order.
A = –L²+544L
Next, we look for the value of the x coordinate
L=272 yards
Plugging L=272 yards into the calculation for area:
A = –L²+544L
A(272)=-272²+544(272)
=73984 square yards
Thus the largest area that could be encompassed would be a square where each side has a length of 272 yards and a width of:
W = 544 – L
= 544 – 272
= 272 yards