Simple..
For area, it's A=l*w
15*2=30
And, perimeter, it's P=l+l+w+w
15+15+2+2=34
Thus, your answer.
l=15 and w=2,or, w=15 and l=2
The smallest positive integer that the intermediate value theorem guarantees a zero exists between 0 and a is 3.
What is the intermediate value theorem?
Intermediate value theorem is theorem about all possible y-value in between two known y-value.
x-intercepts
-x^2 + x + 2 = 0
x^2 - x - 2 = 0
(x + 1)(x - 2) = 0
x = -1, x = 2
y intercepts
f(0) = -x^2 + x + 2
f(0) = -0^2 + 0 + 2
f(0) = 2
(Graph attached)
From the graph we know the smallest positive integer value that the intermediate value theorem guarantees a zero exists between 0 and a is 3
For proof, the zero exists when x = 2 and f(3) = -4 < 0 and f(0) = 2 > 0.
<em>Your question is not complete, but most probably your full questions was</em>
<em>Given the polynomial f(x)=− x 2 +x+2 , what is the smallest positive integer a such that the Intermediate Value Theorem guarantees a zero exists between 0 and a ?</em>
Thus, the smallest positive integer that the intermediate value theorem guarantees a zero exists between 0 and a is 3.
Learn more about intermediate value theorem here:
brainly.com/question/28048895
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Rhombus becasue it has four equal sides but no right angles .
Answer:
x=3
Step-by-step explanation:
Answer:
95:95:19
Step-by-step explanation:
1.Divide 209 by 11.(209÷19)
2.Multiply 19×5 since 5/11 of the coins are nickels.(19×5=95)
3.Multiply 19×5 again since 5/11 of the coins are dimes.(19×5=95)
4.Multiply 19×1 since there would be 1/11 left of the coins which are quarters. (19×1=19)
5.Check your awnser by adding 95+95+19.(95+95+19=209)